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Elastic collision with Conservation of momentum problem?

  1. Nov 11, 2012 #1
    elastic collision with Conservation of momentum problem???

    i need help with part b.

    Two titanium spheres approach each other head-on with the same speed and collide elastically, After the collision, one of the spheres, whose mass (m1) is .3 kg, remains at rest.

    (a) What is the mass of the other sphere?
    (b) Assume that the initial speed of each sphere was 2.0 m/s. What is the speed of the two-sphere center of mass?

    I know i did part a correctly. heres what i did:

    momentum is conserved: m1 * u - m2 * u = m2 * v
    or (m1 - m2) * u = m2 * v
    Also, for an elastic head-on collision, we know that the
    relative velocity of approach = relative velocity of separation
    (from conservation of energy), or, for this problem,
    2u = v
    (m1 - m2) * u = m2 * 2u
    m1 - m2 = 2 * m2
    m1 = 3 * m2
    m1 is the sphere that remained at rest (hence its absence from the RHS), so
    m2 = 0.3kg / 3
    m2 = 0.1 kg

    b) this part confuses me, heres what i did

    (m1 - m2) * u = m2 * v
    (.3kg - .1kg)(2.0m/s) = .1kg * v
    .4 kg = .1 v
    v = 4 m/s

    What my teacher did:
    (.3g - .1g) * 2.0m/s = (.3g + .1g) * v

    I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero??

    v = +1.00m/s

    since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean?

  2. jcsd
  3. Nov 11, 2012 #2
    Re: elastic collision with Conservation of momentum problem???

    Your teacher used this:
    [itex]v_{CM}[/itex] is the velocity of center of mass.
  4. Nov 11, 2012 #3
    Re: elastic collision with Conservation of momentum problem???

    I know he uses that formula but shouldnt m1 be zero since its final v is zero because it's at rest?
  5. Nov 11, 2012 #4
    Re: elastic collision with Conservation of momentum problem???

    If you carefully look at the formula i posted, the numerator is the linear momentum of the system. Here, since the momentum is conserved, the numerator does not change. Your teacher used this and found momentum of system before collision which saved him from doing more steps. Even if you find the final velocities after the collision and substitute it in the formula, you get the same answer as your teacher.
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