A 0,34-kg volleyball is flying west at 1,98 m/s when it strikes a stationary 0,60-kg basketball dead centre. The volleyball rebounds east at

Question

A 0,34-kg volleyball is flying west at 1,98 m/s when it strikes a stationary 0,60-kg basketball dead centre. The volleyball rebounds east at 0,79 m/s. What will be the velocity of the basketball immediately after impact? Choose the east direction as positive.

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Kim Cúc 3 days 2021-07-22T11:47:56+00:00 1 Answers 2 views 0

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    2021-07-22T11:49:13+00:00

    Explanation:

    use the law of conservation of momentum

    M1U1+M2U2=M1V1+M2V2

    m-mass

    u-initial velocity

    v-final velocity

    1-volleyball

    2-basket ball dead center

    so using the above formula

    0.34×1.98+0.64×0=0.34×0.79+0.64×V2

    to get

    0.6732+0 = 0.2686+0.64V2

    0.4046=0.64V2

    divide both sides by 0.64 to get V2 which is the final velocity of basketball

    0.6322m/s=V2

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