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Question

A boy of 50 kg mass is running with a velocity of 2 ms1. 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 ms

Given:
Mass of the boy, m1=50kg
Initial velocity of the boy, u1=2 ms1
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+2v2v1=v2
100=52v2
v2=10052=1.92 ms1

Therefore, velocity of cart after jumping of boy over it is equal to 1.92 ms1. Since, velocity has positive sign, the cart will go in the same direction of boy.


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