Question

A ball of mass m moving with velocity $v_0$ collides a wall as shown in figure. After impact it rebounds with a velocity $\frac{3}{4}{v}_0$. The impulse acting on ball during impact is-

• A. $-\frac{m}{2}{v}_0\hat{j}$
• B. $-\frac{3}{4}m{v}_0\hat{i}$
• C. $\frac{-5}{4}m{v}_0\hat{i}$
• D. $None$ $of$ $these$

Answer 1

Answer: (C)

$\frac{-5}{4}m{v}_0\hat{i}$

Resolving $p_i$ and $p_f$ into $x$ and $y$ components, we get,

$\overrightarrow{p_i}=m{v}_0(\cos {37}^0\hat{i}+\sin {37}^0\hat{j})$

$\overrightarrow{p_f}=\frac{3}{4}m{v}_0(-\cos {53}^0\hat{i}+\sin {53}^0\hat{j})$

Impulse$=\Delta \overrightarrow{p}=m(\overrightarrow{p_f}-\overrightarrow{p_i})$;

$\therefore \Delta \overrightarrow{p}=m{v}_0[(-\frac{3}{4}\cos {53}^0-\cos {37}^0)\hat{i}+(\frac{3}{4}\sin {53}^0-\sin {37}^0)\hat{j}]=$
$\Rightarrow \Delta p=m{v}_0\sqrt{{(-\frac{3}{4}\cos {53}^0-\cos {37}^0)}^2+{(\frac{3}{4}\sin {53}^0-\sin {37}^0)}^2}$
$\Rightarrow \Delta p=m{v}_0\sqrt{\frac{9}{16}+1}=\frac{5}{4}m{v}_0$
$\Rightarrow \Delta \overrightarrow{p}=-\frac{5}{4}m{v}_0\hat{i}$