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Question

Raghu pushes Andy when both are on a frictionless surface. Initially, Andy was moving towards Raghu with a velocity 5 ms1 and Raghu was stationary. After the push, Andy moves away from Raghu with 2 ms1 velocity. Find the velocity of Raghu after he pushes Andy. Take the weight of Raghu to be 70 kg and that of Andy to be 80 kg.

A
8 ms1 away from Andy
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B
8 ms1 towards from Andy
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C
12 ms1 away from Andy
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D
12 ms1 towards Andy
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Solution

The correct option is A 8 ms1 away from Andy
If we take the initial direction of motion of Andy to be negative, then
Initial momentum = 80(- 5) = 400 kg ms1
As there is no external force,
Initial momentum = Final momentum = 400 kg ms1
Final momentum = 80(2) + 70(x)
Where x is the final velocity of Raghu.
160 + 70(x) = - 400
x = 8 ms1
Hence, the final velocity of Raghu is 8 ms1 i.e., 8 ms1 along initial direction of motion of Andy. Since the direction of motion of Andy is reversed after the push, final velocity of Raghu is 8 ms1 away from Andy.

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