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2 Impulse and Momentum related Questions

  1. Apr 11, 2009 #1
    The first Question

    A 2.5 kg ball and a 5.0 kg ball have an elastic collision. The 2.5 kg ball was at rest and the other ball had a speed of 3.5 m/s. What is the kinetic energy of the 2.5 kg ball after the collision?

    2. Relevant equations

    J=p (Impulse = momentum)

    (F)t= m2v2f+m1v1i

    vf(m1 + m2) = m1v1 + m2v2

    1/2mv^2 + m1v1= KE ?

    3. The attempt at a solution

    Well I had attempted this question twice with no avail. Mainly because I keep getting two unknowns and then i recycle the question too many times.

    5 x 3.5 = 17.5

    so 2.5(v) = 17.5

    v = 7 m/s

    kinetic energy which is 1/2 m v^2

    .5 x 2.5 x 7^2

    = 61.25 J

    - This was wrong as it didn't match any of the multiple choice answers provided


    The Secound Question

    A 5.0 kg boy runs at a speed of 10.0m/s and jumps onto a cart. The cart is initially at rest. If the speed, with the boy on the cart, is 2.50 m/s, what is the mass of the cart?

    2. Relevant equations

    Same as the first question

    3. The attempt at a solution

    mv + m2v2 = m'v'

    (5)(10) + m2(0) = (M2 + 5) (2.5)

    50 = 2.5M + 12.5

    M = 15 kg

    -once again this is not one of the choices for the multiple choice....

    Can someone please help me here? I do not know what I am doing incorrectly.

    Thanks
     
  2. jcsd
  3. Apr 11, 2009 #2

    Doc Al

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    Staff: Mentor

    This is true for an inelastic collision.

    ?? Don't add KE and momentum--that's like adding dogs and goldfish.

    That's the initial momentum. Set it equal to the final momentum. Call the final speeds V1 and V2.

    Since you have two unknowns, you'll need a second equation. Luckily it's an elastic collision so you can use conservation of energy. Set the initial KE equal to the final KE.

    This one looks fine. All you needed was momentum conservation.
     
  4. Apr 11, 2009 #3
    Thanks for the feedback, but could you perhaps elaborate a little bit more on the response and to the first question, and what do you mean by "momentum conservation?" I thought the equation I initially used was Momentum Conservation?:confused:
     
  5. Apr 11, 2009 #4

    Doc Al

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    Staff: Mentor

    Momentum conservation would be something like this:
    m1v1 + m2v2 = m1v'1 + m2v'2

    You did some thing like this:
    m1v1 + m2v2 = m2v'2

    (You assumed that only the second ball would be moving after the collision.)
     
  6. Apr 13, 2009 #5
    m1u1^2= m1 [(m1u1- m2V2 )/m1]^2 + m2V2^2
    m12u1^2= [(m1u1- m2V2 )]^2 + m2 m1V2^2
    m12u1^2= (m1u1)^2 - 2(m1u1 m2V2) + ( m2V2 )^2 + m2 m1V2^2

    Assuming that I have to solve for V2 in order to apply that into 1/2mv^2, however once I get to the third line I have difficulties solely isolating V2, have any ideas or help?
     
  7. Apr 14, 2009 #6

    Doc Al

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    Staff: Mentor

    You have a quadratic equation for V2. Solve it.
     
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