A gun of mass `m_1` fires a bullet of mass `m_2` with a horizontal speed
` v_0`. The gun is fitted with a concave mirror of focal length f facing towards a
receding bullet.Find the speed of separations of the bullet and the image just after the gun was fired.

Explanation

Let `v_(1)` be the speed of gun (or mirror) just after the firing of bullet.
From conservation of linear momentum.
`m_(2)v_(0)=m_(1)v_(1)`
or `v_(1)=(m_(2)v_(0))/m_(1)` ...(i)
Now, `(d u)/(d t)=` rate at which distance between mirror and bullet is increasing
`=v_(1)+v_(0)` ...(ii)
`:. (dv)/(d t)=(v^(2)/u^(2)) (d u)/(d t)`
Here `v^(2)/u^(2)=m^(2)=1`
(as at the time of firing bullet is at pole)
`:' (d v)/(d t)=(d u)/(d t)=v_(1)+v_(0)` ...(iii)
Here `(d v)/(d t)` is the rate at which distance between image (of bullet) and mirror is increasing. So if `v_(2)` is the absolute velocity of image (towards right) then,
`v_(2)-v_(1) =(d v)/(d t) = v_(1)+v_(0)`
or `v_(2)=2v_(1)+v_(0)` ...(iv)
Therefore, speed of separation of bullet and image will be,
`v_(r)=v_(2)+v_(0)`
`=2v_(1)+v_(0)+v_(0)`
`=2v_(1)+v_(0)+v_(0)`
or `v_(r)=2 (v_(1)+v_(0))`
Substituting value of `v_(1)` from equation (i) we have.
`v_(r)=2 (1+m_(2)/m_(1)) v_(0)`