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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 the distance covered by the smaller body just before collision is                    [AIEEE 2003]
  • a)
    2.5 R
  • b)
    4.5 R
  • c)
    7.5 R
  • d)
    1.5 R
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Two spherical bodies of mass M and 5M and radii R and 2R respectively ...
Given:
- Two spherical bodies of mass M and 5M and radii R and 2R respectively.
- Initial separation between their centres equal to 12R.
- They attract each other due to gravitational force only.

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

Solution:
1. Let the smaller body of mass M be A and the larger body of mass 5M be B.
2. At any instant, let the distance between their centres be d.
3. By gravitational force, A experiences a force of F = GM(5M)/(d^2), where G is the gravitational constant.
4. By Newton's second law of motion, the acceleration of A is a = F/M = GM(5M)/(Md^2) = G(5M)/(d^2).
5. Let v be the velocity of A at any instant. Then, by kinematics, v^2 = 2ad = 2G(5M)/(d^2) * 12R.
6. Let d1 be the distance covered by A just before collision. Then, at that instant, d = 3R (since the sum of their radii is 3R).
7. Substituting d = 3R in the above equation, we get v^2 = 2G(5M)/(9R^2) * 12R = 80GM/R.
8. At collision, the relative velocity of A and B is v1 = 2v = 2sqrt(80GM/R).
9. Let t be the time taken by A to cover the distance d1. Then, d1 = v1t.
10. To find t, we need to find the distance travelled by B in the same time.
11. The acceleration of B is b = G(5M)/(4R)^2 = 5G(5M)/(16R^2).
12. Let u be the initial velocity of B. Then, by kinematics, d = ut + (1/2)bt^2.
13. At t = 0, d = 12R and u = 0. Substituting, we get 12R = (1/2)(5G(5M)/(16R^2))t^2, or t^2 = 384GM/(5R^3).
14. Substituting t^2 in d = ut + (1/2)bt^2, we get d = (1/2)bt^2 = 15GM/R.
15. Therefore, the distance covered by A just before collision is d1 = v1t = 2sqrt(80GM/R) * sqrt(384GM/(5R^3)) = 48R/5 = 9.6R.

Therefore, the correct option is (c) 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 the distance covered by the smaller body just before collision is [AIEEE 2003]a)2.5 Rb)4.5 Rc)7.5 Rd)1.5 RCorrect answer is option 'C'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 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 [AIEEE 2003]a)2.5 Rb)4.5 Rc)7.5 Rd)1.5 RCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 [AIEEE 2003]a)2.5 Rb)4.5 Rc)7.5 Rd)1.5 RCorrect answer is option 'C'. Can you explain this answer?.
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