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Two spherical bodies of mass M and 5 M and radii R and 2 R 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)
    1.5 R
  • b)
    2.5 R
  • c)
    4.5 R
  • d)
    7.5 R
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Two spherical bodies of mass M and 5 M and radii R and 2 R are release...
Given:
Mass of smaller body, m = M
Mass of larger body, M = 5M
Radius of smaller body, r = R
Radius of larger body, R = 2R
Initial separation, d = 12R

To find: Distance covered by the smaller body before collision

Concepts used:
1. Gravitational force between two bodies of masses m1 and m2 separated by a distance d is given by:
F = G * m1 * m2 / d^2
where G is the universal gravitational constant

2. Law of conservation of energy:
Total mechanical energy (kinetic energy + potential energy) of a system is conserved in the absence of external forces.

Approach:
1. Find the gravitational force acting between the two bodies at a separation of d.
2. Calculate the initial potential energy of the system.
3. Calculate the velocity of the smaller body at the time of collision using conservation of energy.
4. Use the velocity and acceleration due to gravity to find the time taken by the smaller body to cover the distance d/2 (half the initial separation).
5. Use the time taken to find the distance covered by the smaller body before collision.

Calculation:
1. Gravitational force between the two bodies at a separation of d:

F = G * m1 * m2 / d^2
where m1 = M, m2 = 5M, d = 12R

F = G * M * 5M / (12R)^2
F = G * 5M^2 / (144R^2)
F = 5G * M^2 / (144R^2)

2. Initial potential energy of the system:

U = -G * m1 * m2 / d
where m1 = M, m2 = 5M, d = 12R

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

3. Using conservation of energy:
Initial total mechanical energy = Final total mechanical energy

Initial total mechanical energy = U
Final total mechanical energy = (1/2) * m * v^2 + (1/2) * M * V^2
where v is the velocity of the smaller body at the time of collision
V is the velocity of the larger body at the time of collision (which can be assumed to be zero as it is much larger and takes negligible time to come to rest)

U = (1/2) * m * v^2 + (1/2) * M * V^2

Substituting the values:
-5G * M^2 / (12R) = (1/2) * m * v^2
-5G * M^2 / (12R) = (1/2) * M * v^2
(As m = M)

v^2 = -10G * M / (24R)
v^2 = -G * M / (12R)

4. Time taken by the smaller body to cover the distance d/2:

The acceleration due to gravity acting on the smaller body is:
a = F / m
where F is the gravitational force
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Two spherical bodies of mass M and 5 M and radii R and 2 R are release...
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Two spherical bodies of mass M and 5 M and radii R and 2 R are released in free space with initialseparation between their centres equal to 12 R. If they attract each other due to gravitational forceonly, then the distance covered by the smaller body before collision is :a)1.5 Rb)2.5 Rc)4.5 Rd)7.5 RCorrect answer is option 'D'. Can you explain this answer?
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