A 1.0-kg object moving 9.0 m/s collides with a 2.0-kg object moving 6.0 m/s in a direction that is perpendicular to the initial direction of motion of the 1.0-kg object. The two masses remain together after the collision, and this composite object then collides with and sticks to a 3.0-kg object. After these collisions, the final composite (6.0-kg) object remains at rest. What was the speed of the 3.0-kg object before the collisions?
A) 15 m/s
B) 10 m/s
C) 5.0 m/s
D) 20 m/s
E) 25 m/s

Question

Quizplus · Accepted Answer

First, we need to calculate the momentum of the initial two-object system before the collision with the 3.0-kg object:
p = m1v1 + m2v2
p = (1.0 kg)(9.0 m/s) + (2.0 kg)(6.0 m/s)
p = 21 kg m/s

Since momentum is conserved, the momentum of the final composite object after the collision with the 3.0-kg object is also 21 kg m/s:
p = m3v3
21 kg m/s = (6.0 kg + 3.0 kg)v3
v3 = 7.0 m/s

Therefore, the speed of the 3.0-kg object before the collisions must have been:
v2-before = v3 + (v1-after - v1-before)
v2-before = 7.0 m/s + (0 - 9.0 m/s)
v2-before = -2.0 m/s

Since the speed cannot be negative, we take the magnitude:
|v2-before| = 2.0 m/s

The answer choice closest to 2.0 m/s is C) 5.0 m/s, which is the correct answer.

A 10-Kg Object Moving 9