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

Options

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Explanation

To solve the problem of finding the graph between \(\sqrt{E}\) (where \(E\) is kinetic energy) and \(\frac{1}{p}\) (where \(p\) is momentum), we can follow these steps:

### Step 1: Define Kinetic Energy and Momentum
The kinetic energy \(E\) of an object with mass \(m\) moving at velocity \(v\) is given by:
\[ E = \frac{1}{2} mv^2 \]
The momentum \(p\) of the same object is given by:
\[ p = mv \]

### Step 2: Relate Kinetic Energy and Momentum
We can express \(v\) in terms of momentum:
\[ v = \frac{p}{m} \]
Substituting this expression for \(v\) into the kinetic energy formula gives:
\[ E = \frac{1}{2} m \left(\frac{p}{m}\right)^2 = \frac{1}{2} \frac{p^2}{m} \]

### Step 3: Express \(E\) in terms of \(p\)
From the equation derived above, we can rearrange to express \(p\) in terms of \(E\):
\[ p^2 = 2mE \implies p = \sqrt{2mE} \]

### Step 4: Find the Relationship between \(\sqrt{E}\) and \(\frac{1}{p}\)
Now, we want to express \(\sqrt{E}\) in terms of \(\frac{1}{p}\):
\[ \sqrt{E} = \frac{p}{\sqrt{2m}} \]
Taking the reciprocal of \(p\):
\[ \frac{1}{p} = \frac{1}{\sqrt{2mE}} \]
This implies:
\[ \sqrt{E} = \frac{1}{\sqrt{2m}} \cdot \frac{1}{\frac{1}{p}} \implies \sqrt{E} \propto \frac{1}{\frac{1}{p}} \]

### Step 5: Identify the Type of Graph
From the relationship \(\sqrt{E} \propto \frac{1}{p}\), we can see that as \(\frac{1}{p}\) increases, \(\sqrt{E}\) also increases. This indicates a linear relationship between \(\sqrt{E}\) and \(\frac{1}{p}\), which can be represented graphically.

However, since we are plotting \(\sqrt{E}\) against \(\frac{1}{p}\), we can conclude that the graph will be a hyperbola, as the relationship is inversely proportional.

### Conclusion
The graph between \(\sqrt{E}\) and \(\frac{1}{p}\) is a hyperbola.

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