Kinetic Energy: Trial 1 vs. Trial 2

AI Thread Summary
In the discussion about Kinetic Energy in two trials, it is established that in Trial 1, both the block and the plane share the potential energy, resulting in equal final kinetic energy. In Trial 2, all potential energy is transferred to the block, leading to a higher speed for the block compared to Trial 1. The conversation highlights the importance of considering the non-inertial frame of the sliding plane, which requires the inclusion of a pseudo force to accurately analyze the situation. Participants emphasize the need for detailed reasoning when discussing the conservation of momentum and the effects of forces in both trials. The discussion concludes with the suggestion that the initial explanation could be valid if the pseudo force is accounted for in the equations.
Amik
Messages
38
Reaction score
3
Homework Statement
See the picture
Relevant Equations
Conservation of momentum and energy
So, I know the right solution should be like it has same potential energy initially, in the trial one, since the floor is frictionless, the plane will move too and it shares the potential energy with the block. So in trial 1 the potential energy equals final kinetic energy of the block and plane. But in trial 2, all potential energy goes to block and thus in trial 2, the block will have greater speed
But, Can I do it this way?
In both trials, the block will have the same velocity on the plane with respect to the plane(not the ground), However, according to conservation of momentum, In trial 1, when block speed up on the plane, the plane will have horizontal velocity deirected to the left too. So the velocity of the block in trial 1 is always the horizontal component of the velocity of the block minus the horizontal component of the velocity of the plane but in trial 2 the velocity in trial 2 is always the horizontal component of the velocity of the block. So I think the velocity in trial 2 is larger.
Is there any problem with second way to do it?
 

Attachments

  • Annotation 2020-04-13 233515.png
    Annotation 2020-04-13 233515.png
    17.4 KB · Views: 213
Physics news on Phys.org
Amik said:
In both trials, the block will have the same velocity on the plane with respect to the plane
How so?
 
For plane I mean the incline. Because in both senarios blocks are only subject component of force of gravity(that force is parallel to the surface)
I might be wrong.
 
Amik said:
For plane I mean the incline. Because in both senarios blocks are only subject component of force of gravity(that force is parallel to the surface)
I might be wrong.
The frame of reference of the (sliding) plane is non-inertial, so you need to include a pseudo force on the block to represent the plane's acceleration.
 
OK. This is only AP level problem. Dioes that mean my explanation is entirely wrong?
 
Amik said:
OK. This is only AP level problem. Dioes that mean my explanation is entirely wrong?
First, the statement of yours that I quoted in post #2 is not obviously true so you cannot use that argument without more detailed reasoning.
Your explanation in post #3 leans on use of a non-inertial frame, so needs to be particularly careful. In particular, you have omitted the pseudo force that needs to be included in such a frame.
Your explanation might still work if you include the pseudo force and develop the relevant equations.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top