A 0.0140-kg bullet is fired straight up at a falling wooden block that has a mass of 2.42 kg. The bullet has a speed of 555 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.

0.16 s

Explanation:

• Falling from rest (V_block = 0 m/s) the block attains a final velocity V_block before colliding with the bullet. This velocity is given by Equation 2.4 as

V_block (final velocity of block just before hitting) = V_0, block + at

where a is the acceleration due to gravity (a = - 9.8 m/s2) and t is the time of fall. The upward direction is assumed to be positive. Therefore, the final velocity of the falling block is

V_block = at

• During the collision with the bullet, the total linear momentum of the bullet/block system is conserved, so we have that

(M_bullet+M_block) V_f = M_bullet*V_bullet + M_block*V_block

Total linear momentum after collision = Total linear momentum before collision

Here V_f is the final velocity of the bullet/block system after the collision, and V_bullet and V_block are the initial velocities of the bullet and block just before the collision. We note that the bullet/block system reverses direction, rises, and comes to a momentary halt at the top of the building. This means that V_f, the final velocity of the bullet/block system after the collision must have the same magnitude as V_block, the velocity of the falling block just before the bullet hits it. Since the two velocities have opposite directions, it follows that of V_f = -V_block, Substituting this relation and Equation (1) into Equation (2) gives

(M_bullet + M_block) (-at) = M_bullet*V_bullet + M_block (at)

t = - M_bullet*V_bullet/a (M_bullet + 2M_block)

= - (0.0140-kg) * 555 m/s/-9.8 (0.0140-kg+2 (2.42 kg)

=0.16 s