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Linear Momentum Pellet Gun Problem

  1. Oct 12, 2008 #1
    Bored, a boy shoots his pellet gun at a piece of cheese that sits, keeping cool for dinner guests, on a massive block of ice. On one particular shot, his 1.1 g pellet gets stuck in the cheese, causing it to slide 25 cm before coming to a stop. If the muzzle velocity of the gun is 76 m/s and the cheese has a mass of 136 g, what is the coefficient of friction between the cheese and ice?

    Relevant equations:
    Momentum = Mass * Velocity

    This would be an example of an inelastic collision where the two objects stick together.
    The KE is not conserved.

    I converted the mass of the pellet and the cheese to kg.
    Mass of pellet = .0011kg
    Mass of cheese = .136kg

    Momentum of the pellet = .0011kg * 76m/s = .0836 kg m/s
    The momentum will be conserved because there is no external forces.

    This is as far as I got, can someone tell me how should I approach this next?
    Thanks.
     
  2. jcsd
  3. Oct 12, 2008 #2

    Doc Al

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    Staff: Mentor

    Since momentum is conserved, find the speed of the cheese+pellet immediately after the collision. Then apply some kinematics to find the acceleration due to friction.
     
  4. Oct 12, 2008 #3
    So I have mass of the pellet * Vi = mass of pellet + mass of cheese * Vf
    .0011kg * .76m/s = (.0011kg + .136kg) * Vf
    Vf = .61m/s

    The only kinematics equation that do not deal with time is
    Vf^2 = Vi^2 + 2 * A * D
    I converted 25cm to .25m to be consistent

    (.61m/s)^2 = (.76m/s)^2 + 2 * A * (.25m)
    I got A = -.41m/s

    Is this right to have a negative acc.?

    If this is all right, do I continue with Newtons 2nd law to solve for the coefficient of friction?
    F = MA?
     
  5. Oct 12, 2008 #4

    Doc Al

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    Staff: Mentor

    Your formula is correct, but redo your calculation being careful with the decimal point.

    Good.

    Several problems: The speed immediately after the collision becomes your initial velocity for this part of the problem. After the collision, the .76 m/s is no longer relevant. What's the final speed of the block of cheese?

    Sure. That just means the cheese was slowing down.

    Yes.
     
  6. Oct 12, 2008 #5
    Yeah, I put it as .76m/s instead of 76m/s =[ Clumsy me.

    The new final velocity would be 0 since it eventually stops.
    0 = (.61m/s)^2 + 2 * A * (.25m)
    I have A = -.74m/s

    2nd part I have the forces in the Y direction:
    Fn = MG where M = mass of cheese + mass of pellet
    Fn = (.1371)kg*(9.81)m/s = 1.3J

    The forces in the X direction:
    There is just the force of friction correct?
    so is
    -Uk(Fn) = M * A
    -Uk(1.3J) = .1371kg * (-.74m/s)
    Uk = .08

    Am I correct?
     
  7. Oct 12, 2008 #6

    Doc Al

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    Staff: Mentor

    My only suggestion is not to round off intermediate calculations--wait until the last step. Other than that, perfect!
     
  8. Oct 12, 2008 #7
    Thanks Doc for your help =].
     
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