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Question

Find the total kinetic energy (T) of the two particles in the reference frame fixed to their centre of inertia having masses m1 and m2 and velocities v1 and v2 (perpendicular to each other) respectively. Here μ=m1m2m1+m2

A
T=12μ(|v1|2+|v2|2)
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B
T=12μ(|v1|2+2|v2|2)
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C
T=13μ(|v1|2+|v2|2)
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D
T=12μ(2|v1|2+|v2|2)
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Solution

The correct option is A T=12μ(|v1|2+|v2|2)
Velocity of center of inertia frame =m1v1+m2v2m1+m2
IN COM reference frame:

Velocity of m1 relative to this frame =vr1=v1m1v1+m2v2m1+m2=m2(v1v2)m1+m2
Similarly, velocity of m2 relative to this frame =vr2=v2m1v1+m2v2m1+m2=m1(v2v1)m1+m2
Hence the kinetic energy with respect to this frame T=12m1v2r1+12m2v2r2
T=12m1(m2(v1v2)m1+m2)2+12m2(m1(v2v1)m1+m2)2
T=12μ|v2v1|2

As given velocities are perpendicular to each other,
|v2v1|2=(|v1|2+|v2|2)

T=12μ(|v1|2+|v2|2)

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