Question

A 762 g wooden block is initially at rest on a rough horizontal surface when a 15.8 g bullet is fired horizontally into (but does not go through) it. After the impact, the block-bullet combination slides 6.50 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.750, determine the speed of the bullet immediately before impact.

Answer 1

The speed of bullet before impact is 480.95 m/s.

Step-by-step explanation:

mass of block, M = 762 g = 0.762 kg

mass of bullet, m = 0.0158 kg

distance, s = 6.5 m

coefficient of friction = 0.75

Let the velocity of bullet before impact is u and after impact is v.

Use third equation of motion

`v'^2 = v^2 - 2 \mu g s\\\\0 = v^2 - 2 * 0.75* 9.8* 6.5\\\\v =9.77 m/s`

Use conservation of momentum

m x u + M x 0 = (M + m) v

0.0158 x u = (0.0158 + 0.762) x 9.77

u = 480.95 m/s