Question

A bomb of mass 3m is kept inside a closed box of mass 3m length 4L at its centre. It explodes in two parts of mass m and 2m. The two parts move in opposite direction and stick to the opposite sides of the wall.Box is kept at a smooth horizontal surface.Find distance traveled by box during this.

Answer 1

Apply the conservation of momentum principle.

let the velocity of smaller part  m  be = v.    and that of the bigger part = u.
so   3m *0 = m * v + 2m * u   =>   u = - v/2
distance traveled by each part = 2 m

So the smaller part hits the wall first at  t = 2/v seconds after blast.  The bigger part hits the wall at t = 4/v  s.

During this time the smaller part is stuck to the box. 
So    m v = (m+3m) v'
 v' = velocity of box together with part 'm' after collision =  v/4
So distance traveled in time duration 2/v   is   v/4 * 2/v
         = 1/2 meter

Answer 2

Answer:

Explanation:

Hope this will help you....