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?

Expert's answer

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

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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