A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis.Just befor

Question

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis.Just before the collision,one ball,of mass 3.0 kg,is moving up- ward at 20 m/s and the other ball, of mass 2.0 kg, is moving down- ward at 12 m/s. How high do the combined two balls of putty rise above the collision point

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RuslanHeatt 4 years 2021-09-01T15:45:10+00:00 1 Answers 17 views 0

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  1. yenoanh
    0
    2021-09-01T15:46:40+00:00
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    Answer:

    y=2.64m

    Explanation:

    Given data

    Ball one

    mass m₁=3.0kg

    velocity v₁=20 m/s

    Ball second

    mass m₂=2.0 kg

    velocity v₂=12 m/s

    First we need the speed of combined ball.Since the system conserves the linear momentum

    p_{i}=p_{f}\\m_{1}v_{1}+m_{2}v_{2}=m_{t}v_{t}

    So the combined velocity vt is:

    v_{t}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{t}}\\

    Since the two balls 1 and 2 are moving in opposite direction

    So

    v_{t}=\frac{m_{1}v_{1}-m_{2}v_{2}}{m_{t}}\\

    Substitute the given values

    v_{t}=\frac{(3kg)(20m/s)-(2kg)(12m/s)}{(3+2)kg}\\ v_{t}=7.2 m/s

    We have the equation for motion with constant acceleration is given by:

    v^2=v_{o}^2+2g(y-y_{o})\\

    At initial position y₀=0 and vt=v-v₀

    So

    v^{2}=v_{o}^2+2g(y-0)\\ y=\frac{v^2-v_{o}^2}{2g}\\ y=\frac{v_{t}^2}{2g}\\ y=\frac{(7.2m/s)^2}{2(9.8m/s^2)}\\ y=2.64m

       

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