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NEET PHYSICSMedium

Two spherical bodies of masses M and 5M and radii R and 2R 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 before collision is:

A

2.5 R

B

4.5 R

C

7.5 R

D

1.5 R

Step-by-Step Solution

Since the system consists of two bodies attracting each other under mutual gravitational force, there is no external force acting on the system (Fext=0F_{ext} = 0). Therefore, the position of the centre of mass remains stationary.

The initial distance between the centres is 12R12R. Collision occurs when the bodies touch, meaning the distance between their centres becomes the sum of their radii: R+2R=3RR + 2R = 3R. The total distance covered by both bodies is dtotal=Initial SeparationFinal Separation=12R3R=9Rd_{total} = \text{Initial Separation} - \text{Final Separation} = 12R - 3R = 9R.

Since the centre of mass is stationary, the distance moved by each body is inversely proportional to its mass (m1d1=m2d2m_1 d_1 = m_2 d_2). Let d1d_1 be the distance moved by the smaller mass MM and d2d_2 be the distance moved by the larger mass 5M5M. d1=m2m1+m2×dtotald_1 = \frac{m_2}{m_1 + m_2} \times d_{total} Substituting the values: d1=5MM+5M×9R=56×9R=7.5Rd_1 = \frac{5M}{M + 5M} \times 9R = \frac{5}{6} \times 9R = 7.5 R.

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