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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 θ=53\theta = 53 ^ { \circ } to the positive x-axis. Find the fraction of kinetic energy lost in the collision.

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(b)\textbf{(b)} Here given that the mass of puck-1 is m1P=0.2kgm_{1\text{P}} = 0.2\,\mathrm{kg}.

Mass of puck-2 is m2P=0.3kgm_{2\text{P}} = 0.3\,\mathrm{kg}.

Puck-2 initially is at rest, so the initial speed of puck-2 is v2Pi=0ms1\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 v1Pi=2ms1v_{1\text{P}_{\text{i}}} = 2\,\mathrm{m\,s^{-1}}.

After collision the puck-1 moves at an angle θ=53\theta = 53^{\circ} from the +ve x-axis as shown in the figure below. So the final speed of the puck-1 is v1Pf=1ms1v_{1\text{P}_{\text{f}}} = 1\,\mathrm{m\,s^{-1}}.

From part (a), the final speed of puck-2 is v2Pf=1.1ms1\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

Eki=12m1Pv1Pi2+12m2Pv2Pi2\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 v2Pi=0ms1\vec{v}_{2\text{P}_{\text{i}}} = 0\,\mathrm{m\,s^{-1}}, so

Eki=12m1Pv1Pi2\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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