- #1
Amik
- 38
- 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?
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?