## Related questions with answers

A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 2.0 m/s. After the collision, the 0.20-kg puck has a speed of 1.0 m/s at an angle of $\theta = 53 ^ { \circ }$ to the positive x-axis. Find the fraction of kinetic energy lost in the collision.

Solution

Verified$\textbf{(b)}$ Here given that the mass of puck-1 is $m_{1\text{P}} = 0.2\,\mathrm{kg}$.

Mass of puck-2 is $m_{2\text{P}} = 0.3\,\mathrm{kg}$.

Puck-2 initially is at rest, so the initial speed of puck-2 is $\vec{v}_{2\text{P}_{\text{i}}} = 0\,\mathrm{m\,s^{-1}}$.

Puck-1 initially moving along x-axis, so the initial speed of puck-1 is $v_{1\text{P}_{\text{i}}} = 2\,\mathrm{m\,s^{-1}}$.

After collision the puck-1 moves at an angle $\theta = 53^{\circ}$ from the +ve x-axis as shown in the figure below. So the final speed of the puck-1 is $v_{1\text{P}_{\text{f}}} = 1\,\mathrm{m\,s^{-1}}$.

From part (a), the final speed of puck-2 is $\vec{v}_{2\text{P}_{\text{f}}} = 1.1\,\mathrm{m\,s^{-1}}$.

Now as we know the initial kinetic energy of the system containing puck-1 and puck-2 is

$\begin{align*}\\ E_{\text{k}_{\text{i}}} = \frac{1}{2}\,m_{1\text{P}}\,v_{1\text{P}_{\text{i}}}^2 + \frac{1}{2}\,m_{2\text{P}}\,v_{2\text{P}_{\text{i}}}^2\tag{1} \end{align*}$

Where $\vec{v}_{2\text{P}_{\text{i}}} = 0\,\mathrm{m\,s^{-1}}$, so

$\begin{align*}\\ E_{\text{k}_{\text{i}}} = \frac{1}{2}\,m_{1\text{P}}\,v_{1\text{P}_{\text{i}}}^2\tag{2} \end{align*}$

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