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Find the final speed of the heavier particle

  1. Oct 25, 2003 #1
    Two particles, of masses m and 3m, are moving toward each other along the x-axix with the same initial speeds of 9.81 m/s. Mass m is travelling to the left, and mass 3m is traveling to the right. They undergo a head-on elastic collision and each rebounds along the same line as it approached.

    Find the final speed of the heavier particle.

    Not sure how to go about it, if someone could please help!
  2. jcsd
  3. Oct 25, 2003 #2
    Try this...

    In any elastic collision, the linear Momentum and Kinetic Energy are conserved. Thus if v = final velocity of 3m, then

    v = [(2*m)/m+3m]*9.81 = 4.905 m/s

    Hope u have ur problem solved...

  4. Oct 25, 2003 #3
    Nope, that answers wrong.
    I also tried it another way
    I used ur equation plus 9.81(3m - m / m + 3m)
    and that came out as 9.81 which is, again, also wrong.
    I'm wondering if it has something to do with the signs since one is travelling to the left and the other to the right...
  5. Oct 25, 2003 #4


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    As sridhar_n said: in an elastic collision, kinetic energy is preserved (total kinetic energy stays the same). In any collision in which there are not "external" forces, total momentum stays the same.

    You are given a mass, m (kg? Since this is in terms of m, it doesn't really matter), moving to the left with speed 9.81 m/s.
    Taking speed to the left to be negative, the mass has momentum
    -9.81m kg. meters/sec. and kinetic energy (1/2)m (-9.81)2= 48.12m Joules (Notice that the negative speed doesn't matter. Kinetic energy is a scalar, not a vector so this doesn't have a "direction").

    The mass 2m kg. is moving to the right at 9.81 m/s. It has momentum (2m)(9.81)= 19.62m kg m/s and kinetic energy (1/2)(2m)(9.81)2= 96.24m Joules.

    The total momentum before the collision is 19.62m-9.81m= 9.81m kg m/s and the total kinetic energy is 48.12m+ 96.24= 144.36 Joules.

    Call the speed of the mass m after the collision v1 and the speed of the mass 2m v2. The total momentum after the collision is mv1+ 2mv2 and the total kinetic energy after the collision is (1/2)m v12+ (1/2)(2m) v22. Since both total momentum and total kinetic energy are the same before and after collision, we have
    mv1+ 2mv2= 9.81m and
    (1/2)m v12+ (1/2)(2m) v22= 144.36.

    Notice that the "m" terms cancel in both equation.
    v1+ 2v2= 9.81 so v1= 9.81- 2v2.

    Substitute that into the energy equation, solve to find v2 and then find v1.
  6. Oct 25, 2003 #5

    Chi Meson

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    Note that you will be using the quadratic solution to find v2.

    You can always tell if you have done it correctly because one of your two solutions to the quadratic will always be the original velocity.
  7. Mar 5, 2004 #6
    i dont know what to do with the m's when i put v1 into the energy equation...
  8. Mar 5, 2004 #7


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    Did you miss this from my earlier post?

    "mv1+ 2mv2= 9.81m and
    (1/2)m v12+ (1/2)(2m) v22= 144.36.

    Notice that the "m" terms cancel in both equations.
    v1+ 2v2= 9.81 so v1= 9.81- 2v2."
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