Statement I: If a sphere of mass `m`moving with speed `u` undergoes a perfectly elastic head-on collision with another sphere of heavier mass `M` at rest (`M gt m`), then direction of velocity of sphere of mass `m` is reversed due to collision (no external force acts on system of two spheres).
Statement II: During a collision of spheres of unequal masses, the heavier mass exerts more force on the lighter mass in comparison to the force which lighter mass exerts on the heavier one,

Options

A. Both assertion and reason are true and reason is the correct explanation of assertion.
B. Both assertion and reason are true but reason is not the correct explanation of assertion.
C. Assertion is true and reason is false.
D. Assertion is false and reason is true.

Explanation

To solve the problem, we need to analyze both statements given in the question regarding the elastic collision between two spheres of unequal masses.

### Step-by-Step Solution:

**Step 1: Analyze Statement I**
- We have a sphere of mass `m` moving with speed `u` and colliding with another sphere of mass `M` (where `M > m`) that is at rest.
- In a perfectly elastic collision, both momentum and kinetic energy are conserved.
- After the collision, the direction of the velocity of the sphere of mass `m` is reversed. We need to confirm if this is true.

**Step 2: Apply Conservation of Momentum**
- The initial momentum of the system is:
\[ P_{\text{initial}} = mu + 0 = mu \]
- Let the final velocities after the collision be `v1` for mass `m` and `v2` for mass `M`.
- The final momentum of the system is:
\[ P_{\text{final}} = mv1 + Mv2 \]
- By conservation of momentum:
\[ mu = mv1 + Mv2 \quad \text{(1)} \]

**Step 3: Apply Conservation of Kinetic Energy**
- Since the collision is elastic, kinetic energy is also conserved:
\[ \frac{1}{2}mu^2 = \frac{1}{2}mv1^2 + \frac{1}{2}Mv2^2 \]
- Simplifying this gives:
\[ mu^2 = mv1^2 + Mv2^2 \quad \text{(2)} \]

**Step 4: Analyze the Resulting Velocities**
- From the equations (1) and (2), we can derive the velocities `v1` and `v2`.
- For a perfectly elastic collision between two masses where one is initially at rest, the velocities after the collision can be derived as:
\[ v1 = \frac{(m - M)u}{m + M} \quad \text{and} \quad v2 = \frac{(2m)u}{m + M} \]
- Since `M > m`, the term `(m - M)` is negative, which implies that `v1` will indeed be negative, indicating that the direction of the velocity of mass `m` is reversed.

**Conclusion for Statement I:**
- Statement I is **True**.

**Step 5: Analyze Statement II**
- Statement II claims that during the collision, the heavier mass `M` exerts more force on the lighter mass `m` compared to the force that `m` exerts on `M`.
- According to Newton's Third Law, the forces exerted by two colliding bodies on each other are equal in magnitude and opposite in direction.
- Therefore, the statement that the heavier mass exerts more force is **False**.

### Final Conclusion:
- Statement I is **True** and Statement II is **False**.