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Impulse and momentum of a cart down a ramp

  1. Oct 25, 2006 #1
    Question: A 500 g cart is released from rest 1.0 m from the bottom of a frictionless, 30 degrees ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. During collision, the cart and the ramp are in contact for 0.0267 sec. The average force exerted by the ramp on the cart is 100 N. After the cart bounces, how far does it roll back up the ramp?



    To solve this problem, do I need to find magnitude and direction of initial force of the cart first, and then subtract this from the force that the ramp exerts in opposite direction at the bottom to give impulse? I am confused. Any help would be greatly appreciated.
     
    Last edited: Oct 25, 2006
  2. jcsd
  3. Oct 26, 2006 #2

    OlderDan

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    I assume the word ramp is supposed to be block where I noted. You need to know how fast the cart is going when it hits the block, and determine its change in momentum when it bounces. Since gravity is acting the whole time, you really should include that in the total impulse, but it will be a small error if you assume only the block is exerting a force during the collision. It is easy enough to include it, but if you are comparing to an answer key, I can't be sure they included it.
     
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