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. $\theta ={\tan }^{-1}\left(\frac{1}{2}\right)$ to the x-axis
• D. $\theta ={\tan }^{-1}(-\frac{1}{2})$ 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.
$m_1\times 0+m_{2}v=m_1{v}_x+m_2(\frac{v}{2})$
$v_x=\frac{m_{2}v}{m_1}\dots (i)$
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 $v^y=-\frac{m_{2}v}{2m_1}\dots (ii)$
If mass $A$ moves at angle $\theta$ with the $X-$ axis, then
$\tan \theta =\frac{v_y}{v_x}=-\frac{m_{2}v}{2m_1}\times \frac{m_1}{m_{2}v}=-\frac{1}{2}$
$\therefore \theta ={\tan }^{-1}\left(\frac{1}{2}\right)$ to the $x-$ axis