question about momentum and physics?

Question

A bullet of mass 150 grams is fired into a block of mass 4kg with a velocity of 200m/s. The bullet penetrates the wood and stops. What is the speed of the system in m/s just after the bullet stops?

Answer 1

use the conservation of momentum. p(initial)=p(final) and solve for the final velocity of the system.

Answer 2

The simplistic, and incorrect, answer is found from the conservation of momentum: mV + Mv = (m + M)v'; where bullet m = .150 kg, block M = 4 kg, bullet V = 200 mps, and block v = 0 before impact. v' = the velocity after impact, when all the kinetic energy of the bullet is spent. Thus, v' = (mV + Mv)/(m + M) the answer you are looking for...I think.

The reason I say "I think" is because v' would be the wrong answer in real life. It would be wrong because it disregards the immense friction heat energy lost as the bullet comes to a screeching halt in the block. In fact the real v < v' because of the lost energy from friction. But as you gave no friction information, it was ignored in the solution above even though in real life friction would play an important role in the "speed of the system .... just after the bullet stops."

Answer 3

inelastic collision:

pA=m1*v1A+m2*v2A=m1*v1A+0=m1*v1A

pA=impulse (momentum) before collision
m1=150*10^-3 kg
v1A=200 m/s
v2A=0 --->m2*v2A=0
m2= 4 kg

pP=m1*v1P+m2*v2P=m1*vP+m2*vP

pP=impulse(momentum) after collision
v1P=v2P=vP (The bullet is embedded in the block)

The impulse is conserved: pA=pP

m1*v1A=m1*vP+m2*vP
vP(m1+m2)=m1*v1A
vP=m1*v1A/(m1+m2)

vP=[150*(10^-3)kg * 200 m/s]/(150*10^-3 + 4)kg

vP=7.23 m/s
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