If v,p and E denote velocity, lenear momentum and KE of the paritcle respectively, then

Options

A. `p=(dE)/(dv)`
B. `p=(dE)/(dt)`
C. `p=(dv)/(dt)`
D. `p=(dE)/(dv)xx(dE)/(dt)`

Explanation

To solve the problem, we need to establish the relationship between kinetic energy (E), linear momentum (P), and velocity (v) of a particle. Here are the steps to derive the required relationship:

### Step-by-Step Solution:

1. **Define Kinetic Energy**:
The kinetic energy (E) of a particle is given by the formula:
\[ E = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the particle and \(v\) is its velocity.

2. **Differentiate Kinetic Energy with Respect to Velocity**:
To find the relationship between kinetic energy and momentum, we differentiate \(E\) with respect to \(v\):
\[ \frac{dE}{dv} = \frac{d}{dv}\left(\frac{1}{2} mv^2\right) \]

3. **Apply the Derivative**:
Using the power rule of differentiation, we get:
\[ \frac{dE}{dv} = \frac{1}{2} m \cdot 2v = mv \]
Here, the \(2\) from the power of \(v\) cancels with the \(\frac{1}{2}\) in front.

4. **Relate to Linear Momentum**:
We know that linear momentum (P) is defined as:
\[ P = mv \]
Therefore, we can substitute \(mv\) in our previous result:
\[ \frac{dE}{dv} = P \]

5. **Conclusion**:
Thus, we have established that:
\[ P = \frac{dE}{dv} \]
This means that the linear momentum of a particle is equal to the derivative of its kinetic energy with respect to its velocity.

### Final Result:
The correct relationship is:
\[ P = \frac{dE}{dv} \]