Question

a particle of mass m is placed at a point on the line joining two heavy particle of masses m1 and m2 so that the distance from m1 is x1 and from.m2 is x2 if the particle of mass m experiences equal gravitational forces due to m1 and m2 then x1/x2 is​

Answer 1

Answer:

The forces of attraction between the particles are the internal forces. Therefore, the center of mass of the system will have no acceleration. The particles move, but the center of mass will continue to be at the same place. At the time of collision, the two particles are at one place and the center of mass will also be at that place. As the center of mass does not move, the collision will take place at the center of mass. 

Distance of center of mass from M1=M1+M2M1(0)+M2(R)=M1+M2M2R

Distance of center of mass form M2=R−M1+M2M2R=M1+M2M1R

Ratio of the distances=M1M2

I hope I am correct........

Answer 2

Answer:

Answer is root of m1/root of m2

Explanation:

let F1 be the force exerted by m1 and F2 the force exerted by m2

given that gravitational force is equal

hence, F1=F2

F= GMm/r^2

Since both are applying force on the same object, value of M can be neglected. And since G is constant it can also be neglected

Therefore, F1= m1/x1^2    and     F2=m2/x2^2

Since F1=F2

m1/x1^2=m2/x2^2

by cross multiplication,

m1/m2 = x1^2/x2^2

therefore x1/x2 = root of m1/ root of m2

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