Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

When the momentum of a body increases by 100%, its K.E. increases by:
A. 20%
B. 40%
C. 100%
D. 300%

seo-qna
SearchIcon
Answer
VerifiedVerified
382.2k+ views
Hint:To answer the above question, we need to understand the formulas of kinetic energy and momentum and find a relation between both these quantities. Kinetic energy in the energy acquired by a body due to its motion, while momentum is the product of mass and velocity of a body.

Complete step by step answer:
Kinetic energy is the type of energy that an item or particle has as a result of its movement. When work is done on an object by exerting a net force, the object accelerates and gains kinetic energy as a result. Kinetic energy is a property of a moving item or particle that is determined by its mass as well as its motion. The formula for K.E. is as follows:
$KE = \dfrac{1}{2}m{v^2}$
Here $m$= mass of the body and $v$= velocity of the body.

The product of a particle's mass and velocity is called momentum. Momentum is a vector quantity in the sense that it has both a magnitude and a direction. The time rate of change in momentum is equal to the force applied on the particle, according to Isaac Newton's second law of motion. The formula for momentum ($p$) is:
\[p = mv\]

Comparing both equations, we get
$KE = \dfrac{{{p^2}}}{{2m}}$
When momentum is increased by 100%, new kinetic energy $(KE')$ becomes:
$KE' = \dfrac{{{{(2p)}^2}}}{{2m}} \\
\Rightarrow KE'= \dfrac{{4{p^2}}}{{2m}}$
So, percentage increase in KE is,
$\dfrac{{KE' - KE}}{{KE}}\times 100 = \dfrac{{\dfrac{{4{p^2} - {p^2}}}{{2m}}}}{{\dfrac{{{p^2}}}{{2m}}}}\times 100 \\
\therefore \dfrac{{KE' - KE}}{{KE}}\times 100 = 300\% $

Hence,the correct answer is option D.

Note: If a constant force applies on a particle for a particular time, the product of force and time interval (the impulse) equals the change in momentum, according to Newton's second law. The momentum of a particle, on the other hand, is a measure of the time it takes for a constant force to bring it to a stop.