Momentum Concepts: Clearing Up Frictionless Surfaces Questions

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In a discussion about momentum concepts on frictionless surfaces, participants explore the motion of a block sliding down a wedge on a horizontal frictionless surface. The consensus is that the center of mass of the block-wedge system remains stationary due to the conservation of momentum, as there are no net external forces acting horizontally. Additionally, when two bodies of unequal mass are acted upon by equal forces, the body with greater mass will have smaller acceleration, not greater, as dictated by the relationship F = ma. Clarifications were made regarding the motion of the center of mass and the implications of forces acting on the system. The conversation concludes with inquiries about the behavior of the block after it leaves the wedge.
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Can anyone help to clear these up for me ?

If you have a wedge on a horizontal frictionless surface, a block starts from rest and slides down the inclined surface of the wedge, is rough. During the motion of the block the center mass of the block and wedge ---
- would it be moving horizontally with increasing speed ?

and...

Two bodies of unequal mass are at rest on a frictionless surface, and are acted on by equal forces for equal times. Just after these forces are removed, the body with greater mass will...
- greater acceleration or speed, or smaller momentum?
For this one i want to say greater acceleration but can anyone explain if this is right ?
 
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For your first question, yes, the centre of mass of the body would be accelerating horizontally in the opposite direction to the motion of the block.

(2) For this question, think about the relationship F = ma.
 
Hootenanny said:
For your first question, yes, the centre of mass of the body would be accelerating horizontally in the opposite direction to the motion of the block.

(2) For this question, think about the relationship F = ma.

With all due respect I believe your answer for the first question is incorrect. There are no net external forces (in the horizontal direction) acting on the block-wedge system. Thus the momentum of the overall system will be conserved, and the CM will stay where it is.

-Dan
 
topsquark said:
With all due respect I believe your answer for the first question is incorrect. There are no net external forces (in the horizontal direction) acting on the block-wedge system. Thus the momentum of the overall system will be conserved, and the CM will stay where it is.

-Dan

Yes, my apologies, I've misunderstood. Just for my piece of mind, what would happen when the block left the wedge?
 
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