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.

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:

[tex]m_uv_u - m_ov_o = (m_u + m_o)v[/tex](1)

where [tex]m_u = 0.013 kg, v_u = 843 m/s[/tex] are the mass and speed of the bullet prior to the impact. [tex]m_o = 1.58 kg[/tex] 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 [tex]v_t = 0 m/s[/tex]), let the vertical distance travel be h (m). We have the following equation of motion

[tex]v_t^2 - v^2 = 2gh[/tex]

[tex]0^2 - v^2 = 2gh[/tex]

[tex]v = \sqrt{2gh}[/tex](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:

[tex]h = gt^2/2[/tex](3)

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

[tex]v = \sqrt{g^2t^2} = gt[/tex] (4)

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

[tex]0.013*843 - 1.58gt = (0.013 + 1.58)gt[/tex]

[tex]10.959 = 3.173gt[/tex]

[tex]t = \frac{10.959}{3.173*9.81} = 0.352 s[/tex]