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 formed by the plane mirror with respect to the bullet is (Neglect the mass of mirror)

• A. $(1+\frac{m_2}{m_1})v$
• B. $\left(\frac{m_1+m_2}{m_1}\right)v$
• C. $\frac{2(m_1+m_2)v}{m_1}$
• D. $\frac{2(2m_1+m_2)v}{m_1}$

Answer 1

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

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 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,
$(v_{o,m}{)}_{\perp }=-(v_{i,m}{)}_{\perp }$
Taking rightwards +ve direction.
$\Rightarrow (-v-v_m)=-(v_i-v_m)$
or, $-v-v_m=-v_i+v_m$
or, $v_i=v+2v_m$
$\therefore v_i=v+2\left(\frac{m_2}{m_1}\right)v=\left(\frac{m_1+2m_2}{m_1}\right)v$
Thus, the velocity of image w.r.t bullet (object) is,
$v_{i,o}=v_i-v_o$
$\Rightarrow v_{i,o}=v_i-(-v_o)=v_i+v_o$
or, $v_{i,o}=\left(\frac{m_1+2m_2}{m_1}\right)+v$
$\therefore v_{i,o}=\frac{2(m_1+m_2)v}{m_1}$