bullet and block problem?

Question

A 5.50 g bullet is fired horizontally into a 1.10 kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.200. The bullet remains embedded in the block, which is observed to slide 0.230 m along the surface before stopping. What was the initial speed of the bullet?

Answer 1

the answer is ....





4



according to Rodney

Answer 2

Friction does work on the block to slow it to a stop:

work = (force)(distance)
work = (friction force)(0.230m)

The friction force is the normal force (in this case, the weight of the block+bullet), times the coefficient of friction:

friction force = (weight)(μ)
friction force = (M+m)(g)(0.200)

(where "M" is the block's mass and "m" is the bullet's mass).

The amount of work is also the amount of change in kinetic energy:

ΔKE = work
ΔKE = (friction force)(distance)

The block+bullet's initial KE (just after impact) is (M+m)(v_block)²/2; and its final KE is zero. So its change in KE is:

ΔKE = (final KE) – (initial KE) = –(M+m)(v_block)²/2

You can now put all the above equations together to solve for v_block.

Next, use the conservation of momentum principle:

Total initial momentum = total final momentum

m(v_bullet) = (M+m)(v_block)

Since you have already solved for v_block (and you are given "M" and "m"), you can now solve for v_bullet.