If the momentum of a particle increases by 30%, then increase in its kinetic energy is:

CONCEPT:

• Momentum: A property of a body in motion that is equal to the product of the body's mass and velocity is called momentum.

P = mv

where P is the momentum of the body, m is the mass of the body, and v is the velocity of the body.

• Momentum Conservation: When in a system, there is no external force then the total momentum (P) of the system will be conserved.
• Kinetic energy: The energy in a body due to its motion, is known as kinetic energy.

The kinetic energy in terms of momentum is given by:

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

where K is the kinetic energy of the body, P is the momentum of the body and m is the mass of the body.

CALCULATION:

\(K=\frac{P^2}{2M}\)

Let initial value of P = 100, m = 100

After increase of 30% P = 130

Initial kinetic energy \(K_i=\frac{P_i^2}{2m}=\frac{100^2}{2\times 100}=50\)

Final kinetic energy \(K_f=\frac{P_f^2}{2m}=\frac{130^2}{2\times 100}=169/2\)

% Increase in Kinetic energy = \(\frac{(K_f-K_i)}{K_i}\times100 = \frac{(\frac{169}{2}-50)}{50}\times100 = 69\)%

So the correct answer is option 3.