Question

A bullet of mass 4.3 g is fired horizontally into a 7.4 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.53. The bullet stops in the block, which slides straight ahead for 1.7 m (without rotation). (a) What is the speed of the block immediately after the bullet stops relative to it? (b) At what speed is the bullet fired?

Answer 1

(A) after collision,

a = - uk g = - 0.53 x9.8 = - 5.19 m/s^2

Applying vf^2 - v0^2 = 2 a d

0^2 - v^2 = 2(-5.19)(1.7)

v = 4.20 m/s

(B) Applying momentum conservation,

(4.3 v0) + (7400 x 0) = (7400 + 4.3) (4.20)

v0 = 7232 m/s