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 BmA=3mB →Vi=k→ViforVB=0→Vf=→VfforVB=V→Vf(A)=(mA−mBmA+mB)UA+(2mAmA+mB)→Vi(B)∵(2mAmA+mB)→Vi(B)=0⇒→Vf(A)=(mA−mBmA+mB)UA→Vf(B)=(2mAmA+mB)UA(mA−mBmA+mB)=(2mAmA+mB)⇒mA=3mB