Question

a shell at rest on a smooth horizontal surface explodes into fragments of masses m1 and m2 . if just after explosion m1 move with speed u ,then work done by internal forces during explosion is

Answer 1

Let the fragment mass m2 move with speed v just after the explosion.
Apply the law of conservation of momentum.
m2 v + m1 u = (m1+m2)* 0
v = - m1 u/ m2

Work done = K E of m1 and m2.
= 1/2 m1 u^2 + 1/2 m2 v^2
= 1/2 m1 u^2 *( 1 + m1/m2)

Answer 2

Let mass m2 move with velocity v.
applying conservation of linear momentum,
v= -m1u/m2
Now let's apply conservation of kinetic energy,
hence
1/2*mass of shell*velocity=1/2*m1*v1^2 + 1/2*m2*v^2.
Velocity of resting shell is zero and we've already found v, hence
Answer= 1/2m1u^2(1+m1/m2)