In a police ballistics test, a 10.0-g bullet moving at 300 m/s is fired into a 1.00-kg block at rest. The bullet goes through the block almost instantaneously and emerges with 50.0% of its original speed. What is the speed of the block just after the bullet emerges?
A) 1.50 m/s
B) 273 m/s
C) 3.00 m/s
D) 2.97 m/s

Question

In a police ballistics test, a 10.0-g bullet moving at  300 m/s  is fired into a 1.00-kg block at rest. The bullet goes through the block almost instantaneously and emerges with  50.0%  of its original speed. What is the speed of the block just after the bullet emerges?
A)  1.50 m/s
B)  273 m/s
C)  3.00 m/s
D)  2.97 m/s

Quizplus · Accepted Answer

The conservation of momentum principle applies here. Before the collision, the total momentum is the momentum of the bullet since the block is at rest. After the collision, the total momentum is the sum of the momentums of the bullet and the block. The momentum of an object is given by its mass times its velocity (p = mv). Let's denote the mass of the bullet as $m_b = 0.01 \, \text{kg}$, the initial velocity of the bullet as $v_{bi} = 300 \, \text{m/s}$, the final velocity of the bullet as $v_{bf} = 0.5 \times 300 = 150 \, \text{m/s}$, the mass of the block as $m_{block} = 1.00 \, \text{kg}$, and the final velocity of the block as $v_{block}$.Using the conservation of momentum:$$m_b \cdot v_{bi} + 0 = m_b \cdot v_{bf} + m_{block} \cdot v_{block}$$Plugging in the values:$$0.01 \cdot 300 = 0.01 \cdot 150 + 1.00 \cdot v_{block}$$Solving for $v_{block}$:$$3 = 1.5 + v_{block}$$$$v_{block} = 1.5 \, \text{m/s}$$

In a Police Ballistics Test, a 10