Answer to Question #17380 in Mechanics | Relativity for uzair zaidi

Question #17380
An indestructible bullet 2.00 cm long is fired straight through a board that is 10.0 cm thick. The bullet strikes the board with a speed of 420 m/s and emerges with a speed of 280 m/s. (a) What is the average acceleration of the bullet as it passes through the board? (b) What is the total time that the bullet is in contact with the board? (c) What thickness of board (calculated to 0.1 cm) would it take to stop the bullet, assuming the bullet’s acceleration through all parts of the board is the same?
1
Expert's answer
2012-10-30T10:25:21-0400
Total length in collision is 14sm
S=at^2/2+v0t
v1=v0-at
a=(v0-v1)/t
S=-(v0-v1)t/2+v0t=(v0+v1)t/2
0.14=350t
a=140/t
t=0.0004
a=3500 m/s
So, for full stop:
at=v0
t=v0/a
S=at^2/2+v0t
S=v0^2/2a+v0^2/a
S=3v0^2/2a
S=75.6 sm

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