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Suppose we have some ramp on wheels of mass $M$, standing on a frictionless surface. A cart of mass $m$ moves with a certain velocity $v$ towards the ramp. The cart moves up the ramp seamlessly, and there's no friction between them and no loss of energy.

  • Find the maximal height $h$ that the cart will reach after the collision
  • Find the time (starting from the moment of the collision) it will take for the cart to reach the maximal height (given the angle of the incline $\alpha$)

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I have a few questions regarding this tricky question:

1) What happens after the collision (between the moment the cart is getting on the ramp and till the moment it returns to the ground)? Does the cart and the ramp are moving with a common velocity at each point in time (i.e. they both have the same velocity all the time, which is changing due to the deceleration of the cart)?

2) I suppose both the cart and the ramp are moving with the same velocity when the cart reaches the maximal height. Why can I use the conservation of momentum $mv+0=(m+M)V$ here? The conservation of mometum law states that if no external forces are acting on the system then the momentum is conserved. However, during the collision, there IS an external force acting on the cart and that is the gravity.

3) How can I possibly find the time it takes for the cart to get to the maximal height (even if we suppose that the incline is not curved)?

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( +! ) some questions a starter always asks. :) –  ABC May 2 '13 at 1:57

1 Answer 1

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  1. The cart and the ramp have to be able to move at different speeds from one another, otherwise the cart could never move up the ramp or back down it again. So they can't both have the same velocity at all times. However, the centre of mass of the system always keeps moving with the same velocity in the $x$ direction (both before and after the collision), and that's what you'll need in order to work out the answer. It's also useful to think about (i) what are the two $x$-velocities just at the instant of the collision, and (ii) what are the two $x$-velocities just at the very moment the cart reaches its highest point on the ramp?

  2. This is a good question. The answer is that the gravitational force is acting vertically, whereas we're only considering momentum in the horizontal ($x$) direction. Momentum conservation applies separately to the $x$, $y$ and $z$ directions, so as long as there's no external force with a component in the $x$ direction, $x$-momentum is conserved.

  3. It's not an easy question, but perhaps a useful hint is to think about it in terms of how much kinetic energy can get converted into potential energy. If you know $mgh$ then you can work out $h$.

I hope these hints help you to solve your problem...

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Thank you! What about the third question - I thought it could be a simple kinematics, knowing the deceleration due to gravity $g \sin \alpha$ and knowing the length of the hypotenuse (with respect to $h$), however I'm not sure that this is the right way (too simple, and I feel I'm missing here something). –  grjj3 May 2 '13 at 10:48
    
I don't want to give the whole answer away, but here's another hint: you don't need to know anything about the geometry of the ramp. Instead, think about how much kinetic energy there is at the very moment when the cart first touches the ramp, and how much there is at the very moment when it reaches its highest point on the ramp. –  Nathaniel May 2 '13 at 11:11
    
I had no problems with conservation of mechanical energy (I know $h$). The problem is, that the conservation laws tell me nothing about time nor about acceleration. That's why I'm asking - is this a plain kinematics in 3) or something more complex? Thanks in advance. (Maybe you missed my point - I'm not asking about how to find $h$ but about how much time it will take for the body to get there). –  grjj3 May 2 '13 at 18:26
    
Sorry, yes, I had missed that point. What you said sounds like the correct way to go - just remember that both the cart and the ramp are experiencing accelerating forces. –  Nathaniel May 3 '13 at 3:10

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