Question

A shell of mass 200gm is ejected from a gun of mass 4kg by an explosion that generates 1.05kj of energy. The initial velocity of a shell??

Answer 1

Answer:

If v and V represent the initial velocity of the shell of mass m and the recoil velocity of the gun of mass M respectively, we have (from the law of conservation of momentum)

mv + MV = 0 from which V = – (m/M)v = – (0.2/4)v = – v/20 ms–1

The negative sign shows that the recoil velocity of the gun is opposite to the velocity of the shell.

The energy of the explosion is used to impart kinetic energy to the shell and the gun so that we have

½ (mv2 + MV2) = 1.05×103

Or, ½ [0.2 v2 + (4v2/400 )] = 1.05×103

This gives v2 = 104 so that v = 100 ms–1.

Answer 2

Answer:100m/s

Explanation:

Hey

For explanation see the image below