A bullet of mass 04 kg is headed at horizontal speed 130 ms

A bullet of mass 0.4 kg is headed at horizontal speed 130 m/s toward a block of mass 8(0.4 kg) = 3.2 kg resting on a horizontal frictionless surface. The bullet strikes the block and comes to rest relative to the block somewhere deeply embedded inside the block. What average force did the block exert on the bullet if it took 0.06 s to stop the bullet?

Solution

let final speed of block + bullet system be V

then by momentum conservation

0.4*130 = (3.2+0.4) V

V = 14.4444444444m/s

impulse given to bullet by block = change in momentum

= 0.4 (130-14.4444444444)

=impulse = F T

F = (0.4 (130-14.4444444444))/(0.06) =770.370370371 N