Try Magic Notes and save time.Try it free
Try Magic Notes and save timeCrush your year with the magic of personalized studying.Try it free
Question

A particle of mass m1m_{1} with an initial kinetic energy T1{T}_{1} makes an elastic collision with a particle of mass m2m_{2} initially at rest. m1m_{1} is deflected from its original direction with a kinetic energy T1T_{1}^{\prime} through an angle ϕ1{\phi}_{1}. Letting α=m2/m1 and γ=cosϕ1\alpha=m_{2} / m_{1} \text { and } \gamma=\cos \phi_{1}, show that the fractional kinetic energy lost by m1,ΔT1/T1=(T1T1)/T1m_{1}, \Delta T_{1} / T_{1}=\left(T_{1}-T_{1}^{\prime}\right) / T_{1}, is given by ΔT1T1=21+α2γ(1+α)2(γ+α2+γ21)\frac{\Delta T_{1}}{T_{1}}=\frac{2}{1+\alpha}-\frac{2 \gamma}{(1+\alpha)^{2}}(\gamma+\sqrt{\alpha^{2}+\gamma^{2}-1})

Solution

Verified
Step 1
1 of 5

Basically, what we are trying to calculate in this problem is fractional kinetic energy lost by m1m_1

T1T1T1=ΔT1T1=1T1T1r=1r,\begin{align*} \frac{T_1 - T_1'}{T_1} = \frac{\Delta T_1}{T_1} &= 1 - \underbrace{\frac{T_1' }{T_1}}_{\equiv r} = 1 -r , \tag{1} \end{align*}

where T1=12m1v12T_1 = \dfrac{1}{2} m_1 v_1^2 is initial kinetic energy of mass m1m_1, T1=12m1v12T_1' = \dfrac{1}{2} m_1 v_1^{2 \prime} final kinetic energy of mass m1m_1 and r=T1T1=v12v12r= \dfrac{ T_1' }{T_1} = \dfrac{v_1^{2 \prime}}{v_1^2} ration of kinetic energies of scattered particle to incident particle energy. Because this is similar problem that is in textbook (figure 7.6.1.), we can make use of figures in textbook; if we look at figure 7.6.2., we can recognize following relation

v1v1=(v1vcm)(v1vcm),\begin{align*} \overline{v}_1' \overline{v}_1' = (v_1' - v_{ \text{cm}}) \cdot (v_1' - v_{ \text{cm}}), \tag{2} \end{align*}

where v1\overline{v}_1' is scattering speed of particle of mass m1m_1 in the center of mass system, v1v_1' is scattering speed of particle of mass m1m_1 in lab system and vcmv_{ \text{cm}} is speed of center of mass. Then we have from equation (2)

v2=v12+vcm22v1vcmcosϕ1v12=v12vcm2+2v1vcmγ,\begin{align*} \overline{v}^{2 \prime} &= v_1^{2 \prime} + v_{\text{cm}}^2 - 2 v_1' v_{\text{cm}} \cos \phi_1 \\ v_1^{2 \prime} &= \overline{v}_1^{2 \prime} - v_{\text{cm}}^2 + 2 v_1' v_{\text{cm}} \gamma, \tag{3} \end{align*}

where γ=cosϕ1\gamma = \cos \phi_1 (ϕ1\phi_1 is scattering angle of mass m1m_1). With the use of expression (3) and definition of r we have

r=v12v12r=v12v12vcm2v12+2v1vcmγv12.\begin{align*} r &= \frac{ v_1^{2 \prime} }{v_1^2} \\ r &= \frac{ \overline{v}_1^{2 \prime} }{ v_1^2 } - \frac{ v_{\text{cm}}^2 }{ v_1^2 } + \frac{ 2 v_1' v_{\text{cm}} \gamma }{v_1^2 }. \tag{4} \end{align*}

Create an account to view solutions

Create an account to view solutions

Recommended textbook solutions

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics 4th Edition by Randall D. Knight

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th EditionISBN: 9780133942651 (4 more)Randall D. Knight
3,508 solutions
Mathematical Methods in the Physical Sciences 3rd Edition by Mary L. Boas

Mathematical Methods in the Physical Sciences

3rd EditionISBN: 9780471198260Mary L. Boas
3,355 solutions
Analytical Mechanics 7th Edition by George L. Cassiday, Grant R. Fowles

Analytical Mechanics

7th EditionISBN: 9780534494926George L. Cassiday, Grant R. Fowles
302 solutions
Fundamentals of Physics 10th Edition by David Halliday, Jearl Walker, Robert Resnick

Fundamentals of Physics

10th EditionISBN: 9781118230718David Halliday, Jearl Walker, Robert Resnick
8,971 solutions

More related questions

1/4

1/7