A heavy nucleous of masss M kept at rest in laboratory rejects a light particle of mass m .

Options

A. Linear momentum of particle is more than linear momentum of remaining nucleus
B. Velocity of particle is more than velocity of nucleus .
C. Kinetic energy of the partcle is more than kinetic energy of the nucleus .
D. Kinetic energy of the nucleus is more than that of the particle .

Explanation

To solve the problem, we will analyze the situation using the principles of conservation of momentum and kinetic energy.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a heavy nucleus of mass \( M \) at rest.
- It ejects a light particle of mass \( m \).
- After the ejection, the nucleus will have a reduced mass of \( M - m \).

2. **Applying Conservation of Momentum**:
- Initially, the total momentum of the system is zero (since the nucleus is at rest).
- After the ejection, let the velocity of the ejected particle be \( v_1 \) and the velocity of the remaining nucleus be \( v_2 \).
- According to the conservation of momentum:
\[ 0 = mv_1 + (M - m)v_2 \]
- Rearranging gives:
\[ mv_1 = -(M - m)v_2 \]
- This implies:
\[ mv_1 = (m - M)v_2 \]

3. **Finding the Relationship Between Velocities**:
- From the momentum equation, we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = -\frac{(M - m)}{m} v_2 \]
- Since \( M \) is much larger than \( m \), the absolute value of \( v_1 \) will be greater than \( v_2 \):
\[ |v_1| > |v_2| \]

4. **Kinetic Energy Calculation**:
- The kinetic energy of the ejected particle \( KE_1 \) is given by:
\[ KE_1 = \frac{1}{2} m v_1^2 \]
- The kinetic energy of the remaining nucleus \( KE_2 \) is:
\[ KE_2 = \frac{1}{2} (M - m) v_2^2 \]
- We need to compare \( KE_1 \) and \( KE_2 \).

5. **Substituting for \( v_1 \)**:
- Substitute \( v_1 \) into the kinetic energy equation:
\[ KE_1 = \frac{1}{2} m \left(-\frac{(M - m)}{m} v_2\right)^2 = \frac{1}{2} m \frac{(M - m)^2}{m^2} v_2^2 = \frac{(M - m)^2}{2m} v_2^2 \]
- Now, for \( KE_2 \):
\[ KE_2 = \frac{1}{2} (M - m) v_2^2 \]

6. **Comparing Kinetic Energies**:
- To compare \( KE_1 \) and \( KE_2 \):
\[ KE_1 = \frac{(M - m)^2}{2m} v_2^2 \quad \text{and} \quad KE_2 = \frac{1}{2} (M - m) v_2^2 \]
- Dividing \( KE_1 \) by \( KE_2 \):
\[ \frac{KE_1}{KE_2} = \frac{(M - m)}{m} \]
- Since \( M \) is much larger than \( m \), \( \frac{(M - m)}{m} > 1 \), thus:
\[ KE_1 > KE_2 \]

### Conclusion:
- The correct options based on our analysis are:
- **B**: The velocity of the particle is greater than the velocity of the nucleus.
- **C**: The kinetic energy of the particle is greater than the kinetic energy of the nucleus.