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Energy/Momentum related question

  1. Oct 6, 2005 #1
    I got this question that i'm stumped on, was hoping if someone could see if i'm correct and what i did wrong.

    A 5.0kg ball initialy at rests at the edge of 2.0m long, 1.2 m high frictionless table, A hard plastic cube of mass 0.50kg slides across the table at a speed of 26m/s and strikes the ball, causing the ball to leave the table in the direction in which the cube was moving.

    It also gave us Implulse which is 12N/S

    1) Determine the horrizontal velocity of the ball immediatley after the collision
    2) Determine the velocity of the cube immdiately after the collision

    What i did:
    for the 1st question since they gave us impulse which is equal to FT=MV i basically divided the impulse by the mass to get my velocity of 2.4 m/s which i'm not sure if i did right.

    for the second one, i went M1V1 + M2V2 = M1V1(prime) + M2V2(prime) and basically i solved for V2 prime and got 2.0 m/s.

    Did i do them right? if not tell me what i have to do. Thanks
  2. jcsd
  3. Oct 6, 2005 #2


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    Looks OK. Let us know if there are any concepts that are unclear or confusing.
  4. Oct 6, 2005 #3
    well actually theres another question to that, That i cant figure out

    3) Determine the distance between the two points of impact of the objects with the floor.

    Any ideas?
  5. Oct 7, 2005 #4


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    Since you know how high the table is off of the ground, you can find out how long it takes for either object to hit the ground after it gets to the edge of the table and starts to fall. You've already found the post-impact horizontal velocity of both objects, so you can use that to calculate how far away from the table each object will land.
  6. Dec 6, 2007 #5
    old...but still good..

    I know this problem is old, but I'm sure many other people check up on this problem, (as it is a 1995 ap physics c test free response problem.) You get impulse= 12 by counting squares on the diagram. I got the same answers as the first poster, and for the last part, (part e), I got 1.51 m apart by using the kinematic distance equations:

    Delta Y: .5Ayt^2
    Delta X: Vixt

    (some parts of the equation cancel out because there's no horiz. acceleration and there is no initial y velocity.)

    hope this helps.
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