# 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?

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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