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. The speed of separation of the bullet and image just after the gun is fired is given as $a(1+\frac{m}{M})v$. Find $a$.

Answer 1

From law of conservation of momentum between the gun and the bullet:

$mv=-M{v}^{{}^{\prime }}$

or $v^{{}^{\prime }}=\frac{-mv}{M}$

Now, the speed of separation of the bullet and the mirror (gun) may be expressed as:

$\frac{dx}{dt}=v_{bullet}-v_{gun}=(1+\frac{m}{M})v$

For the reflection from a concave mirror, we may write:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

or, $\frac{1}{y}-\frac{1}{x}=\frac{1}{-f}$

or, $y=\frac{-xf}{f-x}$

or, $\frac{dy}{dt}=[\frac{-f}{f-x}+\frac{fx}{{(f-x)}^2}]\frac{dx}{dt}=\frac{dx}{dt}\left[\frac{{-f}^2}{{(f-x)}^2}\right]$

So, speed of separation of the bullet and its image:

$\frac{dy}{dt}-\frac{dx}{dt}=\frac{dx}{dt}\left[\frac{{-f}^2}{{(f-x)}^2}-1\right]$ $=\frac{dx}{dt}\left[\frac{{-2f}^2+2fx-x^2}{{(f-x)}^2}\right]$

$\underset{x\to 0}{Lt}|\frac{dy}{dt}-\frac{dx}{dt}|=\frac{2dx}{dt}=2(1+\frac{m}{M})v$

So, $a=2$