Question

A 716 g wooden block is initially at rest on a rough horizontal surface when a 16.2 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. _____m/s

Answer 1

Let speed of bullet-block combination is v_i and bullet intial speed is v_b

First find the deceleration of the block after the collision , consider that the only force affecting the motion is the frictional force , so :

F = µ_k (mg) = 0.75(mg) = ma

a = 0.75g = 0.75 (9.8) = 7.35 m/sec²

now using netwon’s equation and that deceleration to find speed v of bullet-block just after the collision :

v_f² = v_i² + 2ax

0 = v_i² – 2(7.35) (6.5)

hence v_i = 9.77 m/sec

now we should apply the conservation of linear momentum to obtain v_b at the moment of collision :

Momentum before collision = Momentum after collision

m_bv_b + M_wV_w = (m_b + M_w) v_i      (Mw is block mass and m_b is bullet mass)

0.0162 v_b + 0 = (0.0162 + 0.716) (9.77)

hence v_b = 441.57 m/s