A 480-kg car moving at 14.4 m/s hits from behind a 570-kg car moving at 13.3 m/s in the same direction.If the new speed of the heavier car is 14.0 m/s,what is the speed of the lighter car after the collision,assuming that any unbalanced forces on the system are negligibly small?
A) 13.6 m/s
B) 10.5 m/s
C) 19.9 m/s
D) 5.24 m/s

Question

Quizplus · Accepted Answer

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:$$(480 	ext{ kg})(14.4 	ext{ m/s}) + (570 	ext{ kg})(13.3 	ext{ m/s}) = 14,016 	ext{ kg m/s}$$The total momentum after the collision is:$$(480 	ext{ kg})(14.0 	ext{ m/s}) + (570 	ext{ kg})(v) = 14,016 	ext{ kg m/s}$$where v is the speed of the lighter car after the collision. Solving for v, we get:$$v = \frac{14,016 	ext{ kg m/s} - (480 	ext{ kg})(14.0 	ext{ m/s})}{570 	ext{ kg}} = 13.6 	ext{ m/s}$$Therefore, the speed of the lighter car after the collision is 13.6 m/s.

A 480-Kg Car Moving at 14