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 speed of separation of the bullet and the image just after the gun was fired.

Answer 1

Answer:

the anwer is

$1 \div 2mv ^{2}$

Answer 2

The speed of separation of the bullet and the image just after the gun was fired is given by $\begin{equation}\frac{2(1+m M)}{V}\end$.

Explanation:

Given,

The concave mirror focal length is f, and M is gun mass.

Horizontal Bullet velocity is V.

Let the gear recoil velocity be Vg

We will write with the conservation of linear momentum,

MVg = mV

Vg=mMV

Considering gun and bullet position at time t = t,

For mirror, distance from objects , u = - (Vt + Vgt)

Focal length, f = − f

Image distance, v = ?

We know that

$\begin{equation}\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\end$

$\begin{equation}\frac{1}{v}=\frac{1}{-f}-\frac{1}{u}\end$

$\begin{equation}\frac{1}{v}=-\frac{1}{f}+\frac{1}{(V t+V g t)}\end$

$\begin{equation}\frac{1}{v}=\frac{-(V t+V g t)+f}{(V t+V g t) f}\end$

$\begin{equation}v=\frac{(V t+V g t) f}{-(V t+V g t)+f}\end$    

At time t, the distinction between the image of the bullet and the bullet is given by:

$\begin{equation}v-u=\frac{(V t+V g t) f}{-(V t+V g t)+f}-(V t+V g t)\end$

Differentiating the above equation as to ' t ' we obtain,

$\begin{equation}\frac{d(v-u)}{d t}=\frac{2(1+m M)}{V}\end$

              = $\begin{equation}\frac{2(1+m M)}{V}\end$

Therefore, the pace at which the bullet and the picture are separated just after the weapon was fired is $\begin{equation}\frac{2(1+m M)}{V}\end$.