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Force from Collision

  1. May 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Three billiard balls are placed at rest on a level surface. Each ball has a mass of .15 Kilograms. If Ball 1 (m1) has an initial velocity of 10m/s towards ball 2 (m2), and after the glancing collision ball 1 (m1) moves 3m/s towards ball 3 (m3). Find the average force acting on ball 1 (m1) during the collision if that collision lasts for .025 seconds?


    2. Relevant equations



    3. The attempt at a solution
    I know you have to use the formula Vf=Vi+at. And then maybe use [tex]\sum[/tex]F=ma to get the force. I am having trouble grasping the vector idea of the problem and combining them. Thanks!
     
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  3. May 15, 2008 #2

    Dick

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    You use momentum. Average force is equal to change in momentum over change in time.
     
  4. May 15, 2008 #3
    Do I have to worry about vectors with that equation? What would the equation look like exactly? Thanks so much!
     
  5. May 15, 2008 #4

    Dick

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    Yes, you have to worry about vectors, momentum and force are vectors. The equation is just as I described it (final momentum-initial momentum)/(final time-initial time).
     
  6. May 15, 2008 #5
    so if I use the formula i get: ((.15*3)-(.15*10))/.025 which gives me 42. But I think this is wrong because of the vectors. I am really confused on how to do this problem with the vectors. Thanks.
     
  7. May 15, 2008 #6

    Dick

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    Ok, forget about the vectors. They don't give you enough information to determine the vectors. But concentrate on ball 2. It's initially at rest. What's the initial momentum? After the collision it's moving at 3m/sec. What's the magnitude of the final momentum? Take the difference of the two and divide by time.
     
  8. May 15, 2008 #7
    Ok, the initial mementum of ball 2 is zero. its final momentum is .15*3=.45. Then divide that by time. .45/.025= 18. Would this be that average force that is acting on ball 1 that the problem is looking for? I also know the initial positions of the balls on an xy plane. Ball 1=(-30,0)cm. Ball 2=(0,16)cm. Ball 3=(42,-10)cm. Thanks.
     
  9. May 15, 2008 #8
    You are still basically using F = ma, the accelerartion is found by a = (v - u) /t
     
  10. May 15, 2008 #9

    Dick

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    That's the MAGNITUDE of the average force. If you know the positions of the balls, then you do know enough to find the direction as well. Do you need it? If so then repeat the calculation with vectors. The initial momentum is still (0,0) and the final momentum has magnitude .15*3 and is directed along the vector connecting ball 2 with ball 3.
     
  11. May 15, 2008 #10
    So if 18 is the magnitude of the average force between the two balls, how would I then calculate just the average force acting on ball 1?
     
  12. May 15, 2008 #11

    Dick

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    Use units. It's not '18', it's 18N. You computed the force ball 1 exerts on ball 2. If you want to find the force ball 2 exerts on ball 1 you might want to look at Newton's third law.
     
  13. May 15, 2008 #12
    So the final answer is -18N. Thanks so much. Sorry that it took me so long to understand.
     
  14. May 15, 2008 #13
    How would I include vectors in my answer to make it correct?
     
  15. May 15, 2008 #14

    Dick

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    If you going to put signs on the force, then you need to describe it's direction. I guess here the minus just means opposite to the direction of the other force. You may be expected to provide vectors.
     
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