Does Equal Force Result in Equal Momentum for Different Masses?

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Equal forces applied to two bodies of different masses for the same duration result in equal momentum, as momentum before and after the force application must remain constant. The larger mass does not have greater momentum; both bodies will have the same momentum after the forces are removed. In the second scenario, the impulse experienced by a 0.2kg rubber ball dropped from a height is calculated correctly, with the change in momentum equating to 10 upward. The correct calculation for impulse is 0.2 times the rebound velocity minus 0.2 times the impact velocity, confirming the upward direction. Overall, the concepts of impulse and momentum are applied correctly in both cases discussed.
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Please check over my work

Two bodies of unequal mass, placed at rest on a frictionless surface, are acted on by an equal horizontal forces for equal times. Just after the forces are removed, the body of greater mass will have:
The body of greater mass will have the the greater speed, the greater acceleration, the smaller momentum, the greater momentum, or the same amount as the other body.

Would the answer be the larger body would have the same amount amount of momentum as the other body because P(before)=P(after). The momentum would have to come out the same in order for that to be right.

second question:

A 0.2kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30m/s and rebounds up at 20m/s. The impulse during the collision would be:

my work:
0.2 (-30m/s) - 0.2 (20m/s) = 10 upward

Please let me know if I am on the right track and let me know where I might have gone wrong:redface:
 
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blackout85 said:
Would the answer be the larger body would have the same amount amount of momentum as the other body because P(before)=P(after). The momentum would have to come out the same in order for that to be right.

Yes, but you should use the fact that the impulse of the force equals the change of momentum, and since the impulse is equal for both bodies, you have F*t = m1v1 = m2v2.

blackout85 said:
A 0.2kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30m/s and rebounds up at 20m/s. The impulse during the collision would be:

my work:
0.2 (-30m/s) - 0.2 (20m/s) = 10 upward

Please let me know if I am on the right track and let me know where I might have gone wrong:redface:

Looks good.

Edit: the change in momentum (i.e. the impulse) should actually be: 0.2\cdot 20\vec{j}-0.2\cdot(-30)\vec{j}=10\vec{j}, which means 10 upwards.
 
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