Cart and ramp with same mass(dynamics)

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Homework Statement


On flat surface there is small cart m and big, initally resting ramp with mass m and height h.
a) At what speed should the cart travel to roll on the flat area on top of the ramp and stop relativly to it?
b) Assuming that cart moves with speed greater than calculated in a) what is the distance d between it and ramp when the it hits the ground?

Friction, air resistance and moment of inertia of the wheels of ramp and cart are neglible. Cart does not detach from ramp during rolling.

Image:
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Homework Equations





The Attempt at a Solution


a)
I wanted to use the principle of conservation of energy and came up with equation like this:
[itex]mv^{2}/2=2mv_{0}^{2}/2 + mgh[/itex]
Where v is initial speed of cart (what I need to find knowing only m,h and g) and [itex]v_{0}[/itex] is speed of ramp and cart on it after the cart finish its movment in relation to ramp.
I know I need a second equation here but I can't find it. I was trying to find this equation using principle of conservation of momentum but Im not sure wheter the momentum is same on the beginning and at the end of movement.
b)
Here I need to know what is the speed of the ramp after the cart 'jumps' of it, then I know how to calculate the distance when I know it. But still I need help with finding out the speed of the ramp.
 
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Answers and Replies

  • #2
haruspex
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I know I need a second equation here but I can't find it. I was trying to find this equation using principle of conservation of momentum but Im not sure whether the momentum is same on the beginning and at the end of movement.
Since there are no horizontal forces between the cart+ramp system and anything external to it (such as the ground), horizontal momentum is conserved.
 
  • #3
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So my second equation is:

[itex]mv=2mv_{0} [/itex]

and the solution for a) therefore is:

[itex] v = 2√{gh} [/itex]

But what concerns me isnt there a normal(normal to ramp) force acting on cart when it rolls, which might has horizontal component because ramp is at some angle to surface.
 
  • #4
haruspex
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what concerns me isnt there a normal(normal to ramp) force acting on cart when it rolls, which might has horizontal component because ramp is at some angle to surface.
You are told that when the cart reaches the flat portion at top of ramp the cart and ramp happen to be at the same speed. That will be a consequence of the horizontal force between cart and ramp during ascent being 'just right'. You don't even know whether the ramp is a straight line or a curve - and you don't need to. In short, you only care about the consequence of that horizontal force, not its details.
 
  • #5
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Alright, so for b) I have these two equations to solve for [itex]v_{r}[/itex](ramp speed) and[itex]v_{c}[/itex](cart speed):
[itex]mv^2/2=mv_{r}^{2}/2+mv_{c}^{2}/2+mgh[/itex]
[itex]mv=mv_{c}+m_{r}[/itex]

And then I only have to calculate horizontal projection range and substract distance ramp moved during cart's free fall?
 
  • #6
haruspex
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Alright, so for b) I have these two equations to solve for [itex]v_{r}[/itex](ramp speed) and[itex]v_{c}[/itex](cart speed):
[itex]mv^2/2=mv_{r}^{2}/2+mv_{c}^{2}/2+mgh[/itex]
[itex]mv=mv_{c}+mv_{r}[/itex]

And then I only have to calculate horizontal projection range and substract distance ramp moved during cart's free fall?
Yes.
 

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