If momentumn increase 20 percentage then K.E increase by?

If momentum increases by 20%, the kinetic energy will also increase by 20%. This is because kinetic energy is directly proportional to momentum.

The formula for kinetic energy is:

KE = 0.5 * m * v^2

Where:
KE = Kinetic energy
m = Mass
v = Velocity

Since momentum is given by:

p = m * v

Where:
p = Momentum

If momentum increases by 20%, then the new momentum will be:

1.2p = 1.2 * m * v

Therefore, the new kinetic energy will be:

KE' = 0.5 * m * (1.2v)^2
KE' = 0.5 * m * 1.44v^2
KE' = 1.44 * 0.5 * m * v^2
KE' = 1.44 * KE

So, if momentum increases by 20%, the kinetic energy will increase by 44%.

84%

Explanation

Kinetic energy (K) is given by:

K = 1/2 - m - v^2

where:

m = mass
v = velocity

Momentum (P) is given by:

P = m - v

If momentum increases by 20%, then:

P' = 1.2 - P

Substituting P' into the equation for kinetic energy, we get:

K' = 1/2 - m - (1.2 - v)^2
K' = 1.2^2 - 1/2 - m - v^2
K' = 1.44 - K

Therefore, kinetic energy increases by:

(K' - K)/K - 100% = (1.44 - K - K)/K - 100%
= 0.44 - K/K - 100%
= 44%

However, the question asks for the increase in kinetic energy, so we add the 100% that it already had:

Increase in kinetic energy = 44% + 100% = 84%