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Velocity of ball after it leaves ramp, and after spring it contact is de-compressed? watch

    • Thread Starter


    I don't understand how to work out the velocity of the ball as it hits the floor. As it hits the floor, it's GPE becomes 0..so initial Energy (just spring PE) = final Energy (0 because the spring and the ball both lose their respective potentials, and the ball has no kinetic energy cos it's on the floor right?).

    SO I am stuck :/

    I have calculated the velocity of the ball when it leaves the launching ramp (and answer is ofc. in symbol form).
    I thought KEinitial + PEinitial = KEfinal + PEfinal

    KE + PE = 0.5mv^2 + (mgh + 0.5kx^2)

    Initially, KE = 0 (nothing is moving..everything is at rest), and PE = 0.5kx^2 (the spring has PE because it is compressed).
    So initial total energy = 0.5kx^2

    Finally, KE = 0.5mv^2, and PE = mgh

    So 0.5kx^2 = 0.5mv^2 + mgh

    So (in an easier form) I have: mv^2 = kx^2 - 2mgh. I divide by m and then do square root to get the velocity. But how do I work out the velocity of the ball as it hits the floor?

    Thank you

    Edit: Apparently, the final answer for the velocity of the ball when it LEAVES THE RAMP does NOT depend on h... I am even more confused now

    The initial energy of the ball is not just the pe in the spring. You need to add mgH
    Take zero for gpe at the floor.
    At the start you have eleastic pe + gpe
    At the end you have no elastic pe, no gpe but lots of kinetic energy.

    The velocity as it leaves the ramp is found (via the ke of the ball) from the initial epe in the spring less the gpe gained rising the height y.
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