Question

A 21.9-g bullet is shot horizontally at 1183.6 m/s. It collides with an 3.5-kg block of wood. The bullet embeds in the block, and the block slides along a horizontal surface. The coefficient of kinetic friction between the block and surface is 0.2.

How far will the block slide before friction stops it?

Answer 1

first we will calculate the combined speed V of the bullet and block immediately after collision

(m1*u) = (m1+m2)*V

0.0219*1183.6 = (0.0219+3.5)*V

V = 7.35 m/s

finally the system is at rest

according toi work energy theorem

Work done by the net force = change in kinetic energy

W = Kf-Ki

Kf = 0 (since finnaly at rest)

W = -fk*S = - mu_k*m*g*S = -Ki =

mu_k*m*g*S = 0.5*m*V^2

mu_k*g*S = 0.5*v^2

distance moved is S = 0.5*v^2/(mu_k*g) = (0.5*7.35*7.35)/(0.2*9.81) = 13.76 m