Question

An indestructible bullet 2.00cm long is fired straight through a board that is 10.0cm thick. The bullet strikes the board with a speed of 470 m/s and emerges with a speed of 270 m/s. (To simplify, assume that the bullet accelerates only while the front tip is in contact with the wood.) a). What is the average acceleration of the bullet through the board? ________m/s^2

b). What is the total time that the bullet is in contact with the board? (Enter the total time for the bullet to completely emerge from the board.) _________s
c.) What thickness of board (calculated 0.1 cm) would it take to stop the bullet, assuming that the acceleration through all boards is the same? ________cm

Answer 1

a)
`a=-7.4* 10^(-5)\ m/s^2`

b)
`t=0.27* 10^(-3)\ s`

c)s=14.92 cm

Step-by-step explanation:

Given that

u= 470 m/s

v = 270 m/s

s= 10 cm

a)

We know that

`v^2=u^2+2as`

`270^2=470^2+2* a* 0.1`

`a=-7.4* 10^(-5)\ m/s^2`

b)

v= u + a t

`270=470-7.4* 10^(-5)* t`

`t=0.27* 10^(-3)\ s`

c)

To stop the bullet it means that the final velocity will be zero.

`v^2=u^2+2as`

`0^2=470^2-2* 7.4* 10^(-5) * s`

s=14.92 cm