Question

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

• A. 7.5 R
• B. 1.5 R
• C. 2.5 R
• D. 4.5 R

Hint:

This question is about attraction due to gravitational force. Hence, we apply here the universal law of gravitation to get the answer.

Answer 1

The Correct Option is A

Hint: As per the hint, this question is about the attraction due to gravitational force. Hence, we will apply here the universal law of gravitation to get the answer.

Let's draw the observation as per the question asked:

Fig 1: distance covered by the smaller body before collision

From the figure mentioned above and the details mentioned in the question, we can see:

Distance between the centres of spheres =12R

At the time of the collision the distance between their centres = 3R

So the total distance travelled by both bodies = 12R - (2R + R) = 9R

Since the bodies move under mutual forces, the centre of mass will remain stationary so 

m₁x₁ = m₂x₂
mx = 5 m(9R - x)
x = 45R - 5x
6x = 45 R
x = \(\frac {45}{6}\) R
x = 7.5 R

So, the correct answer is 7.5R would be the distance covered by the smaller body before collision.

Answer 2

Complete step-by-step solution:

• The radius of the first body is R and its mass is M
• The radius of the second body is 2R and its mass is 5m

Though the bodies are spherical so in order to find the distance between the two bodies we need to measure it from the centre of one body to the centre of another. But these are not point objects and hence in order to apply our universal law of gravitation we need to have a separation between the two surfaces.
Distance between their surfaces =12R−R−2R=9R
If x₁ and x₂ are the distance covered by the two bodies, then x₁+ x₂ = 9R
Suppose they both meet at point O, If the smaller sphere moves x distance to reach O, then the bigger sphere will move a distance of (9R−x)
Using the universal law of gravitation,
\(F = \frac{Gm_1m_2}{r_2}\)
And F = ma
So, \(a_{small}=\frac{F}{m}=\frac{G×5m}{(12R−x)^2}\)     -------(1)
\(a_{big}=\frac{F}{5m}=\frac{G×m}{(12R−x)^2}\)             -------(2)
Since initially, both the bodies were at rest, using the second equation of motion,
\(x=\frac{at^2}{2}\)
\(x=\frac{a_{small}t^2}{2}=\frac{5×Gm×t^2}{2(12R−x)^2}\)          ------(3)
\(9R−x=\frac{a_{big}t^2}2=\frac{Gm×t^2}{2(12R−x)^2}\)   ------(4)
Dividing eq (3) by eq (4) we get
x = 45R−5x
6x = 45R
x = 7.5R

So, the correct Option is (A): 7.5 R.

Note: The distance covered by the smaller body just before the collision is 7.5R.