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Impulse and Momentum of a crate

  1. Mar 21, 2004 #1
    Question (I added a picture): The free-rolling ramp has a mass of 40 kg. A 10 kg crate is released from rest at A and slides down 3.5 m to point B. If the surface of the ramp is smooth, determine the ramp's speed when the crate reaches B. Also, what is the velocity of the crate?

    This is what I did for finding the velocity of the ramp:

    1) \sigma mv1 = \sigma mv2

    2) mass ramp * (velocity ramp)1 + mass crate * (velocity crate)1 = mass ramp * (velocity ramp)2 + mass crate *(velocity crate)2

    3) 40*Vr + 0 = 40*Vr2 + 10*Vc2

    Too many unknowns?

    This is what I did for finding the velocity of the crate:

    1) Vc2^2 = Vinitial^2 + 2*a*(S-Sinitial)

    = 0 + 2 * 4.905 * 3.5
    Vc2 = 5.86 m/s

    2) I can substitute this in to the first equation, but I still have too many unknowns. What should I do next?

    Attached Files:

  2. jcsd
  3. Mar 21, 2004 #2
    The crate is released from rest, therefore its initval velocity Vr is zero.

    (Don't forget you can also appy the rules of conservation of energy, even though it's not needed in this case. There are only two kind of forces in the problem - the gravitional force, which is preservative, and the normal force which does no work since it's always perpendicular to the displacement of the objects.)
  4. Mar 21, 2004 #3
    Vr is the velocity of the ramp
    Vc is the velocity of the crate
  5. Mar 21, 2004 #4
    Both [tex]V_R_1[/tex] and [tex]V_C_1[/tex] equal 0 since the system is at rest before it's released.
  6. Mar 21, 2004 #5
    The velocity of the crate and the velocity of the ramp are in opposite directions, right?

    So the equation is
    0 = -40 * Vr2 + 10 * Vb2 and since Vb2 = 5.86 m/s

    Vr2 = 1.465 m/s in the opposite direction of the crate. Is it right?
  7. Mar 21, 2004 #6
    I would think so, unless you made a mistake in some of the calculations. The way is correct.
  8. Mar 21, 2004 #7
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