A tunnel is dug in the earth across one of its diameter. Two masses `m` and `2m` are dropped from the two ends of the tunel. The masses collide and stick each other. They perform `SHM`, the ampulitude of which is `(R =` radius of earth)

Options

A. `R`
B. `R//2`
C. `R//3`
D. `2R//3`

Explanation

To solve the problem step by step, we will analyze the situation where two masses, \( m \) and \( 2m \), are dropped from the ends of a tunnel dug through the Earth. They collide and stick together, performing simple harmonic motion (SHM). The amplitude of this motion is given as \( R \), where \( R \) is the radius of the Earth.

### Step 1: Understand the System
- We have two masses: \( m \) and \( 2m \).
- They are dropped from opposite ends of a tunnel that goes through the diameter of the Earth.
- When they collide, they stick together and perform SHM.

**Hint:** Visualize the Earth and the tunnel. Consider how gravitational force acts on the masses as they fall.

### Step 2: Conservation of Energy
- For mass \( m \):
- Initial potential energy, \( E_{i1} = -\frac{GmM}{R} \) (where \( M \) is the mass of the Earth).
- Final potential energy at the center (where kinetic energy is maximum), \( E_{f1} = 0 \).

- For mass \( 2m \):
- Initial potential energy, \( E_{i2} = -\frac{G(2m)M}{R} \).
- Final potential energy at the center, \( E_{f2} = 0 \).

**Hint:** Use the gravitational potential energy formula and remember that potential energy is negative when considering gravitational attraction.

### Step 3: Calculate Velocities at the Center
- Using conservation of energy for mass \( m \):
\[ E_{i1} = E_{f1} \implies -\frac{GmM}{R} = 0 + \frac{1}{2}mv_1^2 \]
Rearranging gives:
\[ \frac{1}{2}mv_1^2 = \frac{GmM}{R} \implies v_1^2 = \frac{2GM}{R} \]

- For mass \( 2m \):
\[ E_{i2} = E_{f2} \implies -\frac{G(2m)M}{R} = 0 + \frac{1}{2}(2m)v_2^2 \]
Rearranging gives:
\[ \frac{1}{2}(2m)v_2^2 = \frac{G(2m)M}{R} \implies v_2^2 = \frac{2GM}{R} \]

**Hint:** Both masses will have the same speed when they meet at the center due to symmetry.

### Step 4: Conservation of Momentum
- Before the collision, the momentum is:
\[ p_{initial} = mv_1 - 2mv_2 \]
- After the collision, the combined mass moves with velocity \( v' \):
\[ p_{final} = (m + 2m)v' = 3mv' \]
- Setting initial momentum equal to final momentum:
\[ mv_1 - 2mv_2 = 3mv' \]
Since \( v_1 = v_2 \):
\[ mv_1 - 2mv_1 = 3mv' \implies -mv_1 = 3mv' \implies v' = -\frac{v_1}{3} \]

**Hint:** Remember that momentum is conserved in collisions.

### Step 5: Energy Conservation After Collision
- The total energy after the collision is the sum of kinetic and potential energy:
\[ E_{initial} = -\frac{3GM}{R} + \frac{1}{2}(3m)v'^2 \]
Substitute \( v' \):
\[ E_{final} = -\frac{3GM}{R} + \frac{1}{2}(3m)\left(-\frac{v_1}{3}\right)^2 \]

**Hint:** Use the kinetic energy formula and substitute the expression for \( v' \).

### Step 6: Determine Amplitude of SHM
- The amplitude of the SHM is determined from the maximum displacement from the equilibrium position, which is \( R/3 \).

**Final Answer:** The amplitude of the SHM performed by the two masses after colliding is \( \frac{R}{3} \).