A smooth tunnel is dug along the radius of earth that ends at centre. A ball is relrased from the surface of earth along tunnel. Caefficient of restitution for collision between soil at centre and ball is `0.5`. Caculate the distance travelled by ball just second collision at center. Given mass of the earth is `M` and radius of the earth is `R`.

Answer

To solve the problem step by step, we will analyze the motion of the ball in the tunnel and the effects of the collision with the soil at the center of the Earth.

### Step 1: Initial Conditions
The ball is released from the surface of the Earth (height = R) and falls towards the center of the Earth through the tunnel. Initially, the potential energy (PE) at the surface is given by:

\[ PE_i = -\frac{GMm}{R} \]

where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the ball, and \( R \) is the radius of the Earth.

### Step 2: Energy Conservation to Find Velocity at the Center
At the center of the Earth, the potential energy is:

\[ PE_f = -\frac{3GMm}{2R} \]

Using conservation of energy, we equate the initial potential energy to the final potential energy plus kinetic energy (KE):

\[ -\frac{GMm}{R} = -\frac{3GMm}{2R} + \frac{1}{2}mv^2 \]

### Step 3: Solve for Velocity at the Center
Rearranging the equation gives:

\[ \frac{1}{2}mv^2 = -\frac{GMm}{R} + \frac{3GMm}{2R} \]

This simplifies to:

\[ \frac{1}{2}mv^2 = \frac{GMm}{2R} \]

Dividing by \( m \) and multiplying by 2:

\[ v^2 = \frac{GM}{R} \]

Taking the square root:

\[ v = \sqrt{\frac{GM}{R}} \]

### Step 4: Collision at the Center
Upon reaching the center, the ball collides with the soil. The coefficient of restitution \( e \) is given as 0.5. The velocity after the collision is:

\[ v' = e \cdot v = 0.5 \cdot \sqrt{\frac{GM}{R}} \]

### Step 5: Motion After Collision
After the collision, the ball will move back towards the surface. The initial velocity for this upward motion is \( v' \).

### Step 6: Energy Conservation for Upward Motion
Using energy conservation again, we can find the height the ball reaches after the collision. The potential energy at the maximum height \( h \) is given by:

\[ PE = -\frac{GMm}{R - h} \]

Setting the initial kinetic energy equal to the potential energy at the maximum height:

\[ \frac{1}{2}mv'^2 = -\frac{GMm}{R - h} \]

Substituting \( v' \):

\[ \frac{1}{2}m \left(0.5 \cdot \sqrt{\frac{GM}{R}}\right)^2 = -\frac{GMm}{R - h} \]

This simplifies to:

\[ \frac{1}{2}m \cdot \frac{0.25GM}{R} = -\frac{GMm}{R - h} \]

### Step 7: Solve for Maximum Height
Cancelling \( m \) and rearranging gives:

\[ \frac{0.125GM}{R} = \frac{GM}{R - h} \]

Cross-multiplying and simplifying:

\[ 0.125(R - h) = R \]

This leads to:

\[ 0.125R - 0.125h = R \]

Rearranging gives:

\[ -0.125h = R - 0.125R \]

\[ -0.125h = 0.875R \]

Thus:

\[ h = -\frac{0.875R}{-0.125} = 7R \]

### Step 8: Total Distance Travelled
The total distance travelled by the ball before the second collision is:

1. Distance to the center: \( R \)
2. Distance back up to height \( h \): \( 7R \)

Thus, the total distance travelled is:

\[ R + 7R = 8R \]

### Final Answer
The distance travelled by the ball just before the second collision at the center is:

\[ \text{Total Distance} = 2R \]

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