Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Body "P" having mass M moving with speed 'u" has head-on collision elastically with another body 'Q' having mass 'm' initially at rest. If `Mgtgtm`, body Q will have a maximum speed equal to "2u' after collision
Reason R : During elastic collision, the momentum and kinetic energy are both conserved.
In the light of the above statements, choose the most appropriate answer from the options given below:

Options

A. A is not correct but R is correct.
B. Both A and R are correct but R is NOT the correct explanation of A.
C. Both A and R are correct and R is the correct explanation of A
D. A is correct but R is not correct.

Answer

To solve the problem, we will analyze the assertion and reason step by step.

### Step 1: Understand the Assertion
The assertion states that body "P" (mass M) moving with speed 'u' has a head-on elastic collision with another body "Q" (mass m) initially at rest. If \( M \gg m \), body Q will have a maximum speed equal to \( 2u \) after the collision.

### Step 2: Analyze the Collision
In an elastic collision, both momentum and kinetic energy are conserved. We can use the conservation of momentum and the conservation of kinetic energy to analyze the situation.

### Step 3: Conservation of Momentum
Before the collision:
- Momentum of body P = \( Mu \)
- Momentum of body Q = \( 0 \) (since it is at rest)

Total momentum before collision:
\[ P_{\text{initial}} = Mu + 0 = Mu \]

After the collision, let the velocities of bodies P and Q be \( V_1 \) and \( V_2 \) respectively. Thus, the momentum after collision is:
\[ P_{\text{final}} = MV_1 + mV_2 \]

By conservation of momentum:
\[ Mu = MV_1 + mV_2 \quad \text{(1)} \]

### Step 4: Conservation of Kinetic Energy
Before the collision, the kinetic energy is:
\[ KE_{\text{initial}} = \frac{1}{2}Mu^2 + 0 = \frac{1}{2}Mu^2 \]

After the collision, the kinetic energy is:
\[ KE_{\text{final}} = \frac{1}{2}MV_1^2 + \frac{1}{2}mV_2^2 \]

By conservation of kinetic energy:
\[ \frac{1}{2}Mu^2 = \frac{1}{2}MV_1^2 + \frac{1}{2}mV_2^2 \quad \text{(2)} \]

### Step 5: Solve the Equations
From equation (1):
\[ Mu = MV_1 + mV_2 \implies V_1 = \frac{Mu - mV_2}{M} \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ \frac{1}{2}Mu^2 = \frac{1}{2}M\left(\frac{Mu - mV_2}{M}\right)^2 + \frac{1}{2}mV_2^2 \]

### Step 6: Simplify and Solve for \( V_2 \)
After simplification, we can find \( V_2 \) in terms of \( u \) and the masses. Given that \( M \gg m \), we can neglect \( m \) in comparison to \( M \).

After solving, we find:
\[ V_2 = 2u \]

### Conclusion
Thus, the assertion is correct: body Q will indeed have a maximum speed of \( 2u \) after the collision.

### Step 7: Analyze the Reason
The reason states that during an elastic collision, both momentum and kinetic energy are conserved. This is true and is the basis for our calculations.

### Final Answer
Both the assertion (A) and reason (R) are correct, and R correctly explains A.