MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Velocity of COM

Two masses m1...

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.

A

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} (v_{1} - v_{2})^{2}$
$K = \frac{1}{2} m_{1} v_{1 c}^{2} + \frac{1}{2} m_{2} v_{2 c}^{2}$ …… (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).

Suggest Corrections

thumbs-up

0

Similar questions

m_{1}

m_{2}

are moving with velocities

v_{1}

v_{2}

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

m_{1}

m_{2}

are moving with velocities

v_{1}

v_{2}

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

Q. Find the total kinetic energy

(T)

of the two particles in the reference frame fixed to their centre of inertia having masses

m_{1}

m_{2}

_{1}

_{2}

(perpendicular to each other) respectively. Here

μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}

Q. Find the total kinetic energy

(T)

of the two particles in the reference frame fixed to their centre of inertia having masses

m_{1}

m_{2}

_{1}

_{2}

(perpendicular to each other) respectively. Here

μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}

Q. A particle of mass

m_{1}

is moving with a velocity

v_{1}

and another particle of mass

m_{2}

is moving with a velocity

v_{2}

. Both of them the same momentum but their different kinetic energies are

E_{1}

E_{2}

respectively. If

m_{1} > m_{2}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Centre of Mass in Galileo's World

PHYSICS

Watch in App

explore_more_icon

Explore more

Velocity of COM

Standard XIII Physics

Join BYJU'S Learning Program

CrossIcon