Question

Two spheres $A$ and $B$ of masses $m_1$ and $m_2$, respectively, collide. $A$ is at rest initially and $B$ is moving with velocity $v$ along the 𝑥-axis. After collision, $B$ has a velocity $v/2$ in a direction perpendicular to the original direction. The sphere 𝐴 moves after collision in the direction

• A. $\theta ={\tan }^{-1}(-1/2)\text{to the x-axis}$.
• B. $\theta ={\tan }^{-1}\left(1/2\right)\text{to the x-axis}$.
• C. opposite to that of $B$.
  
• D. same as that of $B$.
  

Answer 1

The correct option is A $\theta ={\tan }^{-1}(-1/2)\text{to the x-axis}$.

Step 1: Draw rough diagram for the given situation.


Step 2: Find the direction in which sphere A moves after collision.

Applying conservation of momentum,
along $x$−direction,
$m_{2}v=m_{1}V\cos \theta$
$\Rightarrow V\cos \theta =\frac{m_2}{m_1}v\cdots \left(i\right)$

Along $y$−direction,
$0=m_2\frac{v}{2}-m_{1}V\sin \theta$
$\Rightarrow V\sin \theta =\frac{m_2}{m_1}\frac{v}{2}\cdots \left(ii\right)$

Divide equation $\left(ii\right)$ by equation$\left(i\right)$,
$\tan \theta =\frac{V\sin \theta }{V\cos \theta }=\frac{1}{2}$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{2}\right)$
Since $A$ is moving below the $x$-axis, hence the angle will be ${\tan }^{-1}\left(\frac{1}{2}\right)$ to the axis.

Final answer: (d).