Physics Help Forum Elastic Collision

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 Oct 27th 2009, 07:25 AM #1 Junior Member   Join Date: Sep 2009 Posts: 23 Elastic Collision A 0.24kg object traveling rightward at 3.5m/s collides head-on and elastically with a 0.40kg object traveling leftward at 2.1m/s. What are their velocities after the collision? If the objects are in contact for 0.010s during the collision, what is the magnitude of the average force that they exert on each other? -v2i + v1i = -v1f + v2f Initial Momentum = 0.24kg(3.50m/s) + 0.40kg(2.1m/s) Initial Momentum = 1.68 kg*m/s 1.68kg*m/s = 0.24kg * v1f + 0.40kg*v2f * Now I find a substitution. * -v2i + v1i = -v1f + v2f -v2i + v1i = -2.1m/s + 3.5m/s = 1.40m/s 1.40m/s = -v1f + v2f v2f = v1f + 1.40m/s * Now I substitute it back into the above equation. * 1.68kg*m/s = 0.24kg*v1f + 0.40kg*(v1f + 1.40m/s) 1.68kg*m/s = 0.24kg*v1f + 0.40kg*v1f + 0.56kg*m/s 1.12kg*m/s = 0.64kg*v1f -1.75m/s = v1f (Negative to indicate direction.) * Now I substitute v1f into the original equation. * 1.68kg*m/s = 0.24kg * v1f + 0.40kg*v2f 1.68kg*m/s = 0.24kg * 1.75m/s + 0.40kg*v2f v2f = 3.15m/s I have no idea how to figure out the force exerted on each other. Could someone help here, and also verify that the above answers are correct? Thanks!
 Oct 27th 2009, 09:01 AM #2 Senior Member   Join Date: Oct 2009 Location: india Posts: 409 an easy way to check is to see whether change in momentum for the two bodies are the same. in your case, it is not. please see whether i am correct final speeds of the bodies are the same as they had before, only they are reversed in direction after collision.ie. 1st body has3.5 along left and 2nd one has 2.1 right. change in mom for either body is m(vf- vi)=.24(-3.5-3.5)=.40(2.1+2.1)=1.68 avg force=change in mom/ time=1.68/(.01)=168 N. actually, the initial momentum of the two particle system is coming zero. u missed a negative sign in between.(or maybe your conventions are different) so final momentum of the two body system will also be zero.(no external force on the two body system, so mom conserved) thus u can easily see that the mag of velocities must remain the same, only directions are altered Last edited by r.samanta; Oct 27th 2009 at 12:25 PM.
 Oct 27th 2009, 01:06 PM #3 Banned   Join Date: Aug 2009 Location: UK Posts: 240 I agree with r.samanta answer = zero velocity. take right as positive velocity, left as negative velocity. momentum before collision: (0.24*3.5) + (0.4*(-2.1)) = 0 kg m/s since this is zero i don't need to carry on, lets alter the example so i can show you how it's worked. lets take (-2.1) to be (-1.1) (to the left) momentum before collision: (0.24*3.5) + (0.4*(-1.1)) = 0.4 kg m/s momentum after collision: add the masses, 0.24 + 0.4 0.64*v 0.64*v = 0.4 0.4/0.64 = v = 0.625 m/s to the right Last edited by Paul46; Oct 27th 2009 at 02:45 PM.
 Oct 27th 2009, 03:28 PM #4 Junior Member   Join Date: Sep 2009 Posts: 23 Hmm, I kind of get what you're saying here, but it doesn't correspond to the problem he did similar to this. (Well, he didn't ask for the energy though..)

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