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5 October, 08:55

bumper car A (281 kg) moving + 2.82 m/s makes an elastic collision with bumper car B (209 kg) moving + 1.72 m/s what is the velocity of car A after the collision (unit = m/s) remember: right is (+), left is (-)

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  1. 5 October, 10:24
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    Answer: V1 = 3.559 - 0.744V2

    Explanation: Given that the

    bumper car A has

    Mass M1 = (281 kg) moving with

    Velocity U1 = 2.82 m/s

    bumper car B has

    Mass M2 = (209 kg) moving with

    Velocity U2 = 1.72 m/s

    Where U1, U2 are the initial velocity of the two cars

    Since the collision is elastic, we will use the formula below,

    M1U1 + M2U2 = M1V1 + M2V2

    Substitute the values into the formula

    281*2.28 + 209*1.72 = 281V1 + 209V2

    640.68 + 359.48 = 281V1 + 209V2

    1000.16 = 281V1 + 209V2

    Make V1 the subject of formula

    281V1 = 1000.16 - 209V2

    V1 = 1000.16/281 - 209V2/281

    V1 = 3.559 - 0.744V2

    Therefore, the velocity of car A which

    is V1 after the collision will be expressed as V1 = 3.559 - 0.744V2
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