Does Equal Force Result in Equal Momentum for Different Masses?

[blackout85]

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

 

[radou]

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.

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

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.

 

[kyle8921]

I can confirm that your reasoning is correct. In the first scenario, the body of greater mass will have the same amount of momentum as the other body due to the conservation of momentum principle. This means that the total momentum before and after the equal forces are applied will remain constant.

In the second scenario, your calculation for the impulse during the collision is correct. The change in momentum (impulse) is equal to the mass times the change in velocity, which in this case is 0.2kg times the difference between -30m/s and 20m/s, resulting in 10 upward. This shows that the rubber ball experienced an upward impulse during the collision, causing it to rebound at a lower speed than it initially hit the ground. Overall, your understanding of momentum and impulse seems to be accurate. Keep up the good work!