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Question

A solid spherical planet of mass 2m and radius 'R' has a very small tunnel along its diameter. A small cosmic particle of mass m is at a distance 2R from the centre of the planet as shown. Both are initially at rest, and due to gravitational attraction, both start moving toward each other. After some time, the cosmic particle passes through the centre of the planet. (Assume the planet and the cosmic particle are isolated from other planets)

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Solution

The correct options are

**A** Displacement of the cosmic particle till that instant is 4R3

**B** Acceleration of the cosmic particle at that instant is zero

**C** velocity of the cosmic particle at that instant is √8Gm3R

Applying momentum conservation,

0=mv1−2mv2

⇒v2=v12 -(i)

From energy conservation,

ki+Ui=kf+Uf

0+(−G(2m)2R)m=12mv21+12(2m)v22+(−32G(2m)R)(m) --(ii)

Solving eqn. (i) & (ii) get,

v1=√8Gm3R

(A) COM will be fixed so,

Scm=m1s1+m2s2m1+m20=(m)(x)+(2m)(−(2R−x))m+2m⇒x=4R3

(B) Fnet =0⇒a=0

(D) Wgr=−ΔU↓⇒Wgr=(−G(2m)2R)m−(−32G(2m)R)m

Applying momentum conservation,

0=mv1−2mv2

⇒v2=v12 -(i)

From energy conservation,

ki+Ui=kf+Uf

0+(−G(2m)2R)m=12mv21+12(2m)v22+(−32G(2m)R)(m) --(ii)

Solving eqn. (i) & (ii) get,

v1=√8Gm3R

(A) COM will be fixed so,

Scm=m1s1+m2s2m1+m20=(m)(x)+(2m)(−(2R−x))m+2m⇒x=4R3

(B) Fnet =0⇒a=0

(D) Wgr=−ΔU↓⇒Wgr=(−G(2m)2R)m−(−32G(2m)R)m

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