If a bullet with a mass of 4.2 g and a speed of 650 m/s is fired at a block of wood with a mass of 9.8×10−2 kg . The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 21 m/s .
A) What is the speed of the bullet when it exits the block?
B) Verify Kf<Ki calculating the initial and final kinetic energies of the system.
a] By momentum conservation,
initial momentum of bullet = final momentum of block + bullet
mu = MV + mv
4.2e-3*650 = 9.8e-2*21 + 4.2e-3*v
v = [ 4.2e-3*650 -9.8e-2*21]/4.2e-3
= 160 m/s answer
b] Kf = 0.5MV^2 + 0.5mv^2 = 0.5*9.8e-2*21^2 + 0.5*4.2e-3*160^2 = 75.369 J
Ki = 0.5mu^2 = 0.5*4.2e-3*650^2 = 887 J
Kf<Ki verified
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