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Inelastic collision homework

  1. Nov 14, 2004 #1

    I have the following question to solve, and whilst I've got ideas I have no idea if there right:

    Do we use F=ma ?


    Last edited: Nov 15, 2004
  2. jcsd
  3. Nov 14, 2004 #2
    Use the conservation of momentum

    m1u1 + m2u2 = m1v1 + m2v2

    assign each mass to m1 and m2, u is the initial velocities. this particular example is a completely inelastic collision so there will be only 1 mass (m1 + m2) with a common velocity after the collision. Hope this helps
  4. Nov 14, 2004 #3
    Thanks for that. One more question what is V1 and v2 final velocity which is presumably 0?

    Or do I have to sub in the values and work them out?

    Thanks for your help, it's greatly appreciated

  5. Nov 14, 2004 #4
    since the masses stick together the second half of that equation i wrote would be (m1 + m2) x common velocity of the two. there is no v1 and v2 because they have stuck together and will both be travelling at the same velocity. the final velocity wont be zero though. after they have stuck together they will move off together with a common velocity. does that help?
  6. Nov 15, 2004 #5
    With that in mind is the following right:

    (m1 X u1) + (m2 X u2) = (m1 X v1) + (m2 X v2)
    (3*3) + (2*2) = (3*v1)+(2*v2)
    13 = (m1+m3) * v (is that the right letter to use???)
    13 = 5 * v
    v = 13/5
    v = 2.6

    Is that right, and if not where do you mind explaining where I went wrong please


    Last edited: Nov 15, 2004
  7. Nov 15, 2004 #6
    You've used 2 kg as your mass for one of the trolleys, though above you said it was 24g. You made a typo by putting x on the RHS of your top line, when you meant +, but apart from that it looks ok. Yes, v is the right letter for velocity (were you not taught the usual symbols for things like velocity then?). Also, you don't need to bother putting v1 or v2, just a v in the 2nd line, as you know they're going to move off with the same velocity.
  8. Nov 15, 2004 #7
    Thats what I got, yes hopefully :)

    The letter you use is up to you really, as long as you dont confuse yourself. Make it clear that one side is the initial velocity and the other is the final velocity. i just used v because its what I use for final velocity. In this case I would have noted it as Vc (small c at bottom) simply to remind myself it was a common velocity as the two masses had stuck together and moved off with the same velocity. Does that make it clearer?
  9. Nov 15, 2004 #8
    Sorry I meant 2kg, a typo there. I also mistyped the x on the RHS top line, they should have been +.

    Thanks for all your help, I do not have my exercise book so couldn;t remember the equation for it. In case your wondering I'm an english pupil in year 10 which makes me 15 yrs old, so hopefully this was the right forum to posst in.

    Once again thanks for all your help.

    I have more questions to do so may be back

  10. Nov 15, 2004 #9
    I just knew I'd be back:

    A field gun of mass 1000kg, which is free to move, fires a shell of mass 10 kg at a speed of 200 m/s.

    (b) What is the momentum if the gun just after firing? - Do we use momentum = mass X velocity, and if so what is the velocity it moving back and if so how is that worked out.
    (c) Calculate the recoil velocity of the gun. - Is that an impulse, i.e.g. Ft= mv-mu


  11. Nov 15, 2004 #10
    The conservation of momentum applies to this one again, m1u1 + m2u2 = m1v1 + m2v2

    Initially everything is at rest so the left hand side of that equation will equal 0. (1000*0) + (200*0). Whats the final velocities? Try it from there.

    (Others feel free to correct me if I am leading him astray :) )
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