Bumper car A(281 kg) * m * c moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) * m * c moving – 1.72 m/s . What is the

Question

Bumper car A(281 kg) * m * c moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) * m * c moving – 1.72 m/s . What is the velocity of car A after the collision ?

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Thu Cúc 4 years 2021-09-04T23:12:39+00:00 1 Answers 5 views 0

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  1. MichaelMet
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    2021-09-04T23:14:32+00:00
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    The final velocity of the car A after collision is 1.884 m/s.

    Explanation:

    As per law of conservation of energy, the momentum of objects before collision is tend to be same for the momentum of objects after collision, in case of elastic collision.

    So, from those equation, we can derive the final velocities for both the objects after collision.

    v_{1} = \frac{m_{1}- m_{2} }{m_{1} + m_{2} } u_{1}+ \frac{ 2*m_{2} }{m_{1} + m_{2} } u_{2}

    v_{2} = \frac{2*m_{1} }{m_{1} + m_{2} } u_{1}+ \frac{ m_{2}- m_{1} }{m_{1} + m_{2} } u_{2}

    So if we consider the car A is having a mass m1 which is equal to 281 mc kg with initial velocity before collision as u1 = 2.82 m/s. And then the car B will be having the mass m2 equal to 209 mc kg with the velocity of 1.72 m/s. So if the final velocity of car A after collision is required to determine then we have to solve the formula for v1

    v_{1} = \frac{m_{1}- m_{2} }{m_{1} + m_{2} } u_{1}+ \frac{ 2*m_{2} }{m_{1} + m_{2} } u_{2}

    v_{1} = \frac{281- 209 }{281 + 209} } * 2.82+ \frac{ 2*209 }{281+209}*1.72

    v_{1} = \frac{72 }{490} } * 2.82+ \frac{ 418 }{490}*1.72

    v_{1} = 0.414+1.47=1.884

    So the final velocity of the car A after collision is 1.884 m/s.

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