What is the percentage change in kinetic energy of a body if its momentum is increased by 2?

Free

10 Questions 40 Marks 10 Mins

Concept:

• Kinetic Energy (K): The energy of the body in motion by virtue of motion is called kinetic energy.

\(K = \frac{1}{2}mv^2\)

m is the mass of the body, v is the speed of the body.

• Momentum (P): The product of the mass and velocity of a body in motion is called momentum of the body.

p = mv

• Relationship between Kinetic Energy and Momentum: Kinetic energy is the square of momentum divided by mass. 

\(K = \frac{p^2}{2m}\)

Calculation:

Let initial momentum is p, and kinetic energy is K, then

\(K = \frac{p^2}{2m}\) --- (1)

The momentum is increased by 50 % of its initial value 

New momentum p' = p + 50 % of p = 1.5p

New kinetic energy K'

\(K' = \frac{P'^2}{2m}\)

⇒ \(K' = \frac{(1.5p)^2}{2m}\)

⇒ \(K' = 2.25\frac{p^2}{2m}\) -- (2)

Putting (1) and (2) we get 

⇒ K' = 2.25 K --- (3)

Percentage change 

\(ΔP =\frac{K' - K}{K} \times 100 % = \frac{2.25K - K}{K} \times 100 % \)

\(ΔP =\frac{1.25K}{K} \times 100 % \) = 125 %

So, the Percentage change in kinetic energy is 125 %. 

Let's discuss the concepts related to Work Power and Energy and Kinetic Energy. Explore more from Physics here. Learn now!

India’s #1 Learning Platform

Start Complete Exam Preparation

Daily Live MasterClasses

Practice Question Bank

Mock Tests & Quizzes

Get Started for Free Download App

Trusted by 3.3 Crore+ Students

Text Solution

Solution : Initial momentum , p = mv <br> Initial kinetic energy <br> ` K = 1/2 mv^(2) = (p^(2))/(2m)` <br> Increase in momemtum = 50 % of p =`p/2` <br> Final momentum = `p . = p+p/2 = (3p)/2 ` <br> Final kinectic energy , <br> ` K. = (.p^(2))/(2m) = ((3p//2)^(2))/(2m) = 9/4 (p^(2))/(2m) = 9/4 K ` <br> ` :. ` Increase in kinetic energy <br> ` K. - K = 9/4 K - K = (5K)/4` <br> % Increase = `(K. - K)/((K)) xx 100 = ((5K)/4)/K xx 100 = 500/4 %` <br> = 125 %