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physics I lab: final quiz
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lab 8
one-dimensional collisions
What is your Prediction 1-2? Which would be more effective at closing the door, the superball or the clay ball of the same mass?
the superball will bounce back with the same velocity, whereas the clay blob will have a final velocity of zero
-the superball because the cloy blob would stick and increase the doors mass which would then require more force to close the door in the same amount of time as the superball would
what is the final momentum of the clay ball in activity 1-3?
the final momentum is zero because velocity is zero
how do you find the impulse of a force from the force versus time graph?
FÎ”t = Î”p
why is the collision in activity 2-2 made with a "springy" wall
to create a perfectly elastic collision
how will you measure the impulse in activity 2-2?
F x Î”t = mvf - mvi
In the following question you will find the impulse imparted to an object from a force vs. time graph. What are the units of impulse?
N s
Find the impulse of the force shown on the force-time graph below. You do not need to enter units with your answer, you did that in the question above.
4 N over 3 seconds
12 N s
An object of mass 2.3kg is moving in the negative x direction at a velocity of 2.6m/s. It experiences the force shown above for 3s. What is the final velocity after the object has experienced the impulse?
2.3 x -2.6 = p
p = -5.98
12 - 5.98 = 6.02 (mv, so now divide by mass to find v)
6.02/2.3 = 2.62 m/s
A superball and a clay ball are droped from a height of 10cm above a table top. They have the same mass 0.05kg and the same size. The superball bounces off the table and rises back to the same height. The clay ball sticks to the table. Match the shapes of the forces exerted on the superball and the clay ball by the table as a function of time.
both happen in a fraction of a second but the force vs time graph for the superball will spike higher
For the superball in the previous question, if it was in contact with the table for 31ms, calculate the average force exerted on the ball by the table.
v^2 = v0^2 + 2ad
v^2 = 0 + 2(9.8)(0.1)
v = 1.4
p = mv
p = (0.05)(1.4 - 0)
p = 0.07
p = Ft
0.07 = F (0.031)
F = 2.26 N
For the clay ball in the previous question, calculate the magnitude of the impulse exerted on the clay ball. (Please don't use a sign, which hasn't been defined anyway, and don't enter units, you did that in the first question.)
mass x change in velocity
(0.05)(1.4) = 0.07
If the collision of the clay ball with the table takes the same time as the collision of the superball, compare the average force exerted by the table on the clay ball to that exerted on the superball. Which is larger or are they the same?
The average force on superball is larger.
Suppose that the clay has twice the mass and is dropped from the same height. Compare the impulse exerted on the ball by the table to that with the smaller clay ball.
The heavier clay ball has the larger impulse.
Suppose that the original clay ball is dropped from twice the height. Compare the impulse exerted on the ball by the table to that for the smaller height.
p = mv
v = d/t
v^2 = u^2 + 2ad
The impulse will be 1.41 times larger
In this lab we will explore the forces of interaction between
two objects and study the changes in motion that result from these interactions.
-We are especially interested in studying collisions and explosions in which interactions take place in fractions of a second or less.
-Early investigators spent a considerable amount of time trying to observe collisions and explosions, but they encountered difficulties. This is not
surprising, since the observation of the details of such phenomena requires the use of instrumentation--such as high speed cameras--that was not yet invented.
However, the principles describing the outcomes of collisions were well understood by the late seventeenth century when several leading European scientists including Isaac Newton developed the concept of
quantity-of-motion to describe both elastic collisions in which objects bounce off each other and inelastic collisions in which objects stick together. These days we use the word momentum rather than quantity-of-motion in describing collisions and explosions.
We will begin our study of collisions by exploring the relationship between
the forces experienced by an object and its momentum change.
It can be shown mathematically from Newton's laws and experimentally from our own observations, that the change in momentum of an object is equal to a quantity called
impulse.
Impulse is a quantity which takes into account both
the magnitude of the applied force at each instant in time, and the time interval over which this force acts.
The statement of equality between impulse and momentum change is known as
the impulse-momentum theorem.
In this investigation we are going to develop the concept of momentum to predict the outcome of collisions. But you don't officially know what momentum is because we haven't defined it yet. Let's start by predicting what will happen as a result of a simple one-dimensional collision. This should help you figure out how to define momentum to enable you to describe collisions in mathematical terms.
It's early fall and you are driving along a two lane highway in a rented moving van. Its full of all of your possessions so you and the loaded truck weigh 8000 lbs. You have just slowed down to 15 MPH because you're in a school zone. It's a good thing you thought to do that because a group of first graders are just starting to cross the road. Just as you pass the children you see a 2000 lb sports car in the oncoming lane heading straight for the children at about 80 MPH. What a fool the driver is!
A desperate thought crosses your mind. You just have time to swing into the oncoming lane and speed up a bit before making a head-on collision with the sports car. You want your truck and the sports car to crumple into a heap that sticks together and doesn't move. Can you save the children or is this just a suicidal act?
How fast would you have to be going to completely stop the sports car? Explain the reasons for your prediction.
20 mph
because 2000 * 80 = 8000x
x = 20
Try some head-on collisions with the carts of different mass to simulate the event. Be sure that they stick together after the collision.
Observe qualitatively what combinations of velocities cause the two carts to be at rest after the collision
What happens when the less massive cart is moving much faster than the more massive cart? Much slower? At an intermediate speed?
faster = slows down/less of a reaction
slower = pushed farther back at a faster rate
intermediate = between faster and slower reaction/stops
Based on your prediction and your observations, what mathematical definition might you use to describe the momentum you would need to stop an oncoming vehicle traveling with a known mass and velocity? Should it depend on the mass, the velocity or both? Explain your choice.
p = mv
both because you need a force equal to them both to stop them both in opposite directions
Just to double check your reasoning, you should have come to the conclusion that momentum is defined by the vector equation
pâ‰¡mv
where the symbol â‰¡ is used to designate "defined as."
Originally Newton did not use the concept of acceleration or velocity in his laws. Instead he used the term
"motion" which he defined as the product of mass and velocity (the quantity we now call momentum). Let's examine a translation from Latin of Newton's first two laws with some parenthetical changes for clarity.
Newton's First Two Laws of Motion
1. Every body continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed on it.
2. The (rate of) change of motion is proportional to the motive force impressed: and is made in the direction of the right line in which that force is impressed.
The more familiar contemporary statement of the Second Law is that the net force on an object can be calculated as the product of its mass and its acceleration where the direction of the force and of the resulting acceleration are the same. Newton's statement of the law and the more modern statement are mathematically equivalent.
Now let's test your intuition about momentum and forces.
You are sleeping in your sister's room while she is away at college. Your house is on fire and smoke is pouring into the partially open bedroom door. The room is so messy that you cannot get to the door. The only way to close the door is to throw either a blob of clay or a superball at the door
there isn't time to throw both.
Assuming the clay blob and the superball have the same mass, and that you throw them with the same velocity, which would you throw to close the door--the clay blob which will stick to the door or the superball which will bounce back at almost the same speed as it had before it collided with the door? Give reasons for your choice using any notions you already have or any new concepts developed in physics such as force, energy, momentum, Newton's laws, etc. Remember, your life depends on it!
-superball because of Newton's third law
-the clay blob's velocity will become zero
You can test your prediction by dropping a force probe with a bouncy rubber stopper on its end, and then dropping the force probe from the same height with a clay ball of approximately the same mass on on its end.
We can associate the maximum force read by the force probe with the maximum force a thrown ball can exert on a door. We will later investigate how the maximum force is related to the change in momentum of the ball in each case.
superball mass (kg), height dropped (m), ad maximum force (N)
-0.01379 kg
-0.05 m
-38 N
small sand bag mass (kg), height dropped (m), ad maximum force (N)
-0.389 kg
-0.05 m
-1 N
did your observations agree with your predictions? Which item dropped resulted in a bigger maximum force: the superball or the sandbag?
yes, the superball dropped with the biggest maximum force
Based on your observations, which should you throw at the door: the superball or the sandbag
the superball
prediction: What will happen to the maximum force of you increase the mass of the ball, but allow it to collide with the same velocity (drops it from the same height)?
the maximum force would increase
prediction: What will happen to the maximum force if the velocity just before impact is increased by dropping the ball from a greater height?
the maximum force would also increase in this case
test your predictions by using a more massive sandbag dropped from the same height, so that the mass of the sandbag is about doubled. measure the maximum force and record it
VVV
large sand bag mass (kg), height dropped (m), ad maximum force (N)
-0.2153 kg
-0.05 m
-4 N
test your next prediction by dropping the smaller sandbag from about twice the height (10 cm). Record the height and maximum force
VvVVV
small sand bag dropped at a larger height's mass (kg), height dropped (m), ad maximum force (N)
-0.389 kg
-0.10 m
-2 N
did your observations agree with your predictions? what factors seem to determine the force exerted on the force probe?
yes
the mass of the object and the height the object is dropped from
It would be nice to be able to use Newton's formulation of the Second Law of motion to find collision forces, but it is difficult to measure the rate of change of momentum during a rapid collision without special instruments. However, measuring the momenta of objects just before and just after a collision is not usually too difficult.
This led scientists in the seventeenth and eighteenth centuries to concentrate on the overall changes in momentum that resulted from collisions. They then tried to relate changes in momentum to the forces experienced by an object during a collision.
In the next activity you are going to explore the mathematics of calculating momentum changes for the two types of collisions--the elastic collision where the ball bounces off the door and the inelastic collision where the ball sticks to the door
prediction: Which object undergoes the greater momentum change during the collision with a doorâ€”the clay blob or the superball? Explain your reasoning carefully.
the clay blob will have a greater change in momentum because it goes from a certain velocity to a velocity of zero, whereas the superball still has the same velocity after bouncing back off the door
Recall that momentum is defined as a vector quantity; i.e., it has both magnitude and direction. Mathematically, momentum change is given by the equation
Î”p = pfinal - pinitial
(all vector quantities)
-where pi is the initial momentum of the object just before a collision and pf is its final momentum just after. Remember, in one dimension, the direction of a vector is indicated by its sign.
check your prediction with some calculations of the momentum changes for two of the collisions that you carried out. This is good review of the properties of one-dimensional vectors, too.
Carry out the following calculations for the superball and sandbag of the same mass when dropped from the same height (should have been 5 cm)
general steps to calculating change in momentum
1.) refer to the table to determine the superball's mass and initial height (5 cm)
2.) determine if you need other information to determine its initial velocity just before it struck the circular plate on the force sensor (hint: you will need to recall from kinematics on how to find the velocity of an object in free fall after it has fallen a specific distance, in this case you dropped it so you know something about its velocity before it fell)
3.) now calculate the initial momentum just before it struck the plate on the force sensor and enter that into a table
4.) assume the superball is an ideal superball and it bounced off the plate on the force sensor so that the magnitude of its velocity as it just came up off the plate is equal to the magnitude of its velocity as it struck the plate. Determine the smperball's final velocity as it just came up off the force sensor plate and record in a table
5.) calculate the smperball's final momentum and its change in momentum and record that
6.)repeat the above steps for the small sand bag. It should have a mass that is very nearly equal to the superball, it should have been dropped from the same height as the superball. The only difference here is that you shouldn't't make the assumption about its final velocity that we made in item 4 above about the superball
v =
(2gh) ^1/2
table recordings
superball
-mass: 0.01379 kg
-height: 0.05 m
-initial velocity: 0 m/s
-initial momentum: 0.06 kgâ€¢m/s
-final velocity: 4.48 m/s
-final momentum: -0.06 kgâ€¢m/s
-change in momentum: 0.12 kgâ€¢m/s
sandbag
-mass: 0.0389 kg
-height: 0.05 m
-initial velocity: 0 m/s
-initial momentum: 0.17 kgâ€¢m/s
-final velocity: 4.48 m/s
-final momentum: -0.17 kgâ€¢m/s
-change in momentum: 0.34 kgâ€¢m/s
Compare your calculated changes in momentum to your predictions. Do they agree? Which ball had the larger change in momentum? remember that the sandbag here represents the clay blob
yes, the sandbag here has a larger change in momentum like expected
How does change in momentum seem to be related to the maximum force applied to the ball?
as force increases, so does the change in momentum
If you catch an egg of mass m that is heading toward your hand at speed v what magnitude momentum change does it undergo? Hint: The egg is at rest after you catch it.
mvf - mvi --> 0 - mvi --> negative momentum
Does the total momentum change differ if you catch the egg more slowly or is it the same?
momentum increases
Suppose the time you take to bring the egg to a stop is âˆ†t. Would you rather catch the egg in such a way that âˆ†t is small or large. Why?
large because then it's a smaller momentum
What do you suspect might happen to the average force you exert on the egg while catching it when âˆ†t is small?
greater force
In bringing an egg to rest, the change in momentum is
the same whether you use a large force during a short time interval or a small force during a long time interval. Of course, which one you choose makes a lot of difference in whether the egg breaks or not!
A quantity called impulse has been defined for you in lecture and/or in your textbook. It combines
the applied force and the time interval over which it acts. In one dimension, for a constant force F acting over a time interval âˆ†t, as shown in the graph below,
the impulse J is
J = Fâˆ†t
As you can see, a large force acting over a short time and a small force acting over a long time can have the same impulse.
Note that Fâˆ†t is the area of the rectangle, i.e., the area under the force vs time curve
If the applied force is not constant, then the impulse can be calculated as the area under the force vs. time graph.
It is the impulse which equals the change in momentum. In one dimension
J = âˆ†p
âˆ†p = pf - pi
âˆ†p =
Fâˆ†t
Let's first see qualitatively what an impulse curve might look like in a real collision in which the forces change over time during the collision.
low-friction cart with a spring plunger
-Collide the cart with the wall several times, and observe what happens to the spring plunger.
If friction is negligible, what is the net force exerted on the cart just before it starts to collide?
ma
When is the magnitude of the force on the cart maximum?
when it has the most/highest acceleration
Roughly how long does the collision process take? Half a second? Less time? Several seconds?
fraction of a second
Remembering what you observed, attempt a rough sketch of the shape of the force the wall exerts on the cart Fw as a function of time during the collision.
force vs time graph where it is steady then has a millisecond spike downwards
During the collision the force is not constant. In order to measure the impulse and compare it to the change in momentum of the cart, you must
1) plot a force-time graph, and find the area under it, and 2) measure the velocity of the cart before and after the collision with the wall. Fortunately the force probe, motion detector and MacMotion software will allow you to do this.
The force probe will be mounted on the cart to measure the force applied to the cart. Instead of using a plunger mounted on the cart, you can make the collision nearly elastic, and stretch it out over time by colliding the cart into a "springy wall." You can either collide the cart into a spring mounted on the wall, or into a springy material like foam rubber or a large springy ball.
in a perfectly elastic collision between a cart and a wall, the cart would recoil with
exactly the same magnitude of momentum that it had before the collision. Because your cart;'s spring bumper is not perfect, you can only produce a nearly elastic collision
Does the shape of the force-time graph agree with your Prediction 2-1? Explain.
yes, the force was measured at a fraction of a second downwards
Calculate the change in momentum of the cart. Show your calculations.
(0.40461 kg x -0.3026 m/s) - (0.40461 kg x 0.5994 m/s) =
-0.902kg m/s
use the software to find the area under the force-time graph-the impulse
J = -0.2671 Nâ€¢s
Did the calculated change in momentum of the cart equal the measured impulse applied to it by the wall during the nearly elastic collision? Explain.
yes, close to zero
What would the impulse be if the initial momentum of the cart were larger? What if the collision were inelastic rather than elastic, i.e.. what if the cart stuck to the wall after the collision?
...
It is also possible to examine the impulse-momentum theorem in a collision where the cart sticks to the wall and comes to rest after the collision.
This can be done by replacing the bouncy wall with some clay, and attaching a pointy nail to the end of the force probe.
prediction: Now when the cart hits the wall, it will come to rest. What do you predict about the impulse? Will it be the same, larger or smaller than in the nearly elastic collision. What do you predict now about the impulse and momentum? Will they equal each other, or will one be larger than the other?
-larger, impulse because final momentum is zero
-equal
Compare the force-time curve for the inelastic collision to that for the nearly elastic collision.
-elastic = bounces back with same velocity
-inelastic = stops
both have a similar force
Were the impulse and change in momentum equal to each other for the inelastic collision? Explain why you think the results came out the way they did.
No, clay has some bounce back
Do you think that the momentum change is equal to the impulse for all collisions? Justify your answer.
yes, impulse is always change in momentum
change in p = mvf - mvi
lab 9
Newton's third law and conservation of momentum
definition of momentum
the quantity of motion of a moving body, measured as a product of its mass and velocity
what is your prediction 1-2? does one car exert a larger force on the other or are both forces the same size?
both forces are the same
in activity 1-1. why is the sign of force probe A reversed?
since the push on it is negative
what is your prediction 1-7 when the truck is accelerating? Does either the car or the truck exert a larger force on the other or are the forces the same size?
in both cases (rest and movement) they exert the same force on each other
what makes the collision in activity 2-1 "inelastic"?
the clay will cause the carts to stick together and create one mass instead of 2 separate masses and they will create a velocity of zero
A weight lifter struggles but manages to keep a heavy barbell above his head. Occasionally he slips and the barbell starts to fall downward, but he always recovers. Compare the force exerted by the weight lifter on the barbell to that exerted by the barbell on the weight lifter. Which of the following are correct?
The force exerted by the weight lifter on the barbell is equal to the force exerted by the barbell on the weight lifter.
A bowling ball rolls down the alley and hits a pin. Compare the force exerted by the ball on the pin to the force exerted by the pin on the ball during the collision between the ball and the pin. Which of the following best describe the situation?
A. The magnitudes of the forces are equal.
B. The bowling ball obviously applies a larger force on the pin because the pin moves.
C. The forces are directed in the same direction, that is why the pin moves.
D. The directions of the forces are opposite because the surfaces of the ball and pin push back against each other.
E. The forces should be equal and opposite if the collision is perfectly elastic, but not if it is an inelastic collision.
A and D
A tennis player hits the ball with her racket. Compare the force exerted by the racket on the ball to that exerted by the ball on the racket during the collision between the racket and the ball. Which of the following is correct?
The two forces are equal and opposite at all the times the ball and the racket are in contact.
An automobile of mass 1500kg moving at 25.0m/s collides with a truck of mass 4500kg at rest. The bumpers of the two vehicles lock together during the crash. Compare the force exerted by the car on the truck with that exerted by the truck on the car during the collision.
They are equal to each other in magnitude.
An automobile of mass 1500kg moving at 25.0m/s collides with a truck of mass 4500kg at rest. The bumpers of the two vehicles lock together during the crash. What is the final velocity of the car and truck just after the collision?
Vf = [1500/(1500 + 4500)] * 25 = 6.25m/s
An automobile of mass 1500kg moving at 25.0m/s collides with a truck of the mass 4500kg at rest. The car and the truck bounce off each other completely elastically. Compare the force exerted by the car on the truck with that exerted by the truck on the car during the collision.
They are equal to each other in magnitude.
An automobile of mass 1500kg moving at 25m/s collides with a truck of mass 4500kg at rest. This time, the car and the truck bounce off each other completely elastically. What is the final velocity of the car after the collision? Call the initial direction of the automobile positive.
(1500)(25) + (4500)(0) = 1500v1f + 4500v2f
v2f = (37500-1500v1)/4500
37500 = 1500vf1 + 4500(37500-1500v1)/4500)
vf1 = -12.5 m/s
An automobile of mass 1500kg moving at 25m/s collides with a truck of mass 4500kg at rest. This time, the car and the truck bounce off each other completely elastically. What is the final velocity of the truck after the collision? Call the direction of the automobile before the collision positive.
vf2 = 12.5 m/s
-because found vf1 in previous question to be -12.5, and they are the same magnitude of force in opposite directions according to Newton's third law
Newton's Third Law
To every action there is always opposed an equal reaction, or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.
If you press a stone with your finger, the finger is also pressed by the stone. If a horse draws a stone tied to a rope, the horse (if I may say so) will be equally drawn back towards the stone . . . .
In Lab 8, you looked at the definition of momentum, and examined the momentum changes of objects undergoing collisions. We focused our attention on the momentum change that an object undergoes when it experiences a force that is extended over time (even if that time is very short!). You also looked at the forces acting on objects during collisions, and examined the impulse-momentum law which compares the change in momentum to the impulse.
Since interactions like collisions and explosions never involve just one object, we would like to turn our attention to the mutual forces of interaction between two or more objects. This will lead us to a very general law known as Newton's Third Law which relates the forces of interaction exerted by two objects on each other. Then, you will examine the consequences of this law and the impulse-momentum law which you examined in the last lab, when they are applied to collisions between objects. In doing so, you will arrive at one of the most important laws of interactions between objects, the Conservation of Momentum Law.
As usual you will be asked to make some predictions about interaction forces and then be given the opportunity to test these predictions.
There are many situations where objects interact with each other, for example, during collisions. In this investigation we want to compare the forces exerted by the objects on each other. In a collision, both objects might have the same mass and be moving at the same speed, or one object might be much more massive, and they might be moving at very different speeds. What factors might determine the forces between the objects? Is there some general law which relates these forces?
collision interaction forces
-What can we say about the forces two objects exert on each other during a collision?
Suppose the masses of two objects are the same and that the objects are moving toward each other at the same speed so that m1 = m2 and v1 = - v2 (same speed, opposite direction).
Predict the relative magnitudes of the forces between object 1 and object 2 during the collision.
the objects exert the same force on each other
Suppose the masses of two objects are the same and that object 1 is moving toward object 2, but object 2 is at rest.
m1 = m2 and v1 > 0, v2= 0
Predict the relative magnitudes of the forces between object 1 and object 2 during the collision.
the objects exert the same force on each other
Suppose the mass of object 1 is greater than that of object 2 and that it is moving toward object 2 which is at rest.
m1 > m2 and v1 > 0, v2 = 0
Predict the relative magnitudes of the forces between object 1 and object 2 during the collision.
the objects exert the same force on each other
Provide a summary of your predictions. What are the circumstances under which you predict that one object will exert a greater force on the other object?
according to Newton's third law, every action has an equal and opposite reaction. so in every case, the objects should exert the exact same magnitude of force on each other, just in opposite directions
In order to test the predictions you made you can study gentle collisions between two force probes attached to carts.
You can strap additional masses to one of the carts to increase its total mass so it has significantly more mass than the other. If a compression spring is available you can set up an "explosion" between the two carts by compressing the spring between the force probes on each cart and letting it go.
Two carts of the same mass moving towards each other at about the same speed.
same force exerted on each other
-graph of force A was +35 N and graph of force B was -35 N
Two carts of the same mass, one at rest and the other moving towards it.
same force exerted on each other
-graph of force A was +25 N and graph of force B was -25 N
One cart twice or three times as massive as the other, moving toward the other cart which is at rest.
same force exerted on each other
-graph of force A was +15 N and graph of force B was -15 N
Did your observations agree with your predictions? What can you conclude about forces of interaction during collisions? Under what circumstances does one object experience a different force than the other during a collision? How do forces compare on a moment by moment basis during each collision?
No, all exert the same force on each other in every case
You have probably studied Newton's Third Law in lecture or in your text. Do your conclusions have anything to do with Newton's Third Law? Explain.
-for every action, there is an equal and opposite reaction
-every case had an equal force
How does the vector impulse due to cart 1 acting on cart 2 compare to the impulse of cart 2 acting on cart 1 in each collision? Are they the same in magnitude or different? Do they have the same sign or a different sign?
same magnitude in opposite directions (one positive and one negative)
Interaction forces between two objects occur in many other situations besides collisions. For example, suppose that a small car pushes a truck with a stalled engine, as shown in the picture below. The mass of object 1 (the car) is much smaller than object 2 (the truck).
At first the car doesn't push hard enough to make the truck move. Then, as the driver pushes down harder on the gas pedal, the truck begins to accelerate. Finally the car and truck are moving along at the same constant speed.
predictions of the relative magnitudes of the forces between objects 1 and 2 Before the truck starts moving:
the car and truck exert the same force on each other
predictions of the relative magnitudes of the forces between objects 1 and 2 While the truck is accelerating:
the car and truck exert the same force on each other
predictions of the relative magnitudes of the forces between objects 1 and 2 After the car and truck are moving at a constant speed:
the car and truck exert the same force on each other
push cart 1 toward the right. At first hold cart 2 so it cannot move, but then allow the push of cart 1 to accelerate cart 2, so that both carts move toward the right.
How do your results compare to your predictions? Is the force exerted by cart 1 on cart 2 (reading of force probe 2) significantly different from the force exerted by cart 2 on cart 1 (reading of force probe 1) during any part of the motion? Explain any differences you observe between your predictions and your observations.
same magnitude of force on both
Explain how cart 2 is able to accelerate. Use Newton's second law, and analyze the combined (net) force exerted by all the forces acting on cart 2. Is there a nonzero net force?
there is a nonzero net force because they are both moving with each other in the same direction
our previous work should have shown that interaction forces between two objects are
equal in magnitude and opposite in sign (direction) on a moment by moment basis for all the interactions you might have studied. This is a testimonial to the seemingly universal applicability of Newton's Third Law to interactions between objects.
As a consequence of the forces being equal and opposite at each moment, you should have seen that the impulses of the two forces were always
equal in magnitude and opposite in direction.
You may have seen in lecture or in your text a derivation of the Conservation of Momentum Law by combining these findings with the impulse-momentum theorem (which is really another form of Newton's Second Law since it is derived mathematically from the Second Law). The argument is that the impulse J1 acting on cart 1 during the collision equals the change in momentum of cart 1, and the impulse J2 acting on cart 2 during the collision equals the change in momentum of cart 2:
J1 = Î”p1
J2 = Î”p2
But, as you have seen, if the only forces acting on the carts are the interaction forces between them, then J1 = - J2. Thus, by simple algebra
Î”p1 = -Î”p2
or
Î”p1 + Î”p2 = 0
i.e., there is no change in the total momentum p1 + p2 of the system (the two carts).
If the momenta of the two carts before (initial--subscript i) and after (final--subscript f) the collision are represented as in the diagrams below,
-before the collision: --p1 = m1v1i--> <--p2 = m2v2i--
-after collision: <--p1 = m1v1f-- --p2 = m2v2f-->
then...
pf = pi
where
pi = m1v1i + m2v2i
pf = m1v1f + m2v2f
conservation of momentum in an inelastic collision: velcro cars
prediction: you are going to give the more massive cart 1 a push and collide it with cart 2 which is initially at rest. The carts will stick together after the collision. Suppose that you measure the total momentum of cart 1 and cart 2 before and after the collision. How do you think that the total momentum after the collision will compare to the total momentum before the collision. Explain the basis for your prediction.
there will be an equal momentum after the collision because p = mv and after they collide, the mass doubles because the carts become one mass, but the velocity decreases from the collision force
total momentum before the collision
mass of cart A * velocity of cart A
total momentum after the collision
(mass of cart A + mass of cart B) * velocity of AB
Was momentum conserved during the collision? Did your results agree with your prediction? Explain.
yes and yes
-because they had about the same momentum before and after the collision
Momentum is conserved whether or not the two carts stick to one another after the collision, but to test this you would need to measure
the velocities of both carts after the collision.
Momentum is also conserved if
both carts are moving before the collision
lab 10
two-dimensional motion (projectile motion)
so far we have been dealing separately with motion along a horizontal line and motion along a vertical line.
The focus of this lab is to describe the motion close to the surface of the earth that occurs when an object is allowed to move in both the vertical and horizontal directions
examples are the motion of a baseball or tennis ball after being hit. This type of motion is commonly called projectile motion. To understand this motion, it is helpful to review vertical and horizontal motions separately and then consider how they might be combined
This lab begins with a review of classic kinematic equations that describe the relationships between instantaneous position, velocity, and acceleration of objects that move in one dimension.
In some cases an object moves with constant velocity (zero acceleration). In others, such as the motion of a cart being pushed along by the constant force of a fan unit as in lab 2 or the falling motion of a ball pulled by the constant gravitational force in lab 6, a object moves with a constant acceleration. You have already examined a number of examples of these types of one-dimensional motion
In this lab you will focus on the simultaneous vertical and horizontal motions of objects with a horizontal component of velocity while acted on by
the downward gravitational force of attraction of the earth in the vertical direction
prediction: Suppose you roll a basketball along a horizontal table and measure its position and velocity as functions of time over a short enough distance so that the ball's velocity remains fairly constant. Graph them (position versus time and velocity versus time)
start at zero and increase position constantly over time (positive slope)
starts at zero and increases to a point then remains at a steady straight line over time
describe the shapes of your graphs. Did they agree with your predictions
yes
would you describe this motion with a constant velocity or constant acceleration? explain.
both, because in both cases the increase and then remain at a steady rate over time
Prediction: Now suppose you toss the ball up and catch it on its way down. Graph your predictions for the position, velocity, and acceleration of the ball from the moment it leaves your hands until just before you catch it
velocity: starts positive and has a negative slope down, highest point hits zero, then continues to decrease at the same rate
position: goes up until the highest point (positive) then decrease into the negative area
acceleration: constant negative acceleration
would you describe this motion as having constant velocity or constant acceleration? explain.
constant acceleration, because it stays negative the whole time
the motions of the basketball examined in activities 1-1 and 1-2 can be represented as functions of time by a series of mathematical equations called
the kinematic equations
Kinematic equations that might represent the position, velocity, and acceleration of an object as functions of time
set #1:
x = x0 + v0t
v = v0
a = 0
set #2:
x = x0 + v0t + 1/2at^2
v = v0 + at
a = constant
-x is position
-v is instantaneous velocity
-a is a constant acceleration along the x-axis
-x0 is the initial position at t = 0 (tells you where the object was at the moment you started counting time)
-v0 is initial velocity component at t = 0 (tells you how fast the object was moving at the moment you started counting time
use your results from activity 1-1 to determine which set of kinematic equations could be used to represent the motion of the basketball along the table where the x-axis is assumed to point along the line of motion of the ball. Explain your choice based on the graphs you obtained
set #1:
x = x0 + v0t
v = v0
a = 0
because these can all be used for a case that has a constant 0 acceleration
use your graphs from activity 1-1, and the kinematic equations you chose to determine the values of x0 and v0. Explain how you found there values and what they represent
x = x0 + v0t
v = v0
use your results from activity 1-2 to determine which set of kinematic equations could be used to represent the motion of the basketball tossed straight upward. Explain your choice based on the graphs you obtained
set #2:
x = x0 + v0t + 1/2at^2
v = v0 + at
a = constant
because these can all be used for this case where acceleration is constantly negative
Use your graphs from activity 1-2, and the kinematic equations you chose to determine the values of x0 and v0. Explain how you found there values and what they represent
x = x0 + v0t + 1/2at^2
v = v0 + at
use your graphs from activities 1-1 and 1-2 to determine the values of a for (a) the motion of the ball along the horizontal surface and (b) the motion of the ball tossed vertically upward. Explain how you determined these values
a.) a =
b.) a =
-found by using the kinematic equations
the world is full of phenomena that we can know of directly through our senses-objects moving, pushes and pulls, sights and sounds, winds and waterfalls. A vectors is a mathematical concept-a mere figment of the mathematician's fancy. But, vectors can be used to describe aspects of "real" phenomena such as
positions, velocities, acceleration, and forces. vectors are abstract entities that allow certain rules. For example, in the figure below, the velocity vector of an object is represented by the vector v, drawn relative to different coordinate axes (see pictures on page 221)
as discussed in labs 1 and 2, a vector has 2 key attributes
magnitude and direction
the magnitude of a vector can be represented by
the length of the arrow
its direction can be represent by
the angle theta, between the arrow and the coordinate axes chosen to help describe the vector
in earlier labs you drew vectors in only one dimension. But, vectors are especially useful in representing two-dimensional motion because
they can be resolved into components. In the middle figure on page 221, the velocity vector is resolved into x and y components. Added together, these components are equivalent to the original vector, but they can be analyzed independently of each other. This is one reason it is convenient to use vectors
if a cannonball is shot off a cliff with a certain initial velocity in the x direction, the two-dimensional motion that results is known as
projectile motion
. But will it continue to move forward in that direction at the same velocity and at the same time fall in the y direction as a result of the gravitational attraction between the Earth and the ball?
In this investigation you will examine the motion of a tennis ball that is tossed into the air so that it is moving in both the x and y direction.
The toss of the ball and its trajectory are shown in the photos on page 222
because the motion of the ball is in two dimensions, it is not easy to make measurements using a motion detector. Instead, you will use the method of
video analysis to examine the motion and to determine the mathematical representations of the horizontal and vertical components of the motion
prediction: graph your prediction for how the x coordinate of the ball and the x component of the velocity will vary with time. (assume that positive x direction in the photos is to the right)
position: increase in position over time (positive slope)
velocity: starts off positive, then decreases over time, hits zero at the maximum height, then continues to decrease at the same rate as before (negative slope)
prediction: graph your prediction of how the y coordinate of the ball and the y component of the velocity will vary with time
position: increase in position until it hits the maximum height, then it starts to decrease (looks like a hill)
velocity: starts off positive, then decreases over time, hits zero at the maximum height, then continues to decrease at the same rate as before (negative slope)
describe the shape of the trajectory of the ball
looks like a hill
does the graph for x vs time represent motion at a constant velocity or constant acceleration? How do you know?
constant acceleration
-since the velocity is constantly decreasing over time, then there is a constant acceleration of zero over time
choose the kinematic equation that describes vx versus time for this motion, and model it to find the values of the parameters in that equation, e.g., v0. (again, use the modeling feature in the software to find the equation that best represents the data, and find the parameter(s) from this model. Once again, make an intelligent guess for v0, and then adjust the value to get a mathematical equation that fits the measured data the best)
what is the kinematic equation?
...
does the graph for y versus time represent motion at a constant velocity or constant acceleration? How do you know?
constant acceleration
-since the velocity is constantly decreasing over time, then there is a constant acceleration of zero over time
choose the kinematic equation that describes y versus time for this motion, and model it to find the values of the parameters in that equation, e.g., v0y, y0, and ay. (again, use the modeling feature in the software to find the equation that best represents the data, and find the parameter(s) from this model. Once again, make an intelligent guess for v0, and then adjust the value to get a mathematical equation that fits the measured data the best)
what is the kinematic equation?
...
does the graph for vy versus time represent motion at a constant velocity or constant acceleration? How do you know?
constant acceleration
-since the velocity is constantly decreasing over time, then there is a constant acceleration of zero over time
choose the kinematic equation that describes vy versus time for this motion, and model it to find the values of the parameters in that equation, e.g., v0y, y0, and ay. (again, use the modeling feature in the software to find the equation that best represents the data, and find the parameter(s) from this model. Once again, make an intelligent guess for v0, and then adjust the value to get a mathematical equation that fits the measured data the best)
...
find the value if the y component of the acceleration of the tennis ball
...
use your observations in the two investigations of this lab to justify the statement that projectile motion is a combination of horizontal motion at a constant velocity (zero acceleration) and vertical motion with a constant (gravitational) acceleration.
because the object is moving, it has multiple forces acting upon it (has a net force), so projectile motion will have to involve all of the rates of these forces acting on the ball
lab 10 homework
VVVVV
a heavy ball is released at the origin, and moves at a constant velocity on a horizontal floor along the dashed line in the graph below
graph: starts at zero and has a constant positive slope up
a.) is this projectile motion? explain your answer
b.) sketch the x and y components of the position, velocity, and acceleration of the ball as functions of time
a.) yes, because the only acceleration force on the ball/projectile is gravity
b.)
-position x: positive slope (distance increasing over time)
-velocity x: a straight line (constant velocity over time)
-acceleration x: straight line at zero (no acceleration in the x direction)
-position y: starts at zero and looks like a hill (position increases as it goes up and decreases after its highest point to come down)
-velocity y: starts positive and has a negative slope, reaching zero at its highest point, then continues to decrease in the negative (slows down as it gets to the top, then has a negative velocity as it comes down)
-acceleration y: a negative acceleration of -9.81, a straight line (only affected by gravity in the y direction)
the ball in question 1 begins at rest at the origin. It is tapped repeatedly by 2 students using batons. They alternate tapping in the x and y directions. (Note: these taps provide approximately constant forces in the x and y directions). The ball follows the same dashed path as the ball that rolls freely on the floor.
a.) according to the definition of projectile motion, is the tapped ball undergoing projectile motion? why or why not?
b.) sketch the x and y components of the position, velocity, and acceleration of the ball as functions of time
a.) no it is not projectile motion, because there isn't a acceleration of zero in the x direction and the x and y components aren't occurring at the same time
b.)
-position x: positive slope then negative slope = upside down v (increases then decreases between 2 students
-velocity x: constant
-acceleration x: increases at each hit then goes back to zero
-position y: still has a trajectory motion (hill)
-velocity y: starts positive and has a negative slope, reaching zero at its highest point, then continues to decrease in the negative (slows down as it gets to the top, then has a negative velocity as it comes down)
-acceleration y: only affected by gravity
a pitcher throws a baseball. She releases it so that it is initially moving in the horizontal direction. Is the motion of the ball after it leaves her hand projectile motion? Explain.
yes, because the only acceleration acting on the ball in the y direction is gravity
assuming that air resistance is negligible, sketch the x and y components of the position, velocity, and acceleration of the ball as functions of time. Assume that the origin of the coordinate system is at the point she releases the ball
-position x: positive slope (distance increasing over time)
-velocity x: a straight line (constant velocity over time)
-acceleration x: straight line at zero (no acceleration in the x direction)
-position y: starts at the height of where the girl threw it and decreases once it starts to drop
-velocity y: starts at zero then continues with a negative slope as it starts to fall
-acceleration y: gravity (starts at zero then decreases to -9.81 when it starts to drop)
a ball is moving through the air. The data in the table below is either the x or y position coordinates of the ball as a function of time. (The positive y axis is upward.) Use this data to find velocities as a function of time, and determine from the velocities whether these data are x or y coordinates. Be sure to explain thoroughly how you reached your conclusion
Time (s): 0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.8, 2.0
position (m): 5.00, 7.80, 10.22, 12.24, 13.86, 15.10, 15.94, 16.40, 16.46, 16.12, 15.40
this is the y position coordinates, because velocity is not constant over time, it is changing. Also, the position starts to decrease after 1.6 seconds and it covers less distance as time goes on, which means it is slowing down in the y direction and starting to be pulled back down by gravity
suppose a rocket in outer space is thrust along the y direction with an acceleration of 15 m/s^2 while drifting freely (no applied force) in the x direction. What is the path followed by the rocket?
graph starts at the origin in the top left corner (0,0) and arches down (y) and to the right (x)
is this motion similar to projectile motion? explain.
no, because the acceleration in the y direction is greater than the force of gravity
Lab 12:
conservation of energy
fill in stuff for lab 12
...
Lab 12
Conservation of energy
in lab 11 on work and energy, we defined the kinetic energy associated with the motion of a rigid object
K = 1/2mv^2
where m is the mass of the object, and v is its speed
In this lab we will consider other forms of energy associated with both
gravitational and spring forces.
Suppose you lift an object steadily at a slow constant velocity near the surface of the earth so that you can ignore any change in kinetic energy. You must do work (apply a force over a distance) to lift the object, because
you are pulling it away from the earth. The lifted object now has the potential to fall back to its original height, gaining kinetic energy as it falls. Thus, if you let the object go
It is very useful to define the gravitational potential energy of an object at height y (relative to a height y = 0) as
the amount of work needed to move the object away from the earth at constant velocity through a distance y
If we use the previous definition, then the potential energy of an object is a maximum when
it is at its highest point. If we let it fall, then the potential energy becomes smaller and smaller as it falls toward the earth while the kinetic energy increases as it falls. We can now think of kinetic energy to be 2 different forms of mechanical energy. We define the mechanical energy as the sum of these 2 energies
IS the mechanical energy constant during the time the mass falls toward the Earth?
If it is, then the amount of mechanical energy doesn't change, and we say that the mechanical energy is conserved. If the mechanical energy i conserved in other situations, we might be able to hypothesize a law of conservation of mechanical energy
law of conservation of mechanical energy
In certain situations, the sum of the kinetic and potential energy, called the mechanical energy, is a constant at all times, It is conserved
The concept of mechanical energy conservation raises a number of questions. Does it hold quantitatively for falling masses-Is the sum of the calculated potential and kinetic energies exactly the same number as the mass falls? Can we apply a similar concept to masses experiencing other forces, such as those exerted by springs? Perhaps we can find another definition for elastic potential energy for a mass-spring system. In that case could we say that mechanical energy will also be conserved for an object attached to a spring? Often there are frictional forces involved with motion/ Will mechanical energy be conserved for objects experiencing frictional forces, like those encountered in sliding?
In this lab you will begin by exploring the common definition of gravitational potential energy to see if it makes sense. You will then measure the mechanical energy, defined as the sum of gravitational potential energy and kinetic energy, to see if it is conserved when the gravitational force is the only force acting. Next, you will explore a system where the only net force is exerted by a spring and see the definition of elastic potential energy. You will measure the mechanical energy of this system and see if it is conserved. Finally, you will explore what effects sliding frictional forces or air resistance forces have on systems. You will explore whether or not mechanical energy is still conserved in such systems
based on your knowledge from lab 11, how much work will you do to lift the object through a distance y? Explain.
F = ma
a = gravity
PE = mgy
we choose to define the gravitational potential energy, Ugrav, of an object of mass m to be equal to the work done against the gravitational force to lift it
Ugrav = mgy
Now we can use this equation to calculate the potential energy of a ball
mass of the ball
0.606 kg
do the 2 graphs look similar? does this surprise you? Explain.
they look the same because energy is always conserved
Gravitational potential energy is always measured with respect to a particular height where its value is defined to be zero. In this case, what has been chosen as this reference level? In other words, for what location of the ball would its gravitational potential energy be zero?
when the ball is on the ground
suppose that the ball is dropped and you know its velocity at a certain time. What equation would you use to calculate the kinetic energy of the ball?
KE = 1/2mv^2
prediction: suppose that the ball is dropped from some height. What equation would you use to calculate the mechanical energy (the sum of the gravitational potential energy and the kinetic energy)?
ME = KE + PE
ME = 1/2mv^2 + mgh
prediction: as the ball falls, how will the kinetic energy change? How will the gravitational potential energy change? How will the mechanical energy change?
KE increases
PE decreases
ME stays the same
how did the variation of kinetic energy and gravitational potential energy compare to predictions
KE continued to increases
PE increased then decreased which is close to prediction, but not the prediction pf GPE increasing
what seems to be true about the mechanical energy defined as the sum of the kinetic energy and the gravitational potential energy? Did this agree with your prediction?
ME increased instead of remaining the same, while PE and KE increased then decreased once the ball hit the ground
the gravitational potential energy of the cart which has traveled a distance x up the ramp is mgy, where y is the height of the cart above the table top. find an equation for Ugrav in terms of position x measured by the motion detector along the ramp, the length L of the ramp, and the elevation H of the end of the ramp. (Hint: y = x sinÎ¸)
What is the referance height for the potential energy, i.e., the height where potential energy is zero?
at the bottom (on the table)
measure the values of L and H. Also measure the mass of the cart. Write the equation for Ugrav from the last question in the form of a single constant times x by using the values of L, H (or for sinÎ¸), and the gravitational acceleration, 9.80 m/s^2
L = 1.2 m
H = 0.2 m
mass = .278 kg
Ugrav = mgy
prediction: as the cart rolls down the ramp, how will KE change? How will the gravitational PE change? How will the mechanical energy change?
KE increase
PE decrease
ME increase --> decrease
compare your graphs to those for the falling ball in activity 1-2. How are they similar and how are they different?
mechanical energy doesn't decrease in the graph, only increases
KE does not increase much
What kind of variation is there in the mechanical energy as the cart rolls down the ramp? Does this agree with your prediction? explain
mechanical energy slowly increases then seems to constantly increase, doesn't agree with prediction
the mechanical energy, the sum of the KE and gravitational PE, is said to be conserved for an object moving only under he influence of gravitational force. That is,
the mechanical energy remains constant throughout the motion of the object. This is known as the conservation of mechanical energy
prediction: suppose that the cart is given a push up the ramp and released. It moves up, reverses direction, and comes back down again. How will the KE change? How will the gravitational PE change? How will the ME change? describe in words and sketch your predictions with labeled dashed lines on the axes below
KE will increase then decrease
PE will decrease then increase
ME will stay constant
How does the mechanical energy change as the cart rolls up and down the ramp? Does this agree with your prediction?
the ME increases, then stays constant at the top then increased in the opposite direction
NO
as mentioned in the overview, it is useful to define other kinds of potential energy besides gravitational potential energy. In this investigation, you will look at another common type
the elastic potential energy, which is associated with the elastic force exerted by a spring that obeys Hooke's law
You have seen in lab 11 that the magnitude of the force applied by most springs is proportional to the amount the spring is stretched from beyond its unstretched length. This is usually written
F = kx, where k is called the spring constant
the spring constant can be measure by
applying measured forces to the spring and measuring its extension
you also saw in lab 11 that the work done by a forces can be calculated from
the area under the force vs position graph.
force vs position graph for spring:
positive slope with "kx" written under the graph
-note: k is the slope of this graph
How much work is done in stretching a spring constant k from its unstretched length by a distance x? (Hint: look at the triangle on the force vs distance graph previously mentioned and remember how you calculated the work done by changing force in lab 11).
tanÎ¸ = y/x
If we define the elastic potential energy of a spring to be the work done in stretching the spring, the definition will be analogous to the way we defined Ugrav. In this case, the elastic potential energy would be
Uelas = 1/2kx^2
In this investigation, you will measure the kinetic energy, elastic potential energy, and mechanical energy (defined as the sum of kinetic energy and elastic potential energy) of a mass hanging from a spring when air resistance is very small, and again when it is significant.
You will see if the mechanical energy is conserved
Note: In the activities that follow, you will explore the mechanical energy of a hanging mass oscillating on a spring. The equilibrium position depends on the gravitational force on the mass. However, it can be shown mathematically that the motion of the mass relative to the equilibrium position is only influenced by
the spring force and not by the gravitational forces. Therefore, only Uelas (and not Ugrav) needs to be included in the mechanical energy
to calculate the elastic potential energy of a stretched spring, you need first to determine the spring constant k. since F = kx, this can be found by measuring
a series of forces F and corresponding spring stretches x
-hang the spring an the mass carrier from the force probe and then zero the force probe
-begin graphing. Keep the force value, and enter the value of the hanging mass. Then add 50 g to the carrier, and again record this data point. Repeat until you have added 200 g to the carrier
graph: position (m) vs Force (N)
negative slope, force decreases as position increases
use software to determine spring constant
k = -4.849 N/m
was the force exerted by the spring proportional to the displacement of the spring?
yes, as force was exerted the spring went down
If 2 springs are stretched different amounts by the same mass hung from them, which spring has the larger spring constant, the one that stretches most or the one that stretches the least? Explain.
the higher the weight, the more it stretches out
one that stretches the most
prediction: hang the 200 g mass from the spring. Start it oscillating by pulling it down a small distance and letting it go. As the mass oscillates up and down, what equation would give mechanical energy, including elastic potential energy and kinetic energy? (Note that since the mass oscillates up and down around its new equilibrium position, which is where the spring force just balances out the gravitational force [weight of the mass], you don't need to consider gravitational potential energy)
elastic potential + Kinetic
prediction: as the mass oscillate up and down, how will the kinetic energy change? How will the elastic potential energy change? How will the mechanical energy change?
KE will increase and decrease
PE will do the same
ME wil also increase and decrease
Test you predictions: you will first need to find the new equilibrium position of the spring with the 200 g mass hanging from it
suspend the 200 g mass from the spring and be sure it is at rest
-enter the value of the hanging mass into the formula for kinetic energy and the value of the spring constant into the formula for elastic potential energy. Notice that mechanical energy is now calculated as Uelas + K
start the spring oscillating with an amplitude of about 5 cm, and begin graphing
graphs: Mechanical energy (J) vs time and Uelas and K (J) vs time
rolar coaster hills (up and down and up and down) for both graphs
How did the variations of kinetic energy and elastic potential energy compare to your predictions?
didn't follow predictions
What seems to be true about the mechanical energy defined as the sum of kinetic and the elastic potential energy? did this agree with your predictions?
as the KE and PE went up and down, the ME also went up and down
Heat and thermodynamics lab 1
introduction to heat and temperature
With this lab we begin the study of thermodynamics,
a way of looking at physical phenomena that is very different from studying mechanics. Thermodynamics is the study of heat transfer and the resulting temperature changes. Much of the physics studied in mechanics involves motions that we can see, while many of the changes we will encounter in thermodynamics will not be visible without the help of indirect measuring instruments such as temperature sensors and pressure sensors.
There are several ways of changing the temperature of something.
One way is to put two materials at different temperature together: the hot one will become cooler and the cooler one will become hotter. Over many centuries, people argued about why this happened: some imagined a material substance they called heat moving between the two things. The real physics of what happens only began to come clear when someone noticed that there are other ways of changing the temperature of something: you can do non-conservative work on it (i.e., work where mechanical energy is lost somewhere.) For example, you could rub a piece of metal on emery paper or sandpaper and measure its temperature increase with a thermometer. Also, as you will observe in the next lab, it is possible to produce a temperature increase using an electric heater by supplying electrical energy to it. Observations like these caused physicists and engineers in the middle of the nineteenth century to conclude that heat is just a form of energy, the form that flows when there is a temperature difference between two objects. Today a physicist or engineer would say that heat is a form of heat (or thermal) energy transfer.
In this investigation you will explore ways of raising the temperature of a system by converting other forms of energy, and in particular mechanical work on the system.
If you hold a piece of metal in your hand and rub it back and forth on emery paper or sandpaper, do you expect the temperature of the metal to change? If so, will the temperature increase or decrease?
yes, increase
You heat a cup of coffee or the hot water flowing into your bathtub, so that the water is "hotter" than the room, i.e., the temperature is higher. Even if both have the same temperature, say 100Â°F, it certainly cost more to heat the water for your bath than for the coffee. To understand fully why, and to understand the limitations of any means of transforming one form of energy into another (whether by a gasoline engine, a fuel cell, or your body), we need to use the words
"temperature" and "heat" very carefully, more carefully than we do in everyday life. It is essential to understand the difference between the temperature of an object and the heat transferred to or from the object.
Suppose that you rub the metal back and forth for twice as long a time. Will the temperature change be different from before? If so, how will the temperature change differ?
...
By rubbing with the same force over the same length, the rate of doing work (the power) should be
constant. If this is so then the total work done in rubbing the metal on the emery paper should be proportional to the time interval of rubbing. That is, if you rub for twice as long, you will do twice as much work.
Grasp the nail near its head with your fingers, with the head facing down, and rub it vigorously back and forth on the emery paper for about 20 seconds.
Question 1-3: Describe what you felt. Was there a temperature change?
Question 1-4: Was there any transfer of matter to the nail as the rubbing took place? Was there any evidence of mechanical work being done while the rubbing took place? Explain.
1-3: yes, increase in temperature
1-4: no transfer of matter, but there was evidence of mechanical work because heat energy was added
temperature vs time graph: mechanical work and temperature change
increased from room temperature to about 28ÂºC in almost a minute
table 1-1
change in temperature from 10s to 30s: 0.36ÂºC
change in temperature from 50s to 90s: 1.02ÂºC
Based on your data, does there appear to be a relationship between the temperature change and the work done by rubbing? Explain.
yes, friction increases temperature
As you rub the metal on emery paper, you are doing
mechanical work, but the final result is a change in temperature. Thus it seems appropriate to associate changes in temperature to an energy transfer.
What does the shape of your graph imply about the relationship between the work done and the temperature change?
increase in work = increase in temperature
A thermometer can be based on any property that changes continuously with temperature. The volume of a fixed mass of mercury or alcohol expands when it is heated and contracts when it cools, which gives us the thermometers we are most used to. The ease with which electric currents move through metals and other materials also change with
temperature; this property make them particularly useful thermometers because the results can easily be read by a computer. We used such a thermometer in the last investigation.
Why not hold the metal directly instead of using the foam insulation?
because your body heat will affect the temperature
table 2-1
Room temperature water:
-glass thermometer: 22ÂºC
-sensor 1: 22.3ÂºC
-sensor 2: 22.1ÂºC
-approximate ÂºF: 71.6ÂºF
-calculated K: 95.15 K
Hot water
-glass thermometer: 65.5ÂºC
-sensor 1: 67.1ÂºC
-sensor 2: 67.1ÂºC
-approximate ÂºF: 152.78ÂºF
-calculated K: 340.25 K
Ice water:
-glass thermometer: 2.5ÂºC
-sensor 1: 4.2ÂºC
-sensor 2: 4.0ÂºC
-approximate ÂºF: 39.74ÂºF
-calculated K: 277.15 K
Several different temperature scales are commonly used: In everyday life in the U.S, we use the
Fahrenheit scale, where the range 0-100Â°F covers the range of usual outdoor temperatures pretty well. On this scale, the temperature at which water freezes at sea level is about 32Â°F, while water boils at about 212Â°F. The celsius scale, on which water freezes at about 0Â°C (more exactly, the triple point of pure water, the temperature and pressure at which solid ice, liquid water, and water vapor can coexist is 0.01Â°C) and boils at about 100Â°C, was meant to be more scientifically useful; it is currently the scale used in almost all of the world outside of the U.S. The third major scale is the kelvin scale: it is very similar to the celsius scale, except that 0K is set at "absolute zero", the lowest temperature which is conceivable for ordinary materials by current scientific theories. On this scale the triple point of water is 273.16 K, so that T(K) = T(Â°C) + 273.15 K.
Do the readings of the three thermometers seem to agree? If not, why might they not agree?
pretty much
Calculate the Kelvin temperatures corresponding to the glass bulb thermometer temperatures in Table 2-1 and record them in the last column of the table.
T(K) = T(Â°C) + 273.15
Ice water halfway to room temperature air:
55s
Ice water halfway to room temperature water:
5s
Which reached half-way to room temperature faster? What precautions do these measurements suggest in making temperature measurements?
the one in water
don't add extra work or heat
A temperature sensor always measures its own temperature. When you touch the sensor to something, it takes a few seconds for the sensor and the object to reach a common temperature. After this happens the sensor is said to be in
thermal equilibrium with the object it is touching. To make accurate temperature measurements it is important to wait until the temperature reading of the sensor remains fairly constant (until the temperature sensor and the object are in thermal equilibrium). As you have just seen, for a given thermometer this time lag varies considerably, depending on what system's temperature you are measuring. You also need to beware of cooling by evaporation. Be careful not to measure air temperatures when the sensitive part of the thermometer is wet (especially with alcohol). Evaporation of liquid from the tip of the thermometer will cool it.
Why were you asked the length of time it took to reach halfway, not all the way, to room temperature? (2 hints: How well can you determine when the sensor reaches room temperature? How long would you guess it takes?)
the one in the air would've taken too long
Suppose you take two thermometers out of hot water, dry one, and then wave both around in the air. Will there be any difference in the time it takes them to reach room temperature? (Hint: Imagine getting out of a shower on a dry day. How does perspiration work?)
yes
Hot water to room air (dried):
200s
Hot water to room air (wet):
200s
It is important to distinguish between the concepts of temperature and heat energy transfer as we have refined them in the last two activities. They are summarized:
1. Heat energy is energy in transit between two systems in thermal contact due to temperature difference only, with the hotter system losing heat energy as the cooler system gains it. To remind ourselves of this we will often use the phrase heat energy transfer.
2. Two objects are in thermal equilibrium, and hence have the same temperature if no energy on average is exchanged between them when they are placed in thermal contact.
Which reached room temperature faster? Can you explain why? What precautions do these measurements suggest in making temperature measurements?
neither
We know that when a hotter substance comes into thermal contact with a cooler one the temperatures of the two substances change. This temperature change is easy to observe in situations where substances (e.g., liquids) can be mixed together or come into thermal contact with each other without mixing.
Do the initial temperatures alone allow us to predict the final temperature of the system after the two substances have interacted with each other? In the next activity you will examine more carefully this interaction between hot and cold objects. First make a couple of predictions.
Suppose you have a mass of hot water at 80Â°C in a small, sealed, uninsulated container and a larger mass of water at room temperature (20Â°C) in a cup. The container is submerged in the water in the cup and left there for a long time.
A. What do you predict will be the final temperature of the water in the container? Will it be midway
between 80 and 20Â°C? Or will it be closer to 80Â°C or closer to 20Â°C?
B. Whatdoyoupredictwillbethefinaltemperatureofthewaterinthecup?Willitbemidway between 80 and 20Â°C? Or will it be closer to 80Â°C or closer to 20Â°C? Why?
C. Is it possible, knowing just the initial temperatures of the water (and not how much hot and cool water), to predict exactly what the final temperature will be? Explain.
A.) closer to 80Âº
B.) closer to 20Âº, because it is open, so the heat can escape
C.) No, because if there is more water, then there will be more molecules present, which means it will take longer for the temperature to change because it will take longer for the molecules to move faster (heat up) or slower (cool down)
Quickly do the following. Record the initial temperatures.
Hot container: ____ Cool cup: ____
hot = 69ÂºC
cold = 26ÂºC
temperature over time graph
the cold remained at a straight line of 26Âº
the hot started at 69 and slowly raised to 70ÂºC
table 3-1
Hot cup:
-mass: 303.9 g
-initial temperature: 69ÂºC
-final temperature: 70ÂºC
-change in temperature: 1ÂºC
Cold cup:
-mass: 289.4 g
-initial temperature: 26ÂºC
-final temperature: 26ÂºC
-change in temperature: 0ÂºC
Did the final temperatures of the hot water and cool water agree with your prediction? Does the final temperature seem to depend on anything besides the initial temperature of the water in the container and in the cup? Explain. (Hint: compare the masses of hot and cold water and the change in temperature of each.)
yes
no
What would happen to the final temperature if you changed the relative amounts of water so that there was the same amount of hot water in the film container but less cool water in the cup?
the cool water would reach room temperature quicker
As you have observed in this investigation, inevitably, parts of any thermally isolated (i.e., insulated) system having different temperatures will interact until the entire system is at the same temperature. This occurs even without an exchange of matter. You should expect that the relative masses of the parts of your thermally isolated system affect the value of the final equilibrium temperature. Thus,
the interaction between two parts of a system cannot be explained as a simple temperature exchange. We need to create a new concept to help us understand heating and cooling processes. Scientists have invented the term heat transfer to explain this phenomenon.
As the hot water cooled down, what happened to the room temperature water? Is it likely that an exchange of water between the container and the cup accounts for the temperature changes? If not, how did the two temperatures change even though no hot water mixed with the room temperature water?
room temperature stayed the same
yes
The use of the noun "heat" is misleading, since using this term to explain temperature changes implies the exchange of a substance between two parts of a system. The word "heat" is actually a sloppy shorthand for an interaction process that leads to temperature changes.
As a reminder that we are dealing with a process rather than a substance, we will use the term heat transfer and not simply heat.
We will explore more carefully in the next lab why heat is actually considered to be a form of energy. Thus what people casually refer to as "heating" is actually what scientists refer to as heat energy transfer.
How can you tell when thermal equilibrium has been reached? Is there any evidence that whatever flowed from one container of water to the other has stopped flowing? What is this evidence?
when the temperature stops changing
yes, no change in temperature
Temperature Change and Heat Transfer: other materials: temperature over time graph
...
A material of some kind at 80Â°C is added to a larger mass of water at room temperature (20Â°C) in a cup. The material is submerged in the water in the cup and left there for a long time. Does the final temperature of the water depend on what kind of material you add to the water? Will you get the same result with aluminum, glass, or other building material as with the same amount of water?
yes
no
Does the time it takes for the water to come to an equilibrium temperature depend on the material you add to it?
yes
These relationships mean that Qmaterial =
cmaterial x mmaterial x DTmaterial, where cmat is a constant of the material called the specific heat, which is the amount of heat it takes to change a standard mass of material by 1Â°C.
table 3-2
Aluminum:
-mass of cup and water: 338.8 g
-initial temperature of water: 23.01ÂºC
-final temperature of water: 23.57ÂºC
-mass of material: 31.1 g
-initial temperature of material: 39ÂºC
-final temperature of material: 24ÂºC
Marble:
-mass of cup and water: 351.8 g
-initial temperature of water: 22.63ÂºC
-final temperature of water: 23.57ÂºC
-mass of material: 58.1 g
-initial temperature of material: 39ÂºC
-final temperature of material: 24ÂºC
Hot water (table 3-1):
-mass of cup and water: 303.9 g
-initial temperature of water: 69ÂºC
-final temperature of water: 70ÂºC
The ratio in column 2 is thus
mcool water x DTcool water / mmaterial x DTmaterial = (Qcool water / ccool water) / (Qmaterial /cmaterial) = cmaterial / ccoolwater
the ratio between the specific heat of the material and the specific heat of (cool) water.
Is the change in water temperature the same for each of the two materials you added? Is it the same as when you did the same experiment with the hot water?
no
no
The specific heat of water was originally defined as 1 calorie:
the amount of heat energy it takes to raise 1 g of water by 1 degree. In the next lab, we will add known amounts (in Joules) of energy to translate this into more usual energy units: cwater = 4190 J/kg-Â°C
The amount of energy Qmaterial taken from the aluminum, marble, glass, or warm water to reduce its temperature must be equal to
the amount of energy Qwater put into the cool water to raise its temperature, as long as energy can't escape anywhere else (why do we use Styrofoam?)
The amount of energy put in to a material to change its temperature is proportional to the mass of the material, and also to the change of temperature (as long as that change is small). You can see this by noting that
when the warm and cold materials are the same, as for warm water in the last row, the ratio mcool water x DTcool water / mmaterial x DTmaterial in Table 3-3 is nearly 1; you'll see this more clearly in the
next lab.
Calculate the ratio mwater x DTwater / mmaterial x DTmaterial for each of your materials, and put it in the second column in Table 3-3
...
Fill in the rest of Table 3-3, giving the specific heat of your three materials in both calories and in J/kg.
...
Which material has the highest specific heat? Which material has the lowest one?
highest = hot water
lowest = marble
If you put a cup of hot chocolate at 90C on a table in a room kept at 25ÂºC you know it will cool down. How cool will the hot chocolate get?
25C
table 3-3
Aluminum:
-mwater x DTwater / mmaterial x DTmaterial: 1.22
-specific heat, calories: 1.22
-specifc heat, J/kg: 5.124
Marble:
-mwater x DTwater / mmaterial x DTmaterial: 1.14
-specific heat, calories: 1.14
-specifc heat, J/kg: 4.78
Hot water:
-mwater x DTwater: 303.9
-specific heat, calories: 303.9
-specifc heat, J/kg: 1276.38
c = Q/mDT
just multiply first column by 42
Was there any difference between the two materials in this time it took to reach equilibrium?
yes
In above question, If you put it outside where the temperature is 11C, how cool will it get?
11C
Give at least one example as to how the thermal properties of a material can effect how it is used. (Classic example: why is a hot pad much more important when touching a hot aluminum pan than when touching a hot pyrex pan, even if they have the same temperature? How might cake or bread cook differently in the two materials? What kind of material would make a better frying pan? What kind of material would make a better hot pad?)
the aluminum is a greater conductor of heat
Compare the initial rates of cooling in above questions(a) and (b). Are they same or is one larger?
The initial rate of cooling of when the chocolate is inside at 25ÂºC is smaller than when the chocolate is put outside at 11ÂºC
In the above question, where does the heat go as the hot chocolate cools down?
to the nearby surroundings
Below is the graph of the temperature of the hot chocolate vs. time if the hot chocolate starts at 90 0C and is placed in a room where the temperature is kept at 25 0C. Chose which of the curve best describes the cooling of hot chocolate.
decrease rapidly then slowly decrease until room temp
Which of the choices below best describes the shape of the graph, especially changes in the rate of cooling as the chocolate cools down, in terms of your observations in this lab.
-There is an inverse relationship between the temperature difference (between it and room temperature) and its rate of cooling.
-When the temperature difference between it and room temperature is small the rate of cooling is also small.
-When the temperature difference between it and room temperature is large the rate of cooling is also large.
You have a mass of hot water at 90 0C and twice the mass of cool water at 10 0C. The hot and cold water are mixed together. What will be the final temperature?
Closer to 10ÂºC.
The two containers of water below are completely insulated so that no heat can be transferred in or out. The water in both containers started at room temperature (20 0C), and heat was transferred to both using heating coils until they reached the indicated final temperatures. Which container had more heat transferred on it.
a.) 25ÂºC, 100 g
b.) 25ÂºC, 200 g
Container (B) because it has more water. The amount of heat transferred to change the temperature a given amount depends on the quantity of matter.
The two containers of water below are completely insulated. (No heat can be transferred in or out.) The water in both containers started at room temperature (20 0C), and heat was transferred to both using heating coils until they reached the indicated final temperatures. Which of the following is true?
a.) 25ÂºC, 100 g
b.) 50ÂºC, 10 g
The bigger container had more heat transferred.
A pot of water is heated on a particular stove. After the temperature of the water has increased a certain amount (but not to the boiling point), the temperature stops rising even though heat continues to be transferred to the water. How would it be possible to transfer heat to the water without changing the temperature?
When the water is very hot, much of the energy goes to evaporating the water instead of raising the temperature.
A piece of aluminum (specific heat 0.910kJ/kg0C) of mass 116g at 75ÂºC is dropped into a Styrofoam cup filled with 118ml water at 20ÂºC. What are the final temperatures of the water and the aluminum? Please use C instead of 0C in your answer.
heat lost by aluminium = mass x specific heat capacity x (initial temperature - final temperature)
0.116 x 910 x (75 - Tf) = 105.56(75 - Tf) = 7917 - 105.56Tf
heat gained by water = mass x specific heat capacity x (final temperature - initial temperature)
0.118kg of water
specific heat capacity of water = 4186J/KgC
0.118 x 4186 x (Tf - 20) = 493.948(Tf - 20) = 493.948Tf - 9878.96
heat lost by aluminium = heat gained by water
7917 - 105.56Tf = 493.948Tf - 9878.96 --> 17795.96 = 599.508Tf -->
Tf = 29.68ÂºC
A piece of aluminum (specific heat 0.910kJ/kg0C) of mass 193g at 71ÂºC is dropped into a Styrofoam cup filled with 121ml water at 20ÂºC. What are the final temperatures of the water and the aluminum? Please use C instead of 0C in your answer.
0.193 x 910 x (71 - Tf) = 12469.73 - 175.63Tf
0.121 x 4186 x (Tf - 20) = 506.506Tf - 10130.12
12469.73 - 175.63Tf = 506.506Tf - 10130.12 --> 22599.85 = 682.136Tf -->
Tf = 33.13ÂºC
A piece of aluminum (specific heat 0.910kJ/kg0C) of mass 175g at 74ÂºC is dropped into a Styrofoam cup filled with 183ml water at 20ÂºC. What are the final temperatures of the water and the aluminum? Please use C instead of 0C in your answer.
Hint 1: because the cup is well-insulated, you can assume that all the heat leaving the aluminum as it cools goes to heating the water.
Hint 2: How are the final temperature of the aluminum related to the final temperature of the water?
(Use 4.19kJ/kgoC for the specific heat of water.)
0.175 x 910 (74 - Tf) = 11784.5 - 159.25Tf
0.183 x 4186 x (Tf - 20) = 766.038Tf - 15320.76
11784.5 - 159.25Tf = 766.038Tf - 15320.76 --> 27105.26 = 925.288Tf -->
Tf = 29.29ÂºC
You baked two cakes: one in a light aluminum pan and one in a heavy pyrex pan.
1.) The pan that needs to give up the most heat energy to cool down to room temperature.
2.) The pan that gives up the heat energy the fastest
3.) The pan most likely to burn your fingers if you try to move it too soon.
4.) The pan in which the bottom of the cake is most likely to brown (same temperature oven, same time in oven.)
1.)
2.)
3.)
4.)
HEAT AND THERMODYNAMICS LAB 2:
RELATIONSHIP BETWEEN HEAT ENERGY, TEMPERATURE, AND THE INTERNAL STATE OF SYSTEM.
Heat is energy that is transferred between two systems in thermal contact with each other due to a temperature difference between them. Remember that although the term "heat" is used, even by scientists, it is a shorthand term for an energy transfer process rather than a substance transfer. We will continue to use consistently the term heat energy-transfer to remind you that
the transfer of heat from one system to another does not necessarily involve the exchange of matter.
We can ask very practical questions about heat: How much does the temperature of an object change when you add energy? Do you need to add energy to make an object melt, or vaporize? If so, how much energy does it take?
We can also ask very fundamental questions: Why does it take energy to do these things? Where is the energy going? How is this energy like the mechanical energy that we're used to?
We will begin to answer the second set of questions by answering the first, immediately practical ones.
In the first investigation you will consider something any kitchen sees millions of times: how water changes from a solid to a liquid, heats up to boiling, and changes to steam.
You need energy to do any of these:
â€¢ You will determine how much heat it takes to change the temperature of water 1 degree.
This leads us to consider the relation between energy and heat energy.
â€¢ You will also see how the transfer of heat energy can change a system internally without
changing its temperature. The first process you will examine is the change of a substance from solid (ice) to liquid, called a change of phase. As part of this activity, you will measure the latent heat of fusion of waterâ€”the amount of heat energy transfer required to transform one unit of mass of ice to liquid water.
â€¢ Next you will consider a second example of a process in which heat energy transfer changes a system internally without changing its temperature. This is a phase change from liquid to gas. You will measure the latent heat of vaporization of water-the amount of heat energy transfer required to transform one unit of mass of water from liquid to steam.
As we think about this process: ice to water to steam, we will extend the idea of
heat energy transfer to changes in the motions and internal conditions of atoms that scientists believe make up matter.
Investigation 1
If you transfer equal pulses of heat energy to a perfectly insulated cup of some liquid, what determines how much temperature change DT takes place? What happens to the temperature if you add energy to a liquid/solid mixture?
Prediction 1-1: For a liquid, how does DT depend on
A. The number of pulses of heat energy you transfer (Q)?
B. The mass (m) of liquid in the cup?
C. The kind of liquid you have?
A.) more pulses/energy added = greater increase in energy
B.) greater the mass = longer it takes to change the temperature
C.) the thicker the liquid = longer it takes to change the temperature
Prediction 1-2: For a solid/liquid mixture: What happens as you add energy? Does the temperature increase, the amount of solid change, or both?
temperature increases
Prediction 1-3: On the axes below, sketch a graph of temperature vs. time for a cup of ice water, heated in the following way:
-First 30 s, no heat energy transferred
-Next 10 minutes, heat energy transferred at a constant rate, until the water has partially boiled away.
steady increase over time starting at 20ÂºC
In this investigation you will start with ice water and add heat energy, measuring the temperature, until several minutes after the water starts boiling.
Note: We will add energy with electricity. The immersion heater works the same way that a coffee maker does. The heat pulser allows us to add that energy in pulses. Every time you hit the red pulse button, the heat pulser will turn on for 10 seconds. During that time the heater will add energy at a rate of about 300 W (its power, which depends on the heater), which means it adds energy at a rate of 300 J/s: a 10 second pulse adds 3000 J of energy. Note that when the heater is off (between pulses) it does not add any energy.
temperature vs time graph: Activity 1-1 Heating Ice and Water
steady increase over time
Table 1.1
Power from heater (285 W or 190 W): 285 W
Mass of cup and beaker (kg.): 0.2867 kg
Initial mass of water (kg.): 0.075 kg
Initial mass of ice (kg.): 0.074 kg
Initial mass of ice+water (kg): 0.149 kg
Time when all ice had melted: 260s
# of pulses to melt ice: 13
Total pulse time to melt ice (s) (# pulsesx10s/pulse): 130
# pulses to heat water from freezing to 50Â°C: 23
Total time of these heat pulses (# pulsesx10s/pulse): 230
DT for 4 pulses starting at 20Â°C: 7.7ÂºC
Time at which boiling started: 447s
# pulses after water was boiling: 20
Mass of water minus steam (kg): 0.1161 kg
Mass of steam produced (kg): 0.0329 kg
Question 1-4: Does your graph agree with the prediction you made for an ice and water mixture? If not, describe in words the ways in which the temperature history of the ice and water mixture was different than the one you expected.
yes
Question 1-5:You may find that the graph divides naturally into several regions. If you have to divide it into exactly three, at what temperatures would you put the divides? Why?
0-22ÂºC
22-99ÂºC
99ÂºC-on
In the following activities we will consider each of the three regions separately.
Activity 1-2: heating water, between freezing and boiling
Consider the region well away from boiling and from freezing, from 10-60Â°C.
Question 1-6: Describe the shape of your graph. What does this say about the relationship between the temperature change DT and the quantity of heat energy transferred to the water? (Remember that heat pulses were transferred at a constant rate.) State a mathematical relationship in words and as an equation, based on your measurements
Q = mcDT
Q is directly proportional to T
Question 1-7: Does the temperature change produced by one pulse depend on how warm the water is? Why or why not?
yes, because it just adds more heat/energy to its current state
Prediction 1-8: Suppose you transfer heat energy to a larger mass of water. How will the temperature change?
A. How much did the temperature of your 150 g of water change with 4 heat pulses? Record
the mass, temperature change, # pulses, and temperature change per pulse in the first line of Table 1.2. How many pulses do you think it will take to produce the same temperature increase if you heat twice as much water (300 g)?
B. What will be the temperature increase per pulse if you produce the same temperature increase for twice as much water (300 g)?
temperer will change more slowly
A.) 0.149 kg, 7.7ÂºC, 4 pulses, 0.5ÂºC, twice as much
B.) half as much
table 1.2
total heat energy transferred by heater: Q = mcDT (answer in J)
specific heat: C = Q/m
total heat energy transferred to the water (cal): m(in g) x DT
mechanical equivalent of heat (J/cal): Q/C
Question 1-9: Did the number of pulses required to heat 300 g of water agree with your prediction? Explain.
yes
Question 1-10: Did the rise in temperature per pulse you calculated agree with your prediction? Explain.
yes
Question 1-11: Based on your graphs and data, does the following mathematical relationship (that you used in the last lab) make sense?
Q = cm DT
Q is the heat energy transferred to the water, m is the mass of the water, DT is the change in temperature, and c is a constant characteristic of the liquid. Explain how well your results are described by this relationship.
yes
37050 = c(0.300)(17)
You have seen that the change in temperature is proportional to the amount of heat energy transferred and inversely proportional to the mass of the system. To be more quantitative (e.g., to be able to predict numerical temperature changes), it is necessary to specify what amount of heat energy transfer will produce a one degree change in temperature in unit mass of a material. This quantity is known as the specific heat of the material. It is the value c in the equation in Question 1-6.
specific heat = c = Q / mcDT
The standard units for heat energy (J), mass (kg), and temperature (Â°C), give us the unit for specific heat, J/kg-Â°C.
Question 1-12: Use the data in Table 1-2 to calculate the specific heat in J/kg-Â°C of water. Put this in Table 1-2.
C = Q/m
Question 1-13: How closely did the two values of the specific heat agree with each other? How did the average value agree with the accepted value, cwater = 4190 J/kg-Â°C? What are the possible sources of experimental error that might explain any disagreement?
it was close, we got 4470.5 J/kgâ€¢C, the possible sources of error could be calculating DT from the graph
Before the mid-nineteenth century, heat was regarded as a substance rather than as a form of energy exchange between two substances. Heat was measured in its own special units called calories. By definition,
1 calorie = the quantity of heat that raises the temperature of 1 gram of water by 1 degree Celsius
(Note that the food calorie (also called a kilogram calorie or kilocalorie) is 1000 times larger than this.)
According to the definition of the calorie, the specific heat of water is
Cwater = 1.0 Cal/g-Â°C.
In the mid-nineteenth century, James Joule carried out a series of experiments converting mechanical and electrical energy to heat, demonstrating that heat is a form of energy and not a substance, as you have just seen in converting electrical energy transferred to the heater (supplied through the electrical outlet). This electrical energy is produced by a generator located at a power plant. One of the major achievements of 19th and early 20th century science was to show that chemical potential energy (stored in fossil fuels), nuclear potential energy (stored in nuclear fuels), or gravitational potential energy (stored in water above a dam) could all be turned into mechanical energy of the generator, or into heat.
In the next activity you will find the quantitative mechanical equivalent of heat. Thus you will determine the number of joules that are equivalent to a calorie of heat energy.
Question 1-14: The Energy Equivalent of the Calorie: From your data in table 1.2, calculate the total heat energy transferred to the water in calories using the relationship Q = cm DT, using the specific heat of water 1 Cal/g-Â°C, and enter in the 8th column of the table. Calculate the mechanical equivalent of heat in J/cal (the ratio of the measured heat transferred by the heater in joules to the calculated heat in calories) and enter that in the last column of the table.
...
Question 1-15: How does your measured value of the "mechanical equivalent of heat agree with the accepted value of 4.19 J/cal? By what percentage do the values differ?
it was close for 150g water, but for the second one it was doubled
Question 1-16: Can you think of any reasons why you would expect your value to differ from the accepted value? Was your measured value too small or too large? Do your reasons explain why the value came out this way?
errors in calculating DT in the graph
Activity 1-3: melting ice
Note that the behavior of the temperature as you add energy to the system while the ice is melting is very different from what the temperature does once the ice has stopped melting.
Question 1-17: During the time that the temperature remained constant, what do you think happened to the heat energy you were transferring to the water and ice mixture if it wasn't raising the temperature?
heat energy still transferring to the water, but it was releasing as heat/steam
When heat energy is transferred to a mixture of ice and water in thermal equilibrium at the melting point temperature, all of the heat energy transferred is absorbed by the ice and used to break down the rigid bonds holding the molecules together in the ice crystal structure.
This represents a very significant change in the internal structure, and internal energy of the system. Yet, while this is going on, the temperature of the mixture does not increase.
The heat energy transfer required to melt one kilogram of ice when its temperature is already at its melting point is called the
latent heat of fusion. (This amount of energy is the same as the amount transferred away from one kilogram of water when it freezes.) To calculate the approx- imate latent heat of fusion for the ice to water phase transition using your data in Table 1-1:
Question 1-18. Calculate the total heat energy transferred to the ice/water mixture to melt the ice. Use the power of the heater and the total time of the heat pulses, both in Table 1-1. Show your calculations below. (Remember, 1 watt means that you're adding energy at a rate of 1 joule/second.) Total heat energy transferred to melt the ice: ____ J
Q = mcDT
285 x 130 = 37050 J
Question 1-19. Calculate the latent heat of fusion of the ice. Use the amount of heat energy transferred that went to melting the ice and the mass of the ice to find the amount of heat energy needed to melt one kilogram of ice. Show your calculations. Latent heat of fusion: ____ J/kg
U/m = L
37050/0.149 = 248657.7181 J/kg
Question l-20: Compare the heat energy needed to melt one kilogram of ice with the heat energy needed to raise the temperature of one kilogram of water by one degree Celsius.
Q = mL
Q = 1000 x 80
Q = 80000 cal
Q = mcDT
Q = 1000 x 1 x 1
Q = 1000 cal
Question l-21: Compare your value for the latent heat of fusion to the accepted value, 334 x103 J/kg (or 334 J/g). Discuss the limitations in your experimental method that might account for any differences between these two values.
ours was bigger, because of error in the calculations of DT
Activity 1-4: boiling water
Note that the temperature as you added energy to boiling water behaved very much like the temperature as you added energy to ice water, and very differently from the behavior as you added energy to liquid water away from either freezing or boiling. Indeed, boiling water involves another phase change, as the liquid water changes to a gas (steam).
Question 1-22: During the time that the temperature remained constant, what do you think happened to the heat energy you were transferring if it wasn't raising the temperature?
the energy was still transferred to the water, but it was released as steam
Most substances can exist in three statesâ€”solid, liquid, and gas.
As you have seen, these changes of state or phase changes usually involve a transfer of heat energy. During a phase change, the substance can absorb heat energy without changing its temperature until the phase change is complete. The transferred energy increases the internal energy of the system, as you will explore further in the next investigation.
The amount of heat energy transfer required to transform one kilogram of water at its boiling point into steam is called the
latent heat of vaporization. (This amount of energy is the same as the amount transferred away from one kilogram of steam when it condenses.)
Question 1-23. Calculate the total heat energy transferred to the water after it was boiling. Use
the power of the heater and the total time of the heat pulses from Table 1-1. Show your calculations below. (Remember, 1 W = 1 J/s.)
Total heat energy transferred: ____ J
Q = mcDT
285 x 200 = 56000 J
Question 1-24. Calculate the latent heat of vaporization of the water. Use the amount of heat energy transferred after the water was boiling and the mass of steam produced from Table 1- 1 to find the amount of heat energy needed to convert one kilogram of water to steam. Show your calculations.
Latent heat of vaporization: ____ J/kg
U/m = L
56000 / 150 = 373.3 J/kg
Question 1-25: Compare the heat energy needed to convert one kilogram of water to steam to that needed to convert one kilogram of ice to water. Why do you think that one is much larger than the other?
Q = mL
Q = 1000 x 533
Q = 5.33 x 10^5 cal
Q = mL
Q = 1000 x 80
Q = 8.0 x 10^4 cal
-stronger bond, need more energy to break the bonds
Question 1-26: Compare your value for the latent heat of vaporization to the accepted value, 2.26 x 106 J/kg (or 2260 J/g). Discuss the limitations in the experimental method you used that
might account for any differences between these two values.
errors in calculations of DT
HEAT AND THERMODYNAMICS LAB 2 Homework
VVV
Suppose you transfer a certain amount of heat energy to a known amount of a liquid in a perfectly insulated cup and the temperature changes. Then, you decide to alter the experiment in several different ways. Match the alterations listed below with the best description of the total change in temperature compared to that measured in the original experiment (before the alteration):
1.) Transfer more heat energy
2.) Heat for a longer time but transfer the same total heat energy.
3.) Start with more liquid in the cup.
4.) Increase the starting temperature of the liquid.
5.) Use an equal mass of liquid that has a larger specific heat.
6.) Use the same volume of a denser liquid that has the same specific heat.
1.) The total change in temperature will be larger.
2.) The temperature change will be the same.
3.) The total change in temperature will be smaller
4.) The temperature change will be the same.
5.) The total change in temperature will be smaller
6.)The total change in temperature will be smaller
You have two perfectly insulated cups. One contains water and the other contains an equal volume of another liquid that has half the density of water and two times the specific heat. You heat the water from 10 to 200C and the other liquid from 80 to 900C. Which of the following would describe the amount of heat energy transferred to raise the temperature of the other liquid and the amount transferred to raise the temperature of the water.
Equal amount of heat energy is transferred to the water and to the liquid
A perfectly insulated cup is filled with water initially at 20 0C and standard atmospheric pressure. Heat energy is transferred to the cup by an immersion heater at a steady rate. Match the graph that would best describe the transfer of heat energy as a function of time.
a straight, positive slope from 20 on
Exactly. At standard pressure, and if there is no evaporation or other changes in the water, the amount of heat you need to add to raise the temperature 1 degree should be almost constant up to 100 degrees C.
For the question above, what would be the mathematical relationship between the temperature (T) and the quantity of heat energy (Q) that has been transferred to the water. C is the specific heat capacity of the water and M is the mass of water.
Q= C
M
(T - 200C)
How many calories of heat energy would it take to heat 200ml of water from 20ÂºC to 80ÂºC?
60 x 200 = 12000cal
How long would it take to heat 202ml of water from 20ÂºC to 82ÂºC using a heater with a power rating of 50W (i.e., it puts out 50 Joules of energy every second)?
(For water: use 334kJ/kg for the latent heat of fusion, 2.26X103kJ/kg for the latent heat of vaporization, and 4.19kJ/kgÂºC for the specific heat. Also for water: remember that 1 ml has a mass of approximately 1g)
Q = mcDT
Q = (0.202)(4.19)(62)
Q = 52.47556
50 = 52.47556/t
t = 1.0495112 --> 1050s
Which of the following curves describes what happens to the temperature of a water-ice mixture originally at 0 ÂºC when heat energy is transferred to it at a constant rate, at standard (1atm) pressure (at sea level).
...
How does the internal energy of the ice and water mixture change if the temperature does not rise while the ice is melting?
it increases, because it takes more energy for water molecules to be disordered than to be ordered.
Suppose you start with 200g of ice at 0ÂºC. Calculate the amount of heat energy that must be transferred to the ice to melt it.
(Use 334 kJ/kg for the latent heat of fusion, 2.26X10^3 kJ/kg for the latent heat of vaporization, and 4.19 kJ/kgÂºC for the specific heat of water.)
334 x 200 =
66800 J to melt the ice
--> 66.8 kJ
Suppose you start with 210g of ice at 0ÂºC. Calculate the amount of heat energy that must be transferred to convert the ice to steam at 100ÂºC.
(Use 334kJ/kg for the latent heat of fusion 2.26 X 10^3 kJ/kg for the latent heat of vaporization, and 4.19 kJ/kgÂºC for the specific heat of water.)
Q = mcDT
Q = (0.210)(4.19)(100)
Q = 87.99 kJ
Lab 1: Introduction to Waves and Sound
VVVV
In the last lab activity you investigated how adding or removing heat energy to objects either changed the temperature of the object in question or changed its phase (melting or boiling for example).
In this lab
activity we will begin by looking at the
relationship between pressure, volume, and
temperature of a confined sample of gas.
Next, we will move to waves. Our emphasis will be sound waves, which are simply
pressure variations that travel through the air. Many of the things we experience every day are waves: not just water waves on a pond (see figure 1) or ocean, but sound and light itself. A wave carries energy and momentum from one place to another. In a water or sound wave, this energy is in the movement of individual water or other molecules. However, these molecules may stay almost in place (moving up and down or backwards and forwards), while the wave itself moves forward. A simple wave has a characteristic repetition length, the "wavelength" l, (the distance between high points or low points, for example; see Figure 2), and a characteristic frequency of oscillation f (cycles per second.) For anything traveling, we can relate f and l to its speed v by v= f l. You can see this by imagining a wave with wavelength l moving past you at speed v.
1.) A wave from throwing a pebble in a pond Notice the wavelength between ripples after you leave the center. (source: www.avoxtar.com/images/avox tar_pebble.jpg
2.) You can take a snapshot of a wave in time (at left), to focus on the characteristic wavelength l, or you can measure something in time, at a fixed point (at right), to see the characteristic repeat time or "period" T. The frequency f is defined by f = 1/T.
How many times per second does a wave crest pass you?
This is the "frequency" of the wave. The speed is a characteristic of the material. Sound in air at room temperature and pressure has a speed of about 340 m/s.
If you place a wave within boundaries
(say, fixing the ends of a string),
the wave bounces back and forth between the ends. This sets up a resonance, where the wave oscillates much more strongly with particular wavelengths (and thus frequencies.) This kind of wave is called a "standing wave" (See figure 3.)
3.) A standing wave on a string (from http://www1.union.edu/newmanj/lasers/Light%20as%20a%20Wa ve/light_as_a_wave.htm). Note that the points that pass through "0" on the left of Figure2 are stationary, and that any snapshots would see a wave like in that figure, but with different sizes (Note that at a moment when the string on one side of a noted is "up, that on the other side is "down".)
INVESTIGATION 1: WORK DONE BY A GAS: HEAT ENERGY TRANSFER, INTERNAL ENERGY, AND THE FIRST LAW OF THERMODYNAMICS
One system we will meet often in our study of thermodynamics is a mass of gas confined in a cylinder with a movable piston. The use of such a gas-filled cylinder is not surprising since the development of thermodynamics in the eighteenth and nineteenth centuries was closely tied to the development of the steam engine, which employed hot steam confined in just such a cylinder.
In thermodynamics we are interested not only in heat energy and work, but in how the two interact. For example, if we transfer heat energy to a gas, can we get it to do work? In this investigation, you will begin with some qualitative observations to examine the concept of work done by the gas in a cylinder.
At your lab station you will find a number of syringes that are basically cylinders with movable pistons. By making some simple observations with these syringes, you can begin to appreciate how an expanding gas can do work.
Try compressing the air in the syringe with the end sealed by pushing the piston (the moveable part that you usually press on with your thumb) down against the mouse pad on the table. Then let it go, and see what happens.
Question 1-1: Does it take effort to compress the gas? Do you have to do work on the gas to compress it? (Did you apply a force over a distance?) What happens when you let goâ€”Does the gas spring back?
-yes
-yes
-when you let go it springs back
In thermodynamics, pressure (defined as the component of force that is perpendicular to a given surface for a unit area of that surface) is often a more useful quantity than force alone, since it is independent of the cross-sectional area of the cylinder. It can be represented by the equation
P = F/A
In the next activity you will explore why pressure
is more useful than force in describing the behavior of gases.
You will also extend the definition of work
developed earlier in the course and combine it with this definition of pressure to calculate the work done by a gas on its surroundings as it expands out against the piston with a (possibly changing) pressure P.
From your experience with the syringe, do you expect an expanding gas inside a cylinder to do work? You have probably heard the definition of work in a lecture or seen it in your text. If a force F acts on an object and the object moves a distance Dx, the work is W = F Dx. Using this definition of work and the definition of pressure, you can show that the work done by a gas on its surroundings as it expands out against the piston with a (possibly changing) pressure P can be calculated from
Î”W = PÎ”V
Question 1-2: Show that the above expression for DW follows from DW = F Dx. (Hint: See the preceding diagram.)
F = PA --> A = F/P
Î”V = Î”W/P
Î”V = AÎ”X --> Î”V = (F/P) Î”X --> Î”W/P = (F/P)Î”X --> Î”W = FÎ”X
Suppose you lift a ball of mass m up from the floor through a distance y. The change in the ball's potential energy is DUgrav = mgy. The work done by you against the force of gravity is related to the change in the ball's potential energy so that DUgrav = = - Wgrav. (Keep in mind that the force of gravity on the ball and the displacement y are vectors.) This relationship is true for any system where mechanical energy is conserved. By doing work against gravity, you are storing energy in the form of potential energy. But what about systems where mechanical energy is apparently not conserved?
Is it possible to generalize this relationship for some of these systems? The answer is yes, but we have to give a new meaning to our potential energy. In thermodynamics, U is called the internal energy, and represents any way of storing energy inside a system. The internal energy of a system is the sum of all sorts of energies, including the helter-skelter translational kinetic energies of molecules in a gas, the vibrational energies of gas molecules or atoms in a crystal, and the rotational energies of spinning gas molecules. One way to increase the internal energy of a system is to transfer heat energy to it as you did when you melted ice or produced steam.
For many materials, the pressure P, volume V, and Temperature T are related to each other through what is called an "equation of state." In the next activity, we will exam the relationship between pressure P and volume V at approximately constant temperature T. First make a prediction.
Prediction 1-1: As you compress the air in a syringe by pushing the piston in slowly, what will happen to the pressure? What do you think will be the mathematical relationship between pressure P and volume V?
pressure will increase, P and V are inversely proportional
Activity 1-2: Isothermal Volume Change for a Gas.
The approach to obtaining measurements is to trap a volume of air in the syringe and then compress the air
slowly to both smaller and larger volumes by pushing or pulling the piston. The gas should be compressed slowly so it will always have time to come to equilibrium with the room (and thus be at room temperature) You should take pressure data for about 5 different volumes.
1. Position the piston of the 20-mL syringe at the 10-mL line while the syringe is open to the air, after
positioning the piston connect the end of the syringe to the pressure sensor.
2. Open the experiment file called Pressure vs. Volume (Sound 2A1-2) to display the axes that follow. This
will also set up the software in event mode so that you can continuously measure pressure and decide when
you want to keep a value. Then you can enter the measured volume.
3. Enter the volume of the Pressure sensor, 0.1 cm3, in the second column of Table 1-1.
4. As you pull or squeeze down on the piston slowly, the computer will display the pressure. When the
pressure reading is stable, you can keep that value and then enter the total volume of air from Table 1-1.
5. Repeat this for at least five different volumes of the syringe between 4 and 20 mL.
6. Use the fit routine to find a relationship between P and V.
Table 1-1
Volume of air in syringe (cm3):
20, 15, 10, 5
Volume of sensor (cm3):
0.1
Total volume of air in the system (cm3):
20.1, 15.1, 10.1, 5.1
Pressure (kPa):
100 at 20 cm3
121 at 15 cm3
152 at 10 cm3
235 at 5 cm3
Question 1-4: What is the relationship between P and V? Is it proportional linear, inversely proportional, or something else? Did this agree with your prediction?
inversely proportional
yes it did
Question 1-5: Write down the relationship between the initial pressure and volume (Pi,Vi) and the final pressure and volume (Pf,Vf) for an isothermal (constant-temperature) process.
PiVi = PfVf
The relationship that you have been examining between P and V for a gas with the temperature and amount of gas held constant is known as Boyle's law. If we additionally find the relationship between the Pressure and Temperature with the volume of the gas held constant and the relationship between the volume and temperature with the pressure held constant, we can deduce the ideal gas law.
Question 1-6: Is the relationships you found in the this activity consistent with the ideal gas law PV=constant . T? Explain.
yes, because the pressure and volume were inversely proportional while we kept a constant temperature
-Note that changing the temperature also affects the gas. We don't have time to explore this quantitatively now, but you can still feel it happen:
Activity 1-3: The Heated Syringe
You should have concluded from the last activity that the transfer of heat energy to a system can either cause it to do work on its surroundings or increase its internal energy. What is the relationship between heat energy transfer, changes in a system's internal energy, and the work done by the system? We picture U as the "true" energy of the system. In the theory of thermodynamics, U is called a "state" variable, a quantity that tells us some things we need to know to calculate useful things about a system, such as its temperature
Suppose that the piston of the syringe is clamped in place while the syringe is immersed in hot water. The piston can't move, so no work can be done. However, since the water is initially at a higher temperature than the gas in the syringe, we expect that heat energy is transferred from the water to the gas. This causes the temperature of the gas to increase and the temperature of the water to decrease. The heat energy transfer can be calculated using the equation Q= cm DT, where c is the specific heat and DT is the temperature change of the water.
Assuming that the system is insulated so that no heat energy can be transferred to the surroundings, the transferred heat energy Q must equal the increase in internal energy of the gas. This is based on a belief that energy is conserved in the interaction between the hot water and the gas.
Suppose instead that we release the piston and allow the gas to do work as it expands against the piston. We could calculate the amount of work W the expanding gas did by evaluating DW = P DV for the whole process. Where did the energy to do this work come from? The only possible source is the internal energy of the gas, which must have decreased by an amount W. The total change in the internal energy of the air trapped in the syringe must be given by
Î”U = Q - W
This relationship between transferred heat energy, work done on the surroundings, and the change in internal energy is believed to hold for any system not just a syringe filled with gas. It is known as the first law of thermodynamics.
The first law of thermodynamics has been developed by physicists based on a set of very powerful inferences about forms of energy and their transformations. We ask you to try to accept it on faith. The concepts of work, heat energy transfer, and internal energy are subtle and complex. For example, work is not simply the motion ofthe center of mass of a rigid object or the movement of a person in the context of the first law. Instead, we have to learn to draw system boundaries and total the mechanical work done by the system inside a boundary on its surroundings outside the boundary.
The first law of thermodynamics is a very general statement of conservation of energy for thermal systems. It is not easy to verify in an introductory physics laboratory, and it is not derivable from Newton's laws. Instead, it is an independent assertion about the nature of the physical world.
There are many ways to achieve the same internal energy change DU. To achieve a small change in the internal energy of gas in a syringe, you could transfer a large amount of heat energy to it and then allow the gas to do work on its surroundings. Alternatively, you could transfer a small amount of heat energy to the gas and not let it do any work at all. The change in internal energy, DU could be the same in both processes. DU depends only on Q - W and not on Q or W alone.
Question 1-9: Can you think of any situations where W is negligible and DU =Q? (Hint: Is it necessary to do work on a cup of hot coffee to cool it? Can you think of similar situations?)
melting of ice into water, because volume is held constant
Question 1-10: How could you arrange a situation where Q is negligible and in which DU =-W? Such situations have a special name in thermodynamics. They are called adiabatic processes. (Adiabatic means with no heat energy transferred into or out of the system.)
decreasing volume in a syringe = pressure increase with no temperature change
Investigation 2: Speed of Sound in the Air
You recall that we can also measure the speed of a wave - or anything else moving uniformly - by:
d= vt
With care (you remember The Three Stooges trying to carry long boards? Let's not look like that here!) place a 3.0 m long piece of sewer pipe across your lab table and your neighbor's. (Lab groups at stations 5 and 6, place yours across the room on a couple of chairs.)
Slip a plastic pipe cap over the end away from you to reflect sound waves back.
table 2-1
Distance sound traveled, d (m): 6m (3 down and 3 back
Time for sound to travel, t (s):
0.017,
0.014,
0.013
Speed of sound, v (m/s) average:
414.3 m/s
Question 2-1. Check with your lab instructor to find the temperature in the room, T, in degrees Celsius. Recalling that the speed of sound in air is given by
v = 331 m/s + (0.6 m/soC)T
What is the expected value of the speed of sound?
V = 331 + (0.6 x 25)
V = 346 m/s
Investigation 3: Waves on a string
In this investigation, you will try to oscillate the string at different frequencies. You will find that only a few frequencies lead to waves that are large enough so that you see the wave clearly. You will see parts of the string are nearly stationary ("nodes": see figure 3) while other parts move strongly around equilibrium (the biggest motion is at the "anti-nodes.")
For a string, the velocity of the wave depends almost entirely on properties of the string (material and tension), not on the frequency. This means that every frequency corresponds to a particular wavelength. Only some wavelengths can fit "properly" on a string. You will find that the "proper" wavelengths depend only on how you hold the string. The corresponding frequency is called a "resonance" frequency, because energy can build up at these frequencies until there is enough motion for you to see.
Question 3-1: Draw a still image of a wave below, marking the wavelength. Also mark possible nodes (where the wave doesn't move).
wavelength is top of a crest to top of the next crest (up and up or down and down)
nodes are were the waves pinch in the middle
Question 3-2: How is the distance between nodes related to the wavelength?
shorter the wavelength = shorter distance between nodes
Prediction 3-1: If you fix a string on two sides, draw what the longest "standing wave" could look like. What is the wavelength for this wave, compared to the length of the string?
wavelength is shorter than length
Measure the length of the string between the two attach-points.
L = 79 cm
Table 3.1 Two fixed ends.
Frequency (Hz):
-Longest wavelength = 13
-26
-39
-52
Distance between nodes:
-79
-39.5
-19.75
-9.875
wavelength:
-39.5
-19.75
-9.875
-4.93
Question 3-3: Did your longest wave fit your prediction 3-1?
yes
Question 3-4: Can you find a relation between this "resonant" wavelength and the other ones that you found?
What about the frequencies?
the greater the frequency = the shorter the wavelength
5. Now take a wave with at least 2 nodes, and lightly pinch the string, first at a node and then at an antinode. Question 3-5? What difference did you see? Did pinching the string at the node stop the wave? At the antinode?
-at the node: stopped the top half of the string, while the bottom half had the same wave
-at the antinode: it created a wave with 2 nodes at the bottom half of the string
Prediction 3-2. Now consider a string with one free end. A free end is always at a peak or valley, since that is the only way forces can balance at the end of the string without something holding the other side. Draw the longest resonant wavelength that will fit properly on this string.
one hump up/arc up
The velocity on the string depends on
the tension in the string. To keep an end of the string free to move while keeping the string under tension requires that the "free end" slide on a post holding it in tension, but without any tension. Note that if there is any friction between the post and the string, the string isn't "free. It's not easy to keep the friction low enough with simple equipment, but it is very easy using the computer to simulate the action. In the next activity we are going to use such a simulation.
Activity 3-2: Compare: both ends fixed and one free end using computer simulation. One Free end.
tables 3-2
Question 3-6. Did your longest wave with one free end and one fixed end fit your prediction 3-2?
yes
Question 3-7. Can you find a relation between this "resonant" wavelength and the other ones that you found? What about the frequencies?
the lower the frequency = the longer the wavelength
The resonance wavelengths for any wave depends only on
the conditions at the ends of the waves (boundary conditions: here, one or two fixed ends.) We can find these by just drawing the waves, fitting them in the space, and realizing that any fixed end has to be a node, while a totally free end has to be an anti-node.
Once we figure out the resonance wavelengths, we can figure out the resonance frequencies, as long as we know the relation between wavelength and frequency (by knowing the speed of the wave, for example.)
These ideas hold true not just for waves on a string, where we can visualize the string moving up and down, but also for sound and light waves, where it is harder to visualize what is moving. We'll see how they work for sound in Investigation 4.
INVESTIGATION 4: Sound waves in a tube
The computer simulation you just explored was in fact for sound in a tube. The wave speed was taken to be the speed of sound at 20Â°C. The graph you were looking at was for the motion of particles within that tube: this motion is back and forth along the tube (the direction of the wave), instead of up and down (perpendicular to the wave direction) as on a string. These two types of waves are called "longitudinal" and "transverse" respectively. You can get an idea of the "longitudinal" motion of sound from the picture above the graphs. Note that the picture is very schematic, and doesn't show what happens to air that was originally outside the tube.
Note that at an open end, the air is free to move back and forth as it will (but with the same pressure as the outside air), but at a closed end, the air must be stationary.
Prediction 4-1: What standing wave pattern (displacement of the air) do you expect for the various situations for each of the tubes in each of the configurations given in Table 4-1 (sketch them in the spaces provided in the table)?
Prediction 4-2: You haven't measured the lengths of the tubes use in this activity yet, so use the letter L to represent each particular length. What do you expect the wavelength to be for the various resonant patterns you sketched (right this next to Î»= under each sketch in Table 4-1)?
Prediction 4-3: Consider the short tube, lowest frequency, both one end open and two ends closed. Use your predicted wavelengths for those situations from Prediction 4-2 and the speed of sound as being 343m/s (approximation) to calculate the frequency you would expect for each (you will still have L in these expressions, but later you'll actually measure it and have a number to put in...).
standing wave pattern one end open:
Î» = 4L
standing wave pattern both ends closed:
Î» = 2L
-frequency decreased when decreasing the length, but when closing both ends frequency increased
-when frequency decreased, wavelength increased (and vise versa)
Question 4-1: Did your measured frequencies agree with the lowest frequencies (f0) you predicted in Table 4- 1? If not, what is different? Do you need to reconsider the form of your largest wavelength?
yes, until the short tube with both ends covered. Frequency was higher, so wavelength was shorter
Question 4-2: Did the other measured frequencies agree with your predictions? If not, what is different?
yes, until the short tube with both ends covered. Frequency was higher, because the length of the wave was shortened (cap = wave couldn't lengthen/escape)
Homework
VVV
Gas is in an isolated cylinder and sealed with a piston.
1.) If you push down quickly on the piston and compress the gas, is work done on or by the gas?
2.) The internal energy of the gas increases, decreases, or stays constant.
3.) The temperature of the gas increases, decreases, or stays constant.
1.) the work is done on the gas
2.) increases
3.) increases
Now you transfer heat energy to the gas in the cylinder, but hold the piston so that it can not move
1.) Is work done on or by the gas?
2.) The internal energy of the gas increases, decreases, or stays constant.
3.) The temperature of the gas increases, decreases, or stays constant
1.) No work done
2.) increases
3.) increases
After holding the piston in the previous question for a short while, you let it go.
1.) What happens to the piston?
2.) Is work done on or by the gas?
3.) The internal energy of the gas increases, decreases, or stays constant.
4.) The temperature of the gas increases, decreases, or stays constant.
1.) the piston moves up
2.) The work is done by the gas
3.) decreases
4.) decreases
In a movie, a space ship explodes in outer space, and you hear a big boom a little bit after the very bright light flash. What is wrong with the movie?
Sound can't travel in outer space: it needs air or some other material medium to travel through
You have a light, flexible string. One end (the left in the picture below) is held and oscillated an amount too small to be seen in the pictures, at a carefully chosen frequency (as with the machine you used in lab.) Pick three of the snapshots that could represent the string with the other (right) end fixed.
A, B, F
A: one small arc up
B: squiggle line waves
F: one arc up and one down ~
You have a light, flexible string. One end (the left in the picture above) is held and oscillated an amount too small to be seen in the pictures, at a carefully frequency (as with the machine you used in lab.) Pick three of the snapshots that could represent the string with the other (right) end free to move up and down.
CDG
C: one small arc down
D: one arc up, one arc down, half arc up
G: almost flat ish arc up then down (small)
In the picture above: Which line corresponds to the lowest frequency, supposing that the right hand end is fixed?
A: one small arc up
In the picture above: Which line corresponds to the lowest frequency, supposing that the right hand end is free to move up and down?
C: one small arc down
You have a tube, with a white-noise (random) vibrator at one end. The other end is open. Pick three of the snapshots that could represent the motion of the molecules back and forth along the tube.
CDG
C: one small arc down
D: one arc up, one arc down, half arc up
G: almost flat ish arc up then down (small)
You have a tube, with a white-noise (random) vibrator at one end. The other end is closed. Pick three of the snapshots in the figure above that could represent the motion of the molecules back and forth along the tube.
A, B, F
A: one small arc up
B: squiggle line waves
F: one arc up and one down ~
Your tube is half a meter long, with one end closed and the other open. The speed of sound is 344m/s. What is the lowest resonant frequency (largest wavelength)?
Please give the units in Hz.
172Hz
because
V = n/z
V = 344/(4L)
V = 344/(4 x 0.5)
V = 172
Your tube is a half meter long, with both ends closed. The speed of sound under current conditions (maybe different from the last experiment) is 344m/s. What is the lowest resonant frequency (largest wavelength)?
Please give units in Hz.
344Hz
because
V = n/z
V = 344/(2L)
V = 344/(2 x 0.5)
V = 244
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