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 $v/2$ along x-axis. After collision $B$ has a velocity in a direction perpendicular to the original direction. The mass $A$ moves after collision in the direction

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

Answer 1

Answer: (D)

Here, mass of $A=m_{1}$, mass of $B=m_{2}$
Initial velocity of $A, U_{1}=0$
Initial velocity of $B, u_{2}=v$, along $x-$ axis. Itbr.
After collision, let final velocity of $A$ along $x-$ axis be $v_{x}$ and final velocity $A$ along $y-$ axis be $v_{y}$.
Final velocity of $B=\frac{v}{2}$ along $Y-$ axis.
Applying principle of conservation of linear momentum along $x-$ axis.
$m_{1} \times 0+m_{2} v= m_1v_x+m_2(\frac{v}{2})$
$v_{x}=\frac{m_{2} v}{m_{1}} \ldots(i)$
Applying principle of conservation of linear momentum along $Y-a \xi s$,
$m_{1} \times 0+m_{2} \times 0=m_{1} v_{y}+m_{2}\left(\frac{v}{2}\right)$
or $v^{y}=-\frac{m_{2} v}{2 m_{1}} \ldots(i i)$
If mass $A$ moves at angle $\theta$ with the $X-$ axis, then
$\tan \theta=\frac{v_{y}}{v_{x}}=-\frac{m_{2} v}{2 m_{1}} \times \frac{m_{1}}{m_{2} v}=-\frac{1}{2}$
$\therefore \theta=\tan ^{-1}\left(\frac{1}{2}\right)$ to the $x-$ axis