A 10-g bullet moving horizontally with a speed of 1.8 km/s strikes and passes through a 5.0-kg block initially at rest on a horizontal frictionless surface. The bullet emerges from the block with a speed of 1.0 km/s. What is the kinetic energy of the block immediately after the bullet emerges?
A) 8.0 J
B) 6.4 J
C) 5.3 J
D) 9.4 J
E) 10 J

Question

Quizplus · Accepted Answer

The law of conservation of momentum can be used to find the velocity of the block after the bullet emerges. The initial momentum of the system (bullet + block) is equal to the final momentum of the system. Let's denote $m_b$ as the mass of the bullet, $v_{bi}$ as the initial velocity of the bullet, $m_{bl}$ as the mass of the block, $v_{bf}$ as the final velocity of the bullet, and $v_{bl}$ as the velocity of the block after the bullet emerges.Initial momentum = Final momentum$$m_b \cdot v_{bi} = m_b \cdot v_{bf} + m_{bl} \cdot v_{bl}$$Given: $m_b = 0.01$ kg (10 g converted to kg), $v_{bi} = 1800$ m/s, $v_{bf} = 1000$ m/s, $m_{bl} = 5.0$ kg$$0.01 \cdot 1800 = 0.01 \cdot 1000 + 5.0 \cdot v_{bl}$$$$18 = 10 + 5v_{bl}$$$$8 = 5v_{bl}$$$$v_{bl} = 1.6 \, \text{m/s}$$The kinetic energy ($KE$) of the block is given by $KE = \frac{1}{2} m_{bl} v_{bl}^2$.$$KE = \frac{1}{2} \cdot 5.0 \cdot (1.6)^2$$$$KE = 2.5 \cdot 2.56$$$$KE = 6.4 \, \text{J}$$Therefore, the kinetic energy of the block immediately after the bullet emerges is 6.4 J.

A 10-G Bullet Moving Horizontally with a Speed of 1