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

Answer:

a)$a=-7.4\times 10^{-5}\ m/s^2$

b)$t=0.27\times 10^{-3}\ s$

c)s=14.92 cm

Explanation:

 Given that

u= 470 m/s

v = 270 m/s

s= 10 cm

a)

We know that

$v^2=u^2+2as$

$270^2=470^2+2\times a\times 0.1$

$a=-7.4\times 10^{-5}\ m/s^2$

b)

v= u + a t

$270=470-7.4\times 10^{-5}\times t$

$t=0.27\times 10^{-3}\ s$

c)

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

$v^2=u^2+2as$

$0^2=470^2-2\times 7.4\times 10^{-5} \times s$

s=14.92 cm