The momentum of a body is increased by 50%. The percentage increase in the kinetic energy of the body is ______%
Correct Answer: 125
Solution and Explanation
Step 1: Relation between kinetic energy and momentum - Kinetic energy \[ K = \frac{P^2}{2m}, \] where \(P\) is momentum and \(m\) is mass.
Step 2: Calculate new kinetic energy - Initial momentum \(P_1\), kinetic energy \[ K_1 = \frac{P_1^2}{2m}. \] New momentum \[ P_2 = P_1 + 0.5P_1 = 1.5P_1. \] New kinetic energy \[ K_2 = \frac{P_2^2}{2m} = \frac{(1.5P_1)^2}{2m} = \frac{2.25P_1^2}{2m} = 2.25K_1. \]
Step 3: Calculate percentage increase in kinetic energy - Percentage increase is given by: \[ \frac{K_2 - K_1}{K_1} \times 100 = \frac{2.25K_1 - K_1}{K_1} \times 100 = 125\%. \]
Final Answer: The percentage increase in kinetic energy is 125%.
Top Questions on momentum
- An object of mass \( m \) is projected from the origin in a vertical \( xy \)-plane at an angle \( 45^\circ \) with the x-axis with an initial velocity \( v_0 \). The magnitude and direction of the angular momentum of the object with respect to the origin, when it reaches the maximum height, will be:
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Knowing the initial position \( x_0 \) and initial momentum \( p_0 \) is enough to determine the position and momentum at any time \( t \) for a simple harmonic motion with a given angular frequency \( \omega \).
Reason (R): The amplitude and phase can be expressed in terms of \( x_0 \) and \( p_0 \).
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