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

A, B

Explanation

Let `v_1`be the speed of gun (or mirror ) just after the firing of bullet . From conservation of linear momentum,
` m_2v_0=m_1v_1`
or `v_1=(m_2v_0)/(m_1)` ...(i)
Now , `(du)/(dt)` is the rate at which distance vetween mirror and bullet is increasing `=v_1+v_0` ...(ii)
We have already read in extra points that:
` :. (dv)/(dt)=((v^2)/(u^2))(du)/(dt)`
Here, `(v^2)/(u^2)=m^2=1` (as at the time of firing,bullet is at pole.)
`:. (dv)/(dt)=(du)/(dt)=v_1+v_0`

Here, `(dv)/(dt)` 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=(dv)/(dt)`
`=v_1+v_0`
or `v_2=2v_1+v_0`
Therefore, speed of separation of bullet and image will be
` v_r=v_2+v_0`
`=2v_1+v_0+v_0`
`v_(r)=2(v_(1)+v_(0))`
Substituting value of `v_1` from Eq. (i) we have
` v_(r)=2(1+(m_(2))/(m_(1)))v_(0)`.