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Question

A body of mass 2 kg has an initial velocity vi=(^i+^j)ms1. After the collision with another body its velocity becomes vf=(5^i+6^j^k)ms1. If the impact time is 0.02 s, the average force of impact on the body (in newton) is

A
50(4^i+5^j+^k)
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B
50(4^i5^j^k)
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C
100(4^i+5^j^k)
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D
100(4^i+5^j^k)
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Solution

The correct option is D 100(4^i+5^j^k)
Change in momentum = m(vfvi)=2 kg(5^i+6^j^k) ms1 2 kg(^i+^j) ms1=2(4^i+5^j^k)
Average force = change in momentumtime of impact
=2(4^i+5^j^k)0.02 s=100(4^i+5^j^k)
Hence the correct choice is (d).

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