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Question

Particle A makes perfectly elastic collision with another particle B at rest. They fly apart in opposite direction with equal speeds. If their masses are mA and mB respectively. Then

A
2mA=mB
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B
mA=3mB
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C
4mA=mB
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D
3mA=mB
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Solution

The correct option is B mA=3mB
Vi=kViforVB=0Vf=VfforVB=VVf(A)=(mAmBmA+mB)UA+(2mAmA+mB)Vi(B)(2mAmA+mB)Vi(B)=0Vf(A)=(mAmBmA+mB)UAVf(B)=(2mAmA+mB)UA(mAmBmA+mB)=(2mAmA+mB)mA=3mB

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