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(C) The initial velocity vector is $\vec{v}_i = v_0 \cos 37^{\circ} \hat{i} + v_0 \sin 37^{\circ} \hat{j} = \frac{4}{5} v_0 \hat{i} + \frac{3}{5} v_0

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$-\frac{5}{4}mv_0\hat{i}$

Vedclass · Answer

(C) The initial velocity vector is $\vec{v}_i = v_0 \cos 37^{\circ} \hat{i} + v_0 \sin 37^{\circ} \hat{j} = \frac{4}{5} v_0 \hat{i} + \frac{3}{5} v_0 \hat{j}$.The final velocity vector is $\vec{v}_f = -(\frac{3}{4} v_0) \cos 53^{\circ} \hat{i} + (\frac{3}{4} v_0) \sin 53^{\circ} \hat{j}$.Using $\cos 53^{\circ} = \frac{3}{5}$ and $\sin 53^{\circ} = \frac{4}{5}$,we get $\vec{v}_f = -\frac{3}{4} v_0 (\frac{3}{5}) \hat{i} + \frac{3}{4} v_0 (\frac{4}{5}) \hat{j} = -\frac{9}{20} v_0 \hat{i} + \frac{3}{5} v_0 \hat{j}$.The impulse $\vec{J}$ is the change in momentum: $\vec{J} = m(\vec{v}_f - \vec{v}_i)$.$\vec{J} = m [(-\frac{9}{20} v_0 \hat{i} + \frac{3}{5} v_0 \hat{j}) - (\frac{4}{5} v_0 \hat{i} + \frac{3}{5} v_0 \hat{j})]$.$\vec{J} = m [(-\frac{9}{20} - \frac{16}{20}) v_0 \hat{i} + (\frac{3}{5} - \frac{3}{5}) v_0 \hat{j}] = m [-\frac{25}{20} v_0 \hat{i}] = -\frac{5}{4} mv_0 \hat{i}$.

$A$ ball of mass $m$ moving with velocity $v_0$ collides with a wall as shown in the figure. After impact,it rebounds with a velocity $\frac{3}{4}v_0$. The impulse acting on the ball during the impact is

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