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.
Two spherical bodies of mass M and 5M and radii R and 2R respectively ...