AP PHYSICS acceleration

Question

1)An indestructible bullet 2.35 cm long is fired straight through a board that is 12 cm thick. The bullet strikes the board with a speed of 455 m/s and emerges with a speed of 299 m/s . What is the average acceleration of the bul let through the board ?

2) How long is the bullet in contact with the board ? Answer in units of .

3)what minimum thickness of boards would it take to stop the bullet assuming the deceleration through all boards is the same.

Answer 1

The bullet is in contact with the board at the tip of impact to the tail of the exit (surface traction). This a contact length of 14.35 cm and a reduction in speed of 156 m/s.

Using the distance equation we can estimate the time of impact.

d = v*t

0.1435 m = 156 m/s * t --> t = 0.0009198 seconds or 0.92 ms (Answer B)

Using the time we can calculate the average acceleration.

Aavg = dV/dt = -156 [m/s]/0.0009198 [s] = -16.958 km/s (Answer A)

Using the X-acceleration equation we can find the depth of boards needed to stop the bullet.

x(t) = x0 +v0*t +1/2 Aavg *t^2

First derivative test to find time until stopping distance

0 = 0 +v0 +Aavg*t --> t(stop) = 0.005366 seconds or 5.366 ms

Applying this to the distance equation

d(stop) = 455[m/s] *0.005366[s] =2.4388[m]

knowing we need the length of the bullet in addition to our boards we can find the number of boards required

(n*board_length +bullet_length) = d(stop)

n*12cm +2.35cm = 243.88cm --> n = 20.1275 (Answer C)

Assuming we cannot have a fractional amount of stacks so we need a minimum of 21 boards to stop the bullet.