A 0.0130-kg bullet is fired straight up at a falling wooden block that has a mass of 1.58 kg. The bullet has a speed of 843 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 occurs. 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.

0.352 s

Explanation:

Let g = 9.81 m/s2. Then the speed of the block after it's fall down a time t (seconds) before the collision is:

(0)

Using the law of momentum conservation, the total momentum of the system after the collision must be same as before the collision. Let the upward be the positive direction:

(1)

where are the mass and speed of the bullet prior to the impact. is the mass of the block. v is the speed of the system after the impact. We will focus on v for the next part:

As the system raise and come to a momentarily halt on top of the building (speed at top ), let the vertical distance travel be h (m). We have the following equation of motion

(2)

As h is the same vertical distance that the block has fallen before the collision, we can solve for h in term of t:

(3)

If we plug eq. (3) into (2):

(4)

And plug eq (4) and (0) into eq (1), with all the numbers:

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.