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Question

Two spherical bodies, having a mass of M and 5M and radius of 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

A
2.5R
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B
4.5R
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C
7.5R
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D
1.5R
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Solution

The correct option is C 7.5R
From Newton's 3rd Law, the force of attraction will be the same on both bodies, say, F.
Thus, the acceleration of the smaller body will be a=FM and A=F5Mof the bigger, respectively.
Until the collision, the smaller body travels s=12at2 and the bigger one S=12At2
The criterion of the collision is
s+S+R+2R=12R

By taking the ratio of S and s, we get,
Ss=F5MFM=15
S=s5
By substituting S=s5 into s+S+R+2R=12R, we get,
s+s5+R+2R=12R
65s=(1212)R=9R
s=9R×56=152R=7.5R

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