Question

A point mass $m$ moving with velocity $v_0$ hits a ball of same mass $m$ at rest elastically. After the collision, point mass moves with velocity $v_0/2$ at an angle $\theta$ from initial direction of motion. Then value of $\theta$ is

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${90}^0$

Answer 1

The correct option is C ${60}^0$


Let $v=$Velocity of ball
Applying conservation of linear momentum in the direction of motion,
$m{v}_0=\frac{\mathrm{m.}{v}_0}{2}\cos \theta +\mathrm{m.}v\cos \alpha ...(1)....Applying\; conservation\; of\; linear\; momentum\; in\; the\; direction\; perpendicular\; to\; the\; direction\; of\; motion,$
$m\frac{V_o}{2}\sin \left(\theta \right)=mV\sin \left(\alpha \right)$ ....(2)
Applying conservation of kinetic energy,
$\frac{1}{2}m{V}_o^2=\frac{1}{2}m{\left(\frac{V_o}{2}\right)}^2+\frac{1}{2}m{V}^2\frac{3}{4}{V}_o^2=V^2\Rightarrow V=\frac{\sqrt{3}}{2}V.....\left(3\right)$
solving (1), (2) and (3), we get
$\theta ={60}^0$