Question

A 0.026 kg bullet is fired straight up at a falling wooden block that has a mass of 5.0 kg. The bullet has a speed of 750 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occured. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.

Answer 1

Answer:

0.198 s

Explanation:

Consider the motion of the block  before collision

$v_{o}$ = initial velocity of block as it is dropped = 0 m/s

$a$ = acceleration = - g

$t$ = time of travel

$v_{f}$ = final velocity of block before collision

Using the kinematics equation

$v_{f} = v_{o} + at \\v_{f} = 0 + (-g)t\\v_{f} = - gt$

$m$ = mass of the bullet = 0.026 kg

$v_{b}$ = velocity of block just before collision = 750 m/s

$M$ = mass of the block = 5 kg

$V$ = final velocity of bullet block after collision = gt

Using conservation of momentum

$m v_{b} + Mv_{f} = (m + M) V\\(0.026) (750) + (5) (- gt) = (0.026 + 5) (gt)\\19.5 - 49 t = 49.2548 t\\t = 0.198 s$