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Modified Newtons cradle and the Conservation of Momentum

  1. Dec 2, 2011 #1
    1. The problem statement, all variables and given/known data
    2 balls hang from a modified Newtons Cradle, on of mass 80g, the other of mass 200g. Assuming gravity is 9.81ms^-2 ball A is pulled out so it reaches a height of 10cm then released. After impact ball 2 moves at a speed of 0.80ms^-1... calculate the velocity of sphere A immediately after impact.

    A total of 9 marks

    2. Relevant equations
    E grav = mgh
    E trans = 0.5mv^2
    The law of conservation of momentum

    3. The attempt at a solution

    (a) m1 = 0.08kg m2 = 0.2kg v2 = 0.8ms-1 v1 = ? height = 0.1m

    Neglecting air resistance when ball A is pulled up to a height of 0.10m where its gravitational potential energy is at a maximum:

    Egrav = mgh = 0.08kg x 9.81ms-1 x 0.1m = 0.078 Joules

    The ball is then released and Egrav is transferred to kinetic energy Etrans fully when the ball reaches the bottom of the swing so Egrav will be at a minimum and Etrans will be at a maximum.

    So if Etrans = 0.5mv^2 then assuming there is no energy loss then Etrans = 0.078J

    Therefore rearranging Etrans for v then the velocity at the point of impact is:

    v1 = 1.4ms-1 at the point of impact.

    The kinetic energy of m2 after being struck is Etrans = 0.5 = 0.5 x 0.2 x 0.82 = 0.064 J

    So the difference in energy would be 0.078J - 0.064J = 0.014J

    so the velocity of m1 after the collision v = √(0.014J) / (0.5 x 0.08kg) = 0.6ms-1

    ----------------
    My answer just seems a long winded way of doing it but for 9 marks I can't see where I've gone wrong if I have. It feels wrong because v1 of m1 is just the difference between the initial velocity of the m1 and the final velocity of m2...

    Any pointers would be greatly appreciated.
     
  2. jcsd
  3. Dec 2, 2011 #2

    PeterO

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    Homework Helper

    With a traditional Newtons cradle we usually get an elastic collision so this may be correct, but we can be certain that momentum is conserved so you could double check that way.
     
  4. Dec 2, 2011 #3
    thanks for the reply. Using the conservation of momentum and p1 = m1v1 should equal p2 = m2v2 in an elastic collision...

    P1 = m1v1 = 0.08kg * 1.4ms-1 = 0.112
    P2 = m2v2 = 0.2kg * 0.8ms-1 = 0.16

    If you take the difference P2-P1 = 0.048

    Rearrange P=mv for v: v=P/m = 0.048/0.08 = 0.6ms-1

    So again the same answer????
     
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