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Homework Help: Some momentum questions

  1. Nov 3, 2008 #1
    Here's a couple of mcs I'm trying to use to study, but they got me stuck and I don't have the answers, much less the explanations. A little help here would be greatly appreciated.

    1. A 2kg ball collides with the floor at an angle  and rebounds at the same angle and speed as shown above. Which of the following vectors represents the impulse exerted on the ball by the floor?



    The left and right cancels, leaving only up, so I'm guessing E is the answer?



    22. What is the total kinetic energy of the two-puck system before the collision?
    (A) √13 J
    (B) 5.0 J
    (C) 7.0 J
    (D) 10 J
    (E) 11 J

    For this one I used the kinetic energy equation on the 2 particles and added them, getting 5.0 J so B is the answer.

    3. What is the magnitude of the total momentum of the two-puck system after the collision?
    (A) 1.0 kg.m/s
    (B) 3.5 kg.m/s
    (C) 5.0 kg.m/s
    (D) 7.0 kg.m/s
    (E) 5.5 √5 kg.m/s

    I solved for momentum of both, then used pythagorean theorem and got C as the answer


    4. As shown above, two students sit at opposite ends of a boat that is initially at rest. The student in the front throws a heavy ball to the student in the back. What is the motion of the boat at the time immediately after the ball is thrown and, later, after the ball is caught? (Assume that air and water friction are negligible.)

    After the Throw After the Catch

    (A) Boat moves forward Boat moves forward
    (B) Boat moves forward Boat moves backward
    (C) Boat moves forward Boat does not move
    (D) Boat moves backward Boat does not move
    (E) Boat moves backward Boat moves forward

    This one I'm really clueless. What difference does it make if the ball is moving or caught?

    I knoew (or think I know) that the boat would move right when the ball is caught, but what happens immeidately after the ball is thrown? I'd try this if I had a heavy ball, a buddy and a boat but I don't live close to any of those.

    Please help me understand all this stuff.
  2. jcsd
  3. Nov 3, 2008 #2


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    1) is correct, but maybe the wrong reason. The x component of velocity is unchanged, so only the y component would be affected. Since it was reversed by the bounce ...

    2) bad math mv2/2

    3) total m*v before = ...

    4) for every action, there is a reaction.
  4. Nov 4, 2008 #3
    Huh, for #2 I used mv2/2 on the 2 objects, added them but keep on getting 5 as the answer? What went wrong?

    Ok, so the p after the collision would be the sum of p before the collision, and the answer is 7 (D)?

    For 4 I'm guessing that the boat does not move when it's immediately thrown because it's an internal force and moves forward after it's caught because center of mass does not move.
  5. Nov 4, 2008 #4
    I think your answer for #2 is fine:
    I found the kinetic energy of the first particle to be 0.5*1.5 kg*(2.0 m/s)^2 = 3.0 J
    and the kinetic energy of the second to be: 0.5*4.0 kg*(1.0 m/s)^2 = 2.0 J
    So I think you've applied the kinetic energy equations correctly (finding the energies separately and then adding to result in 5.0 J, because energy is not a vector).

    For #3, are you using momentum as a vector? I think you've found the magnitudes of their momentum correctly, but when you find the net momentum you will have to add these as vectors!

    For #4, think of the instants when the ball and the boat are separated, and then when they join (examine the momentum): You could even try to do this yourself on a skateboard (or an office chair), tossing a heavy object from one hand to the other.
  6. Nov 4, 2008 #5
    #3 asked for total momentum, so I guess it would be the same as before collision.

    I unfortunately don't have any toys to test #4. My second guess is that the boat moves forward and then stop when it's caught???
  7. Nov 4, 2008 #6


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    Sorry I thought you checked A. I didn't read your answer below that.
  8. Nov 4, 2008 #7
    After much thought into #4...

    I'm guessing that the boat would move in the opposite direction that the ball moves. I'm thing of recoil. After the student in the back catches the ball, the boat would move opposite (recoil again?)

    So the answer would be B, boat moves forward immediately after throw and boat moves backward after the catch. Is this correct?
  9. Nov 4, 2008 #8


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    Before the throw, how much momentum is there in the system?

    During the throw then how much momentum in the system? The boat forward plus the ball backward is what?

    After the catch how much momentum?
  10. Nov 4, 2008 #9
    Momentum is conserved. When the ball is thrown, the ball moves backward, which I'll refer to as left from now on. The boat moves right.

    When the person catches the ball, the ball stops and the person (plus boat) moves left.

    I think that's how it goes, unless I have major conceptual errors.
  11. Nov 4, 2008 #10


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    When the person catches the ball what is the forward momentum of the boat just before he catches it? Isn't it equal to the backward momentum of the ball in motion?

    And when it is caught and the ball has no momentum, hasn't its momentum gone to exactly cancel the momentum of the boat moving forward? If the boat has no momentum what is the velocity again?
  12. Nov 4, 2008 #11
    Oh, so the boat doesn't move after it's caught?

    Does it then mean that the answer is C, the boat moves forward after thrown and does not move after it's caught?
  13. Nov 4, 2008 #12


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    That's the way I see it.
  14. Nov 4, 2008 #13
    Thanks for the help. This is one of those problems that I have to spend hours to figure out how it works.
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