The graph betwee `sqrt(E)` and `(1)/(p)` is (E=kinetic energy and p= momentum)

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Answer

To analyze the relationship between the square root of kinetic energy (E) and the reciprocal of momentum (p), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Kinetic Energy and Momentum**:
- The kinetic energy (E) of an object is given by the formula:
\[ E = \frac{1}{2} mv^2 \]
- The momentum (p) of an object is defined as:
\[ p = mv \]

2. **Expressing Kinetic Energy in terms of Momentum**:
- From the momentum equation, we can express velocity (v) as:
\[ v = \frac{p}{m} \]
- Substituting this expression for v into the kinetic energy formula:
\[ E = \frac{1}{2} m \left(\frac{p}{m}\right)^2 \]
- Simplifying this gives:
\[ E = \frac{1}{2} m \cdot \frac{p^2}{m^2} = \frac{p^2}{2m} \]

3. **Rearranging the Equation**:
- We can rearrange the equation to express E in terms of p:
\[ E = \frac{1}{2m} p^2 \]
- This shows that E is proportional to the square of momentum.

4. **Finding the Relationship Between \(\sqrt{E}\) and \(\frac{1}{p}\)**:
- Taking the square root of both sides:
\[ \sqrt{E} = \sqrt{\frac{1}{2m}} \cdot p \]
- Rearranging gives:
\[ \sqrt{E} = K \cdot p \quad \text{(where \( K = \sqrt{\frac{1}{2m}} \))} \]
- Now, substituting \( p = \frac{1}{\frac{1}{p}} \):
\[ \sqrt{E} = K \cdot \frac{1}{\frac{1}{p}} \]

5. **Final Form of the Equation**:
- If we let \( Y = \sqrt{E} \) and \( X = \frac{1}{p} \), we can write:
\[ Y = K \cdot \frac{1}{X} \]
- This is the equation of a rectangular hyperbola.

6. **Conclusion**:
- The graph of \(\sqrt{E}\) versus \(\frac{1}{p}\) will be a rectangular hyperbola.