how far from the table will the block land?

Question

A block of wood with a mass of 2.32 kg sits on the edge of a table that is 1.43 m tall. A bullet with a mass of 53.4 g traveling with a speed of 390 m/s is shot into the block and is lodged in it. Assuming that no energy is lost to friction, how far from the table will the block land?

Answer 1

We have to divide this problem into two parts:
(i) an inelastic collision of the bullet and the block
(ii) a horizontal hit of the block from the table

(i) The bullet is shot into the block and is lodged in it, moving after this together with the block. This is an example of the inelastic collision. In all kind of collisions, a (linear) momentum is conserved:
p = p'
A momentum before the collision consists only from the momentum of the bullet since the block is at rest:
p = m(bullet) x v(bullet)
A momentum after the collision is the product of the net mass of the bullet and the block M and their velocity v':
p' = (m(bullet) + m(block)) x v' = Mv'
A velocity of the system bullet-block after the collision is therefore:
v' = m(bullet) x v(bullet) / M = 0.0534 kg x 390 m/s / (0.0534 kg + 2.32 kg) = 8.77 m/s

(ii) The horizontal hit is a combination of the free fall in the vertical and the motion at constant velocity in the horizontal direction.
The horizontal displacement of the block D can be written in terms of its velocity v' and the time it takes to fall the height of the table: D = v' x t
From the equation for the free fall (uniformly accelerated motion) we have y = H = 1/2 x g x t2. We solve for t:
t = sqrt(2H / g) = sqrt(2 x 1.43 m / 9.8 ms-2) = 0.54 s
We substitute this into the equation for D:
D = 8.77 m/s x 0.54 s = 4.7 m