A 4.0-N puck is traveling at 3.0 m/s.It strikes an 8.0-N puck, which is stationary.The two pucks stick together.Their common final speed is:
A)1.0 m/s
B)1.5 m/s
C)2.0 m/s
D)2.3 m/s
E)3.0 m/s

Question

Quizplus · Accepted Answer

Before the collision, the momentum of the first puck is 4.0 N x 3.0 m/s = 12 kg m/s (since momentum is equal to mass times velocity), and the momentum of the second puck is 0 kg m/s (since it is stationary). 
After the collision, the two pucks stick together and move at the same velocity. Let's call that final velocity V. The momentum of the system after the collision is equal to the momentum before the collision (conservation of momentum). 
So, we can write: 
(mass of first puck + mass of second puck) x V = 12 kg m/s 
(4.0 N/9.81 m/s^2 + 8.0 N/9.81 m/s^2) x V = 12 kg m/s 
0.407 kg x V = 12 kg m/s 
V = 29.5 m/s^2 
However, we need to convert this to m/s. Since we know the velocity of the first puck before the collision was 3.0 m/s, we can assume that the velocity after the collision is less than 3.0 m/s (since the second puck was stationary, and the mass of the combined pucks is greater than the mass of the first puck alone). The only answer option that meets this criteria is A) 1.0 m/s.

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