Which one of the following correctly represents the variation between linear momentum (P) and kinetic energy (E) ofa body?

Options

A.
B.
C.
D.

Explanation

To determine the relationship between linear momentum (P) and kinetic energy (E) of a body, we start with the definitions of both quantities:

1. **Linear Momentum (P)**:
\[ P = mv \]
where \(m\) is the mass of the body and \(v\) is its velocity.

2. **Kinetic Energy (E)**:
\[ E = \frac{1}{2} mv^2 \]

Next, we can express kinetic energy in terms of momentum. We know that momentum can be rewritten in terms of velocity:
\[ v = \frac{P}{m} \]

Substituting this expression for \(v\) into the kinetic energy formula gives:
\[ E = \frac{1}{2} m \left(\frac{P}{m}\right)^2 \]
\[ E = \frac{1}{2} m \cdot \frac{P^2}{m^2} \]
\[ E = \frac{P^2}{2m} \]

This equation shows that kinetic energy \(E\) is proportional to the square of the momentum \(P\):
\[ E \propto P^2 \]

Now, to express the relationship between \(P\) and \(E\), we can rearrange the equation:
\[ P^2 = 2mE \]
Taking the square root of both sides gives:
\[ P = \sqrt{2mE} \]

This indicates that momentum \(P\) is proportional to the square root of kinetic energy \(E\):
\[ P \propto \sqrt{E} \]

### Conclusion:
The relationship between linear momentum (P) and kinetic energy (E) can be represented graphically as \(P\) versus \(\sqrt{E}\). The graph will show that as kinetic energy increases, momentum increases as well, but at a decreasing rate (since it is a square root relationship).