Energy problem with 2 objects (a mass flying vertically off of a ramp)

AI Thread Summary
The discussion focuses on a physics problem involving energy and momentum conservation for a mass flying off a ramp. The initial scenario describes a fixed platform where the object reaches a maximum height H after being launched, while the second scenario involves a movable platform, affecting the object's trajectory and height. Participants emphasize the importance of momentum conservation, noting that the platform gains some kinetic energy from the object, which alters the final velocities. The final horizontal component of the object's velocity remains non-zero when it leaves the ramp, contradicting initial assumptions about momentum transfer. Understanding these principles is crucial for solving the problem effectively.
PedroPicapiedra
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Homework Statement


Consider a platform (mass: M) which horizontal surface AB s smoothly joined to vertical surface CD as shown in the figure below. Initially, the platform is fixed in place on a horizontal floor. A small object (mass: m) is placed on AB and given an initial speed of v in the horizontal direction so that it travels along CD, flies vertically off the platform, and reaches a maximum height of H from AB. Friction between the platform and the object is negligible. Next, the platform is kept at rest on the horizontal floor, but is no longer fixed in place. Again, the same object is placed on AB and given the same initial speed of v. The object travels along CD, floes off platform, and reaches a maximum height of from AB. Friction between the platform and the floor is negligible.

What is h/H?
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Homework Equations


K1 + U1 = K2 + U2

The Attempt at a Solution


So I got H form
½ mv^2 + 0 = 0 + mgH
H = v1^2/2g
However I'm not sure on how to get h.
I do know that some of the kinetic energy is transferred to the platform and also momentum should be conserved.

Thanks.
 

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Momentum is conserved in the second case, where the platform is free to move. You use momentum conservation to get the velocity of the platform after the small object has interacted with it. All the momentum of the small object is transferred to the platform. This gives you the velocity, and kinetic energy, of the platform. The remaining energy goes into the upward motion of the small object.
 
Gene Naden said:
All the momentum of the small object is transferred to the platform.
Careful. For the second case, the final horizontal component of velocity of the small object is not zero.

PedroPicapiedra said:
View attachment 227000
I do know that some of the kinetic energy is transferred to the platform and also momentum should be conserved.
What do you learn from conservation of momentum?

How does the final horizontal component of velocity of the small object compare to the final velocity of the ramp? Hint: Section CD is vertical.

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Another approach to the problem is to go to the center of mass reference frame. EDIT: What I mean here is to choose an inertial frame such that the horizontal component of velocity of the center of mass remains zero.
 
Last edited:
TSny said:
Careful. For the second case, the final horizontal component of velocity of the small object is not zero.

I see that the platform curve is vertical where the small object exits. That is why I said all the (horizontal) momentum is transferred to the platform. Was I mistaken?
 
Gene Naden said:
I see that the platform curve is vertical where the small object exits. That is why I said all the (horizontal) momentum is transferred to the platform. Was I mistaken?
When the small object leaves the ramp at point D, it is moving vertically with respect to the ramp but not with respect to the ground. So, the small object does not lose all of its horizontal momentum with respect to the ground.
 
What I must to know to resolve this exercise? Thanks
 
eyespy said:
What I must to know to resolve this exercise? Thanks
Please post an attempt, per forum rules.
 
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