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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 distance covered by the smaller body just before collision is
  • a)
    2.5R
  • b)
    4.5R
  • c)
    1.5R
  • d)
    7.5R
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Two spherical bodies of mass M and 5M and radii R and 2R respectively...
Given:
- Mass of smaller body = M
- Mass of larger body = 5M
- Radius of smaller body = R
- Radius of larger body = 2R
- Initial separation between their centers = 12R

To find:
The distance covered by the smaller body just before collision.

Concept:
When two bodies attract each other due to gravitational force, they accelerate towards each other. The acceleration of each body can be found using Newton's law of gravitation:

F = G * (m1 * m2) / r^2

Where:
- F is the gravitational force
- G is the universal gravitational constant
- m1 and m2 are the masses of the bodies
- r is the distance between their centers

Since the force is the same for both bodies, we can equate the accelerations:

a1 = a2

The distance covered by each body can be found using the equation of motion:

s = ut + (1/2)at^2

Where:
- s is the distance covered
- u is the initial velocity (which is zero in this case)
- a is the acceleration
- t is the time taken

Calculation:
1. Initial separation between the centers = 12R
2. Distance covered by the smaller body just before collision = s

Acceleration of the bodies:
- Acceleration of the smaller body (a1) = G * (M * 5M) / (12R)^2
- Acceleration of the larger body (a2) = G * (5M * M) / (12R)^2

Since a1 = a2, we can equate the two expressions:

G * (M * 5M) / (12R)^2 = G * (5M * M) / (12R)^2

Simplifying, we get:

5M^2 = 5M^2

This shows that the accelerations of both bodies are the same.

Distance covered by the smaller body:
Using the equation of motion, we can find the distance covered by the smaller body (s):

s = ut + (1/2)at^2

Since the initial velocity (u) is zero, the equation simplifies to:

s = (1/2)at^2

We need to find the time taken (t) for the smaller body to reach the larger body.

Time taken:
The initial separation between the centers is 12R. The distance covered by the smaller body just before collision is s. Therefore, the distance traveled by the smaller body is (12R - s).

Using the equation of motion, we can find the time taken (t):

12R - s = (1/2)at^2

Simplifying, we get:

t^2 = (2(12R - s)) / a

Since a1 = a2, we can substitute the value of a:

t^2 = (2(12R - s)) / (G * (M * 5M) / (12R)^2)

Simplifying further:

t^2 = (2(12R - s) * (12R)^2) / (G * (M * 5M))

t^2 = (2
Free Test
Community Answer
Two spherical bodies of mass M and 5M and radii R and 2R respectively...
Let the spheres collide after time t, when the smaller sphere covered distance x1 and bigger sphere covered distance x2.
The gravitational force acting between two spheres depends on the
distance which is variable quantity.
The gravitational force,
Acceleration of smaller body,
Acceleration of bigger body,
From equation of motion,
We know that x1 + x2 = 9R
Therefore, the two spheres collide when the smaller sphere covers a distance of 7.5R.
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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 distance covered by the smaller body just before collision isa)2.5Rb)4.5Rc)1.5Rd)7.5RCorrect answer is option 'D'. Can you explain this answer?
Question Description
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