Two object are initially at rest on a frictionless surface. Object 1 has a greater mass than object 2. The same constant force start to act on each object. The force is removed from each object after it accelerates over a distance d. after the force is removed from both objects, which statement is correct (p:momentum, K: kinetic energy)?

Options

A. `p_1ltp_2`
B. `p_1gtp_2`
C. `K_1gtK_2`
D. `K_1ltK_2`

Explanation

To solve the problem, we need to analyze the motion of two objects (Object 1 and Object 2) subjected to the same constant force over the same distance on a frictionless surface.

### Step-by-Step Solution:

1. **Define the Masses:**
Let the mass of Object 1 be \( m_1 \) and the mass of Object 2 be \( m_2 \). We know that \( m_1 > m_2 \).

2. **Apply Newton's Second Law:**
The force \( F \) acting on each object can be expressed using Newton's second law:
\[ F = m \cdot a \]
For Object 1:
\[ F = m_1 \cdot a_1 \implies a_1 = \frac{F}{m_1} \]
For Object 2:
\[ F = m_2 \cdot a_2 \implies a_2 = \frac{F}{m_2} \]

3. **Calculate Final Velocity After Distance \( d \):**
Using the kinematic equation \( V^2 = U^2 + 2aS \) (where \( U \) is the initial velocity, \( a \) is acceleration, and \( S \) is distance):
Since both objects start from rest, \( U = 0 \):
For Object 1:
\[ V_1^2 = 0 + 2a_1d = 2\left(\frac{F}{m_1}\right)d \implies V_1 = \sqrt{\frac{2Fd}{m_1}} \]
For Object 2:
\[ V_2^2 = 0 + 2a_2d = 2\left(\frac{F}{m_2}\right)d \implies V_2 = \sqrt{\frac{2Fd}{m_2}} \]

4. **Calculate Momentum:**
Momentum \( p \) is given by \( p = mv \).
For Object 1:
\[ p_1 = m_1 V_1 = m_1 \sqrt{\frac{2Fd}{m_1}} = \sqrt{2Fd \cdot m_1} \]
For Object 2:
\[ p_2 = m_2 V_2 = m_2 \sqrt{\frac{2Fd}{m_2}} = \sqrt{2Fd \cdot m_2} \]

5. **Compare Momentums:**
Since \( m_1 > m_2 \):
\[ p_1 = \sqrt{2Fd \cdot m_1} > \sqrt{2Fd \cdot m_2} = p_2 \]
Thus, \( p_1 > p_2 \).

6. **Calculate Kinetic Energy:**
Kinetic energy \( K \) is given by \( K = \frac{1}{2}mv^2 \).
For Object 1:
\[ K_1 = \frac{1}{2} m_1 V_1^2 = \frac{1}{2} m_1 \left(\frac{2Fd}{m_1}\right) = Fd \]
For Object 2:
\[ K_2 = \frac{1}{2} m_2 V_2^2 = \frac{1}{2} m_2 \left(\frac{2Fd}{m_2}\right) = Fd \]

7. **Compare Kinetic Energies:**
Since both \( K_1 \) and \( K_2 \) equal \( Fd \):
\[ K_1 = K_2 \]

### Conclusion:
After the force is removed, we conclude that:
- The momentum of Object 1 is greater than the momentum of Object 2 (\( p_1 > p_2 \)).
- The kinetic energy of both objects is equal (\( K_1 = K_2 \)).

### Final Statements:
- **Correct Statement:** \( p_1 > p_2 \) and \( K_1 = K_2 \).