Question

A bullet of mass 6.00 g is fired horizontally into a wooden block of mass 1.29 kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.210. The bullet remains embedded in the block, which is observed to slide a distance 0.290 m along the surface before stopping.

What was the initial speed of the bullet?

Answer 1

v = speed of bullet

V = speed of bullet -block combination

M = mass of block

m = mass of bullet

using conservation of momentum

mv = (m + M) V

(0.006) v = (0.006 + 1.29) V

V = (0.006) v /(0.006 + 1.29)                        eq-1

frictional force acting on the block-bullet combination is given as

F_f = u_k (m + M)g

d = stopping distance = 0.29 m

using conservation of energy

kinetic energy of bullet-block combination after collision = work done by frictional force

(0.5) (m + M) V² = F_f d

(0.5) (m + M) V² = u_k (m + M)gd

V² = 2u_k gd

using eq-1

((0.006) v /(0.006 + 1.29))² = 2 x 0.210 x 9.8 x 0.29

v = 236 m/s