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 AT=12μ(|→v1|2+|→v2|2) Velocity of center of inertia frame =m1→v1+m2→v2m1+m2 IN COM reference frame:
Velocity of m1 relative to this frame =→vr1=→v1−m1→v1+m2→v2m1+m2=m2(→v1−→v2)m1+m2 Similarly, velocity of m2 relative to this frame =→vr2=→v2−m1→v1+m2→v2m1+m2=m1(→v2−→v1)m1+m2 Hence the kinetic energy with respect to this frame T=12m1v2r1+12m2v2r2 T=12m1(m2(→v1−→v2)m1+m2)2+12m2(m1(→v2−→v1)m1+m2)2 T=12μ|→v2−→v1|2
As given velocities are perpendicular to each other, |→v2−→v1|2=(|→v1|2+|→v2|2)