A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2 m/s in the opposite direction on a horizontal frictionless surface. If the block is brought to rest by the collision, what is the kinetic energy of the bullet as it emerges from the block?
A) 0.51 kJ
B) 0.29 kJ
C) 0.80 kJ
D) 0.13 kJ
E) 20 kJ

Question

Quizplus · Accepted Answer

Explanation:Momentum is conserved in the collision, so the total momentum before the collision is equal to the total momentum after the collision. The total momentum before the collision is:$$(10	ext{ g})(2.0	ext{ km/s}) + (4.0	ext{ kg})(-4.2	ext{ m/s}) = -16.8	ext{ kg m/s}$$The total momentum after the collision is:$$(10	ext{ g})v + (4.0	ext{ kg})(0	ext{ m/s}) = 10v	ext{ kg m/s}$$Setting these two quantities equal to each other, we get:$$-16.8	ext{ kg m/s} = 10v	ext{ kg m/s}$$Solving for v, we get:$$v = -1.68	ext{ km/s}$$The kinetic energy of the bullet as it emerges from the block is:$$K = \frac{1}{2}(10	ext{ g})(1.68	ext{ km/s})^2 = 0.14	ext{ kJ}$$

A 10-G Bullet Moving Horizontally with a Speed of 2