Question

A gun of mass $M$ fires a bullet of mass $m$ with a horizontal speed $v$. The gun is fitted with a concave mirror of focal length $f$, facing towards the receding bullet. Find the separation speed of the bullet and its image just after the gun is fired.

• A. $2(1+\frac{m}{M})v$
• B. $(1+\frac{m}{M})v$
• C. $(1+\frac{2m}{3M})v$
• D. $\left(\frac{2m}{3M}\right)v$

Answer 1

The correct option is A $2(1+\frac{m}{M})v$


From linear momentum conservation :
$mv=M{v}^{\prime }$ where $v^{\prime }\to$ recoil speed of gun
$\Rightarrow v^{\prime }=\frac{mv}{M}$

At any time $t$, position of bullet w.r.t mirror $x=v^{\prime }t+vt$
$=\frac{m}{M}vt+vt=vt\underset{\text{= k}}{\underset{}{(1+\frac{m}{M})}}=vt\times k$

Position of image [w.r.t mirror] at time $t$ is given by
$\frac{1}{x^{\prime }}+\frac{1}{x}=\frac{1}{f}\Rightarrow \frac{1}{x^{\prime }}=\frac{1}{(-f)}-\frac{1}{(-vkt)}[x=vkt]$
$\Rightarrow x^{\prime }=\frac{kvtf}{f-kvt}$
where $x^{\prime }=$ image postion w.r.t mirror

Velocity of image
$v_I=\frac{d{x}^{\prime }}{dt}=\frac{d}{dt}\left(\frac{kvtf}{f-kvt}\right)=\frac{(f-kvt)\left(kvf\right)-\left(kvtf\right)(-kv)}{(f-kvt{)}^2}$
$\Rightarrow v_I=\frac{kv{f}^2}{(f-kvt{)}^2}$

Thus, velocity of separation between image and object $=v+v^{\prime }+v_I$
[since mirror itself moving with $v^{\prime }$]
$=v+\frac{mv}{M}+\frac{kv{f}^2}{(f-kvt{)}^2}$

Just after the gun is fired, i.e at $t=0$,
Velocity of separation between bullet and its image
$=v+\frac{mv}{M}+kv$
$=kv+kv=2kv$
[substituting $k=1+\frac{m}{M}$]
$=2(1+\frac{m}{M})v$