A 3.0 kg block is sitting at rest at the edge of a cliff. A 400 gram bullet is fired at the block with an initial velocity of 500m/s. The bullet goes through the block and has a speed of 400m/s after it leaves the block. The block lands 50m from the base of the cliff. The pavement at the base of the cliff is flat (horizontal) but it does have friction. The block slides an additional 30m before it comes to a stop. What is the coefficient of friction of the pavement? How high was the cliff?

Answer 1

Momentum is conserved.
.4*500=.4*400+3*V solve for v, the initial horizontal velocity of the block.
V*time=50
so you now can find the time to fall from the cliff.
but h=1/2 g t^2, so solve for height of cliff.
now, the sliding
vf^2=vi^2+2ad
0=V^2+2*forcefriciton/mass * distance
you know V, distance. Force friction= massblock*mu*g
solve for mu

Answer 2

.4 * 500 = 200 = initial bullet momentum
.4 * 400 = 160 = final bullet momentum
so
200 - 160 = 40 = block momentum
so
40 = 3 u
u = 40/3 = horizontal speed of block in the air.
goes 50 m horizontal
so time in air = t = 50 /(40/3) = 150/40 = 15/4 seconds
h = (1/2) g t^2 = 4.9(225/16) = 68.9 meters high

initial slide speed = 40/3 = 13.33 m/s
final = 0
F = m a = -m * 9.81 *mu
average speed during stop = 13.33/2 = 6.67 m/s
so time to stop = 30 /6.67 = 4.5 seconds
acceleration = change in v /time = -13.33/4.5 = -2.96 m/s^2
so
2.96 = mu g = 9.81 mu
mu = 0.302