As shown below, a 5 g bullet moving at 400 m/s is fired directly into and passes
ID: 1292226 • Letter: A
Question
As shown below, a 5 g bullet moving at 400 m/s is fired directly into and passes through a 1 kg block. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with a force constant of 900 N/m. The block moves 5 cm before being brought to rest by the spring.
a. With what speed does the bullet emerge from the block?
b. During the collision, how much kinetic energy is converted into internal energy in the bullet-block system?
Please explain in detail to get all points.
Explanation / Answer
a)
to find speed (V) of block after collision
0.5*K*x^2 = 0.5*M*V^2
V^2 = k*x^2/M = (900*0.05*0.05)/1
V = 1.5 m/s
from momentum conservation
initial momentum of bullet + block = initial momentum of bullet + block
m*vo = M*V + m*v
(0.005*400) = (1*1.5)+(0.005*v)
v = 100 m/s < -------answer
b) KEi = 0.5*m*vo^2 = 0.5*0.005*400*400 = 400J
KEf = 0.5*M*V^2 + 0.5*m*v^2
KEf = (0.5*1*1.5*1.5)+(0.5*0.005*100*100) = 26.125 J
loss = KEf - KEi = -373.875 J < ------answer
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