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Which car has greater momentum at the cliff's edge?

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Two cars are initially at rest on a horizontal parking lot at the edge of a steep cliff. Car 1 (HC) has twice as much mass as car 2 (TLC). Equal and constant forces are applied to each car and they accelerate across equal distance to the cliff. WE ignore the effects of friction. When they reach the far end of the lot, the force is suddenly removed, whereupon they sail through the air and crash to the ground below. Which car has the greater momentum at the cliff's edge?


    2. Relevant equations
    Impulse= force x time
    Momentum= mass x velocity

    3. The attempt at a solution
    Since velocity is a factor of momentum, I would need to figure out the acceleration? However, I don't see that there is enough information to do that. My guess is that since car 1 has twice the mass, but the same force and time, it has more momentum. Car 2, has 1/2 the mass of car one, but the same force and time, so less momentum? This looks right. Anyone know for sure??
     
    Last edited: Nov 10, 2008
  2. jcsd
  3. Nov 10, 2008 #2
    You are correct that the change in momentum equals the force multiplied by the time the force is applied. So, to which car is the force applied for a longer time? That is, which car takes longer to cover the same distance?

    That will answer your question. On the other hand, unlike what you wrote, mass certainly does enter the definition of momentum: p=mv.
     
  4. Nov 10, 2008 #3
    The same force is applied to both cars for the same amount of time. With that being said, since I don't know the velocity, do I assume that since car 1 has twice the mass, it therefore has twice the momentum??
     
  5. Nov 10, 2008 #4
    I guess, I'm not sure how to find the velocity without the information necessary; direction and speed. help please. I know this is probably easy but it is overwhelming to me!
     
  6. Nov 10, 2008 #5
    No, the force is not applied for the same time, but over the same distance.
     
  7. Nov 10, 2008 #6
    The final velocities of the two cars will not be the same....
     
  8. Nov 10, 2008 #7
    Maybe I'm starting to understand, since car 1 has twice the mass therefore, twice the inertia. It has the equal and constant force applied for a longer time period because it is slower to accelerate than car 2, with 1/2 the mass? Therefore, car 2, with half the mass, would have a greater velocity. So, even though the force applied is equal and constant, it is only until the cars reach the end of the cliff? Which will be faster for car 2.
     
  9. Nov 10, 2008 #8
    That's right! So, car 2 experiences the same force for a shorter time. Hence...
     
  10. Nov 10, 2008 #9
    So, car 2 has less momentum and car 1 has more momentum!
     
  11. Nov 10, 2008 #10
    Alrighty!
     
  12. Nov 10, 2008 #11
    Thank you for your help! I REALLY appreciate it. I am envious of your understanding of this foreign language.
     
  13. Nov 10, 2008 #12
    It took many years :-)
     
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