Two masses $m_{1}$ and $m_{2}$ $(m_{1} < m_{2})$ are released from rest from a finite distance. They start under their mutual gravitational attraction-

Acceleration of

m_{1}

is more than that of

m_{2}

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

Acceleration of

m_{2}

is more than that of

m_{1}

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

Centre of mass of system will remain at rest in all the reference frame

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

Total energy of system remains constant

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

Open in App

Solution

The correct options are
B Acceleration of $m_{1}$ is more than that of $m_{2}$ .
D Total energy of system remains constant

Mutual force of attraction is same on two bodies, but acceleration $a = \frac{F}{m}$

Therefore, $a \propto \frac{1}{m}$

As $m_{1} < m_{2}$ , $\Rightarrow a_{2} < a_{1}$

Here, Potential energy is getting converted to kinetic energy.

Therefore, no energy loss.

Suggest Corrections

Similar questions

Two masses $m_{1}$ and $m_{2}$ ( $m_{1}$ $<$ $m_{2}$ ) are released from rest from a finite distance. They start under their mutual gravitational attraction-

Q. Two masses

m_{1}

and

m_{2} (m_{1} < m_{2})

are released from rest from a finite distance. They start to move under their mutual gravitational attraction. Identity the incorrect statement.

Q. Two masses

m_{1}

and

m_{2}

(

m_{1}

m_{2}

) are released from rest from a finite distance between them. They start moving under the mutual gravitational attraction.

Which of the given options is correct?

Two spheres of masses m1 and m2 (m1>m2) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is: