```markdown Q5. Single Correct A massive planet of radius $R$ has a diametrical hole as shown in the figure, such that the hole does not affect the sphericity of the planet. The planet has uniform density. Two equal masses $A$ and $B$ $(m \ll M)$ are simultaneously released from the positions shown. If the collisions are elastic, then the total distance traveled by mass $A$ after being released at the time of 6th collision is: \([ M \rightarrow \text{mass of planet}, \, m \rightarrow \text{mass of } A \text{ and } B \text{ neglect the gravitational interaction between mass } A \text{ and mass } B.]\) Options: (1) $8.5 \, R$ (2) $5 \, R$ (3) $6.5 \, R$ (4) $9.5 \, R$ ```

1. Gravity inside a uniform sphere

For a planet of radius $R$ and uniform density $\rho$, the mass enclosed within a radius $r$ is
$M(r)=\frac{4}{3}\pi r^{3}\rho$
The gravitational field magnitude at distance $r$ is
$g(r)=\frac{G M(r)}{r^{2}}=\frac{4\pi G\rho}{3}\,r$
Hence the acceleration is directly proportional to $r$. That is the hallmark of SHM with angular frequency
$\omega = \sqrt{\frac{4\pi G\rho}{3}} = \sqrt{\frac{G M}{R^{3}}}$

2. Setting up the SHM of each mass

Take the centre of the planet as the origin. Choose the right direction as positive.

• Initial position of $A$ : $x_{A0}=a\,(=0.5R)$, with zero initial velocity.
• Initial position of $B$ : $x_{B0}=-R$, also with zero initial velocity.

Therefore $x_A(t)=a\cos(\omega t),\qquad x_B(t)=-R\cos(\omega t)$ Both reach the centre whenever $\cos(\omega t)=0$, i.e. at
$t_k=\frac{(2k-1)\pi}{2\,\omega}\;\;(k=1,2,3,\dots)$ Each of these instants is called a collision (they occupy the same point).

3. What happens in an elastic head-on collision of identical masses?

For identical masses an elastic head-on collision simply exchanges their velocities. A useful visualisation: they ‘pass through’ each other if you do not label them; but if you keep the labels, each marble behaves as if it is reflected at the centre.

4. Distances travelled between successive collisions

• Before 1ˢᵗ collision: $A$ moves from $x=+a$ to $x=0$ → distance $=a$.

After the 1ˢᵗ collision, $A$ picks up the speed that $B$ had. That speed corresponds to amplitude $R$. Hence the subsequent travel segments alternate:

5. Total distance up to the 6ᵗʰ collision

Add the six portions (one before the 1ˢᵗ collision and five between collisions): $\text{Total}=a+3(2R)+2(2a)=5a+6R$ The diagram in the paper shows $a=\tfrac{R}{2}$. Plugging in: $\text{Total}=5\left(\frac{R}{2}\right)+6R=\frac{5R}{2}+6R=8.5R$ Therefore option (1) 8.5 R is correct.

What is happening?

Imagine you have a very big orange (the planet) and a straw (the tunnel) that goes straight through the centre. Now put two tiny marbles in the straw:

• Marble A is placed half-way between the centre and the right end of the straw (so it is only halfway out).
• Marble B is placed at the far left end of the straw (just touching the peel).

Because the orange is perfectly round and has the same thickness everywhere, the pull of gravity inside works like invisible rubber bands: the farther a marble is from the centre, the stronger the band pulls it back. That kind of pull makes the marbles move to-and-fro in Simple Harmonic Motion (SHM), just like a swing.

Both marbles start from rest and are let go at the same time. Every time they meet at the centre they bump elastically (they swap speeds just like two identical balls on a pool table).

The question simply asks: ‘How far has Marble A travelled altogether by the moment the 6ᵗʰ bump happens at the centre?’

After adding up each little trip, the total comes out to 8.5 times the planet’s radius (8.5 R).

Step-by-step Calculation

1. Gravitational SHM parameters
   $\omega = \sqrt{\frac{G M}{R^{3}}},\qquad T=\frac{2\pi}{\omega}$

2. Initial amplitudes
   $A_0 = a\;(=0.5R),\qquad B_0 = R$ $x_A(t)=a\cos(\omega t),\quad x_B(t)=-R\cos(\omega t)$

3. Time of successive collisions (centre encounters): $t_k=\frac{(2k-1)T}{4}\quad(k=1,2,3,\dots)$

4. Distances travelled by $A$

   • Segment 0 (start → 1ˢᵗ collision): $d_0 = a$

   • After every collision the amplitudes swap. Therefore the sequence of path lengths is $d_1=2R,\;d_2=2a,\;d_3=2R,\;d_4=2a,\;d_5=2R$

5. Total until 6ᵗʰ collision $D_{\text{total}} = d_0+d_1+d_2+d_3+d_4+d_5$ $\boxed{D_{\text{total}} = a + 3(2R) + 2(2a) = 5a + 6R}$ Taking $a = \dfrac{R}{2}$ (as indicated in the figure): $D_{\text{total}} = 5\left(\frac{R}{2}\right) + 6R = 8.5\,R$

Hence the correct choice is (1) 8.5 R.