When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be
- \(500\%\)
- \(600\%\)
- \(6\%\)
- \(60\%\)
The Correct Option is A
Approach Solution - 1
The kinetic energy \( K \) is given by:
\[K = \frac{P^2}{2m}\]
Therefore, momentum \( P \) can be expressed as:
\[P = \sqrt{2mK}\]
If the final kinetic energy \( K_f \) is 36 times the initial kinetic energy \( K_i \), we have:
\[K_f = 36 K_i\]
Thus, the final momentum \( P_f \) will be:
\[P_f = \sqrt{2m \cdot 36K_i} = 6P_i\]
The percentage increase in momentum is:
\[\text{Percentage increase} = \frac{P_f - P_i}{P_i} \times 100\%\]
\[= \frac{6P_i - P_i}{P_i} \times 100\%\]
\[= \frac{5P_i}{P_i} \times 100\% = 500\%\]
Approach Solution -2
To solve this problem, we need to understand the relationship between kinetic energy and momentum.
The kinetic energy \((K)\) of a body is given by the formula:
\(K = \frac{1}{2}mv^2\)
where \(m\) is the mass and \(v\) is the velocity of the body.
The momentum \((p)\) of a body is given by:
\(p = mv\)
Given that the kinetic energy becomes 36 times its original value:
\(K' = 36K\)
Let's express the new kinetic energy with respect to the new velocity \(v'\):
\(K' = \frac{1}{2}m(v')^2 = 36 \times \frac{1}{2}mv^2\)
This simplifies to:
\((v')^2 = 36v^2\)
Taking the square root of both sides:
\(v' = 6v\)
Now, let's find the new momentum \(p'\):
\(p' = mv' = m(6v) = 6mv = 6p\)
Thus, the new momentum is 6 times the original momentum.
The percentage increase in momentum is calculated as follows:
\(\text{Percentage Increase} = \left(\frac{p' - p}{p}\right) \times 100\%\)
Substituting the values, we have:
\(\text{Percentage Increase} = \left(\frac{6p - p}{p}\right) \times 100\% = \left(\frac{5p}{p}\right) \times 100\% = 500\%\)
Therefore, the percentage increase in the momentum of the body is \(500\%\).
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