Question

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.

Answer 1

0.16 s

Step-by-step 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