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March 24, 2017

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A m = 1.4 kg object moving at v = 15 m/s collides with a stationary 2.0 kg object. If the collision is perfectly inelastic, how far along the inclined plane (37degrees) will the combined system travel? Neglect friction.

  • Physics - ,

    I have never know what a perfectly inelastic collision is. I do know what an inelastic collision is: Energy is NOT conserved. Ask your teacher this question, see if they know.

    Momentum applies:
    Now on this statement, conservation ofm momentum applies
    momentum before=momentum after
    along the path of the moving object..

    1.4*15m/s+2*0=(6kg*V) assuming the objects are stuck together, the problem did not state that.

    Now, you can consider energy:
    initialKE after collision= change PEnergy

    1/2 (6)V^2= (6)g height

    where distancealongplane=hSin37

  • Physics - ,

    oops...

    distancealongplane=h/sin37

  • Physics - ,

    where are you getting 6kg from?

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