Two masses m1 and m2 are moving with velocities v1 and v2 respectively. Find the

Question

Two masses

m_{1}

and

m_{2}

are moving with velocities

v_{1}

and

v_{2}

respectively. Find their total kinetic energy in the reference frame of centre of mass.

Options:

A . m1m2m1+m2(v1−v2)2

B . 12m1m2m1+m2(v1−v2)2

C . 12m1m2m1+m2(v1+v2)2

D . 12(m1m2)(v1−v2)2

Answer: Option B
:
B

K = \frac{1}{2} m_{1} v_{1 c}^{2} + \frac{1}{2} m_{2} v_{2 c}^{2}

…… (i)
Where

v_{1 c}

and

v_{2 c}

are velocities relative to the CM.

v_{1 c} = v_{1} - v_{C M} = v_{1} - (\frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}) = (\frac{m_{2} (v_{1} - v_{2})}{m_{1} + m_{2}})

v_{2 c} = v_{2} - v_{C M} = v_{2} - (\frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}) = (\frac{m_{1} (v_{2} - v_{1})}{m_{1} + m_{2}})

Putting these in Equation (i),

K = \frac{1}{2} \frac{m_{1} m_{2}}{(m_{1} + m_{2})} (v_{1} - v_{2})^{2}

Note: For a system of two particles of masses

m_{1}

and

m_{2}

, the total kinetic energy is

k = (1 / 2) μ v^{2} + (1 / 2) (m_{1} + m_{2}) v_{C M}^{2}

where

μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}

(known as reduced mass of the system and

\to v = {\to v}_{1} - {\to v}_{2}

(relative velocity).

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