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Momentum and Collisions

  1. Nov 8, 2003 #1
    I just needed help on these problems because I got stuck on them. Anyone know how to start off these problems?

    An amoeba of mass 1.0 x 10^-12 kg propels itself through the water by blowing a jet of water through a tiny orifice. Suppose the amoeba ejects water with a speed of 1.0 x 10^-4 m/s and at a rate of 1.0 x 10^-13 kg/s. Assume the water is continuously replenished so the mass of the amoeba stays the same. If there were no force on the amoeba other than the reaction force caused by the emerging jet, what would be the acceleration of the amoeba? If the amoeba moves with constant velocity through the water, what is the force exerted by the surrrounding water (exclusive of the jet) on the amoeba?

    A bullet of mass (m) and speed (v) passes copmletely through a pendulum bob of mass (M). The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length l and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical ciricle?
  2. jcsd
  3. Nov 9, 2003 #2


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    1. Assuming the blob starts at rest, use consevation of momentum:
    0 = mv + MV

    (lower case = water, upper case = amoeba)

    For this problem, m is a function of time given by mdot*t
    (mdot is the mass flow rate of the water and t is the time measured from the start of the flow)

    Solve for V as a function of time

    when they ask for acceleration, just differentiate V with repect to time.

    For constant velocity, the "other forces" (like viscous drag) must counteract that acceleration, so they should equal Ma.

    2. Assume negligible frictional loss due to bullet impact.

    You can first figure out how fast you want the bob to be going after the bullet gets through, let's call it V:

    To just barely make a complete circle, it should have ~0 kinetic energy at the top of the circle. Define the low point of the bob as your datum, so that at the high point, the potential energy due to gravity is U = Mgh = Mg(2*l) (we're also ignoring the mass of the hole removed by the bullet).

    This U must be equal to the initial kinetic energy, .5MV^2, so set those equal to find an expression for V.

    V is determined by the conservation of momentum:

    momentum before the collision = mv (to the right)
    momentum after the collision = MV + mv/2 (to the right)

    set those equal and solve for v in terms of V. Then plug in your solution for the minimum V and that's it.
  4. Nov 11, 2003 #3
    1a) 1.0 x 10^-5 m/s^2
    b) 1.0 x 10^-17 N

    On the 2nd problem I got stuck
    I set Mg(2 x l) and .5MV^2 equal.

    V^2=g(2 x l)/.5 = (g(2 x l)^1/2) / .5

    After setting those two equations equal and solving for v in terms of V I got this.

    mv= MV + mv/2

    After that, where do I plug in the v equation?
  5. Nov 11, 2003 #4


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    With the equations:

    V^2=g(2 x l)/.5



    you can plug in V from the first equation: V = 2√(g*l)

    v = 4M(√(g*l))/m
  6. Nov 12, 2003 #5
    Thanks for the help! I knew I had the answer, plugging in V to v totally slipped my mind.
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