Question

A bullet of mass $m_2$ is fired from a gun of mass $m_1$ with horizontal velocity $v$. A plane mirror is fixed to the gun facing towards the bullet. The velocity of the image of the bullet with respect to a frame fixed to ground is:

• A. $\frac{2(m_1+m_2)}{m_1}v$
• B. $\left(\frac{m_1+2m_2}{m_1}\right)v$
• C. $\left(\frac{2m_1+m_2}{m_1}\right)v$
• D. $\left(\frac{2m_1+m_2}{m_2}\right)v$

Answer 1

The correct option is B $\left(\frac{m_1+2m_2}{m_1}\right)v$

Since mirror is mounted at the gun, the velocity of mirror $\left(v_m\right)$ and gun will be same.
Applying linear momentum conservation for the bullet and gun system,

$m_1$ is mass of gun and mirror.
Initial momentum of the system $=0$
$\Rightarrow m_1{v}_m+m_2(-v)=(m_1+m_2)\times 0$
$\Rightarrow m_1{v}_m=m_{2}v$
$\therefore v_m=\left(\frac{m_2}{m_1}\right)v$

Applying,
${\left(v_{o,m}\right)}_{\perp }=-{\left(v_{i,m}\right)}_{\perp }$

Taking rightwards as positive direction.
$\Rightarrow (-v-v_m)=-(v_i-v_m)$
or, $-v-v_m=-v_i+v_m$
or, $v_i=v+2v_m$
$\therefore v_i=v+2\left(\frac{m_2}{m_1}\right)v=\left(\frac{m_1+2m_2}{m_1}\right)v$