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.

Open in App

Solution

Given,
The focal length of the concave mirror is f and M is the mass of the gun. Horizontal speed of the bullet is V.

Let the recoil speed of the gun be V_g
Using the conservation of linear momentum we can write,
MV_g = mV
$\Rightarrow V_{g} = \frac{m}{M} V$
Considering the position of gun and bullet at time t = t,
For the mirror, object distance, u = − (Vt + V_gt)
Focal length, f = − f
Image distance, v = ?
Using Mirror formula, we have:

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{- f} - \frac{1}{u} \Rightarrow \frac{1}{v} = - \frac{1}{f} + \frac{1}{(V t + V_{g} t)} \Rightarrow \frac{1}{v} = \frac{- (V t + V_{g} t) + f}{(V t + V_{g} t) f} \Rightarrow v = \frac{- (V + V_{g}) t f}{f - (V + V_{g}) t}$

The separation between image of the bullet and bullet at time t is given by:

$v - u = \frac{- (V + V_{g}) t f}{f - (V + V g) t} + (V + V_{g}) t = (V + V_{g}) t [\frac{f}{f - (V + V_{g}) t} + 1] = 2 (1 + \frac{m}{M}) V t$

Differentiating the above equation with respect to 't' we get,
$\frac{d (v - u)}{d t} = 2 (1 + \frac{m}{M}) V$
Therefore, the speed of separation of the bullet and image just after the gun was fired is $2 (1 + \frac{m}{M}) V$ .

Suggest Corrections

Similar questions

^{'} m_{1}^{'}

^{'} m_{2}^{'}

^{'} v_{0}^{'}

M

m

v

f

a (1 + \frac{m}{M}) v

a

m_{2}

m_{1}

v

m_{2}

m_{1}

v

m_{2}

m_{1}

v

Join BYJU'S Learning Program