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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.

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#### 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`

⇒ `V_g = m/M `V

Considering the position of gun and bullet at time

*t*=

*t*,

For the mirror, object distance,

*u*= − (

*Vt*+

*V*

_{g}

*t*)

Focal length,

*f*= −

*f*

Image distance,

*v*= ?

Using Mirror formula, we have:

`1/v + 1/u = 1/f`

⇒ `1/v + 1/u = 1/f `

⇒ `1/v = 1/-f - 1/u`

⇒ `1/v = -1/f + 1/ ( Vt + V_ g t)`

⇒ `1/v = (-(Vt + V_g t) + f)/( (Vt + V_g t) f`

⇒ `v = (-(Vt - V _g t ) tf) / ( f - (V + Vg) t)`

The separation between image of the bullet and bullet at time

*t*is given by:

`v = u -(( V + V_g )tf )/ ( f- ( V + V g) t + ( V + Vg ) t`

`= (V + Vg ) t [ f/ (f-(V + Vg )t) + 1]`

`= 2( 1 + m/M )Vt`

Differentiating the above equation with respect to '*t*' we get,

`d ( v -u ) = 2 ( 1 + m/M ) V`

Therefore, the speed of separation of the bullet and image just after the gun was fired is

`2 ( 1 + m/M ) V`.

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