Abullet with a mass =12.5 g and speed =86.4 m/s is fired into a wooden block with =113 g which is initially at rest on a horizontal surface. the bullet is embedded into the block. the block-bullet combination slides across the surface for a distance before stopping due to friction between the block and surface. the coefficients of friction are =0.753 and =0.659. (a) what is the speed (m/s) of the block-bullet combination immediately after the collision? (b) what is the distance d (m)?

a) 8.61 m/s, b) 5.73 m

Explanation:

a) During the collision, momentum is conserved.

mv = (m + M) V

(12.5 g) (86.4 m/s) = (12.5 g + 113 g) V

V = 8.61 m/s

b) After the collision, energy is conserved.

Kinetic energy = Work done by friction

1/2 (m + M) V² = F d

1/2 (m + M) V² = N μk d

1/2 (m + M) V² = (m + M) g μk d

1/2 V² = g μk d

d = V² / (2g μk)

d = (8.61 m/s)² / (2 × 9.8 m/s² × 0.659)

d = 5.73 m

Notice we used the kinetic coefficient of friction.  That's the friction when an object is moving.  The static coefficient of friction is the friction on a stationary object.  Since the bullet/block combination is sliding across the surface, we use the kinetic coefficient.