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}

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Acceleration of

m_{2}

is more than that of

m_{1}

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

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Total energy of system does not remain constant

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Solution

The correct option is A Acceleration of $m_{1}$ is more than that of $m_{2}$
$A n s w e r :$ A
$a_{1}$ acceleration of $m_{1}$
$a_{2}$ acceleration of $m_{2}$
Since on both particles same force of gravitation is acting so using ( $F = m a$ ) $m_{1} < m_{1}$
Hence, $a_{1} > a_{2}$

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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 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-

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:

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Two masses m1 and m2(m1<m2) are released from rest from a finite distance. They start under their mutual gravitational attraction.

The correct option is A Acceleration of m1 is more than that of m2Answer: Aa1 acceleration of m1a2 acceleration of m2Since on both particles same force of gravitation is acting so using (F=ma ) m1<m1 Hence, a1>a2

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.

The correct option is A Acceleration of $m_{1}$ is more than that of $m_{2}$
$A n s w e r :$ A
$a_{1}$ acceleration of $m_{1}$
$a_{2}$ acceleration of $m_{2}$
Since on both particles same force of gravitation is acting so using ( $F = m a$ ) $m_{1} < m_{1}$
Hence, $a_{1} > a_{2}$