Question

# A boy of 50 kg mass is running with a velocity of 2 msâˆ’1. He jumped into a stationary cart of 2 kg while running. Find the velocity of the cart after the boy jumped into it.

A

1.92 m/s

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B
2.3 ms
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C
1.92 ms
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D
3.25 ms
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Solution

## The correct option is C 1.92 msGiven: Mass of the boy, m1=50kg Initial velocity of the boy, u1=2 ms−1 Mass of the cart, m2=2 kg Initial velocity of the cart u2=0 Let the final velocity of cart be v2 Since, boy jumped over cart, the final velocity v1 of the boy will be equal to that of the cart. Therefore, v1=v2 From law of conservation of momentum: m1u1+m2u2=m1v1+m2v2 (50×2)+(2×0)=50v1+2v2 100=50v2+2v2∵v1=v2 100=52v2 v2=10052=1.92 ms−1 Therefore, velocity of cart after jumping of boy over it is equal to 1.92 ms−1. Since, velocity has positive sign, the cart will go in the same direction of boy.

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