Remember that when dealing with collisions in 2-dimensions, momentum is conserve
ID: 1325182 • Letter: R
Question
Remember that when dealing with collisions in 2-dimensions, momentum is conserved in each dimension.
Two pucks are sliding on a frictionless tabletop. Block A (mass 4.52 kg) is moving to the East (+x direction) at 1.70 m/s. Block B (mass 5.42 kg) is moving to the North (+y direction) at 2.04 m/s. Assume the system to be both Block A and Block B. (Note: It could be useful to make a drawing of the situation.)
1. What is the x-component of the total momentum of the system before the collision?
2.What is the y-component of the total momentum of the system before the collision?
3. What is the x-component of the total momentum of the system after the collision?
4.What is the y-component of the total momentum of the system after the collision?
Explanation / Answer
Only block A is moving along x. Thus,
ptotxi = mA vAx
= 4.52 kg * 1.70 m/s
= 7.68 kg * m/s [ANSWER, PART 1]
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Only block B is moving along y. Thus,
ptotyi = mB vBy
= 5.42 kg * 2.04 m/s
= 11.1 kg * m/s [ANSWER, PART 2]
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Momentum is conserved. Thus, after th collision, the total x component of the momentum is still the same,
ptotxf = 7.68 kg*m/s [ANSWER, PART 3]
The same is true for the y component,
ptotyf = 11.1 kg*m/s [ANSWER, PART 4]