If the error in the measurement of momentum is 20 % , then the error in the calculation of kinetic energy is (assume the error in measurement of m as zero)

Options

A. `20%`
B. `44%`
C. `40%`
D. `200%`

Answer

To solve the problem step by step, we will analyze the relationship between momentum and kinetic energy, and how errors propagate through these calculations.

### Step 1: Understand the relationship between momentum and kinetic energy.
The momentum \( p \) of an object is given by the formula:
\[ p = mv \]
where \( m \) is the mass and \( v \) is the velocity of the object.

The kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} mv^2 \]

### Step 2: Express kinetic energy in terms of momentum.
We can express kinetic energy in terms of momentum. Squaring the momentum gives:
\[ p^2 = (mv)^2 = m^2v^2 \]
From the kinetic energy formula, we can express \( K \) as:
\[ K = \frac{p^2}{2m} \]

### Step 3: Determine the relative error in kinetic energy.
To find the error in kinetic energy, we need to consider how the error in momentum affects kinetic energy. The relative error in kinetic energy \( \frac{\Delta K}{K} \) can be derived from the formula for \( K \):
\[ \frac{\Delta K}{K} = \frac{\Delta (p^2)}{p^2} + \frac{\Delta m}{m} \]
Since we are assuming the error in mass \( m \) is zero (\( \Delta m = 0 \)), the equation simplifies to:
\[ \frac{\Delta K}{K} = \frac{\Delta (p^2)}{p^2} \]

### Step 4: Apply the error propagation rule.
For a quantity raised to a power, the relative error is multiplied by that power:
\[ \frac{\Delta (p^2)}{p^2} = 2 \frac{\Delta p}{p} \]
Thus, we have:
\[ \frac{\Delta K}{K} = 2 \frac{\Delta p}{p} \]

### Step 5: Substitute the given error in momentum.
We are given that the error in the measurement of momentum is 20%. Thus:
\[ \frac{\Delta p}{p} = \frac{20}{100} = 0.2 \]
Substituting this into our equation gives:
\[ \frac{\Delta K}{K} = 2 \times 0.2 = 0.4 \]

### Step 6: Convert the relative error to percentage.
To convert the relative error into a percentage, we multiply by 100:
\[ \text{Percentage error in } K = 0.4 \times 100 = 40\% \]

### Conclusion
The error in the calculation of kinetic energy is **40%**.