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.

Options

A. `(1+(m_(2))/(m_(1)))v_(0)`
B. `2(1-(m_(2))/(m_(1)))v_(0)`
C. `2(1+(2m_(2))/(m_(1)))v_(0)`
D. `2(1+(m_(2))/(m_(1)))v_(0)`

Answer

D

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)`
`v_(1)=(m_(2)v_(0))/(m_(1))` (i)
Now rate at which distance between mirror and bullet is increasing,
`(du)/(dt)=v_(1)+v_(0)` (ii)
`(dv)/(dt)=((v^(2))/(u^(2)))(du)/(dv)`
Here, `(v^(2))/(u^(2))=m^(2)=1`
(as at the itme of firing, the bullet is at pole).
`(dv)/(dt)=(du)/(dt)=v_(1)+v_(0)` (iii)
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)` (iv)
Therefore, speed of separation of the bullet and the 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)`