Question #28405

Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is..

a) 1.5R b)2.5R c)4.5R d)7.5R

a) 1.5R b)2.5R c)4.5R d)7.5R

Expert's answer

Due to the 3rd Newton's Law the force of interaction willbe the same for both bodies, say, F.

Thus, the acceleration of the smaller body will be a = F/Mand of the bigger, respectively, A = F/5M.

Until the collision, the smaller body will travel s =a*t^2/2 and the bigger one S = A*t^2/2.

The criterion of the collision is s + S + R + 2R = 12R.

Thus, from this system we'relooking for s.

We can find a proportion between s and S.

S = s*A/a = s*(F/5M)/(F/M) =s/5.

s*6/5 = (12-1-2)*R = 9R;

s = R*9*5/6 = R*15/2 = 7.5R.

Answer: d, 7.5*R.

Thus, the acceleration of the smaller body will be a = F/Mand of the bigger, respectively, A = F/5M.

Until the collision, the smaller body will travel s =a*t^2/2 and the bigger one S = A*t^2/2.

The criterion of the collision is s + S + R + 2R = 12R.

Thus, from this system we'relooking for s.

We can find a proportion between s and S.

S = s*A/a = s*(F/5M)/(F/M) =s/5.

s*6/5 = (12-1-2)*R = 9R;

s = R*9*5/6 = R*15/2 = 7.5R.

Answer: d, 7.5*R.

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