Question

When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be

• A. \(500\%\)
• B. \(600\%\)
• C. \(6\%\)
• D. \(60\%\)

Answer 1

The Correct Option is A

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:
\[\text{Percentage increase} = \frac{P_f - P_i}{P_i} \times 100\%\]
\[= \frac{6P_i - P_i}{P_i} \times 100\%\]
\[= \frac{5P_i}{P_i} \times 100\% = 500\%\]

Answer 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 \(500\%\).