If the momentum of a body is increased by 50%, then the percentage increase in its kinetic energy is[1995]a)50%b)100%c)125%d)200%Correct answer is option 'C'. Can you explain this answer?

Question

Accepted Answer

Initial momentum (p₁) = p; Final momentum

(p₂) = 1.5 p and initial kinetic energy (K₁) = K.

Kinetic energy

or, K₂ = 2.25 K.
Therefore, increase in kinetic energy is 2.25 K – K = 1.25 K or 125%.

Answer

Explanation:

Let the initial momentum (P) of the body be P and its initial kinetic energy (K) be K.

According to the question, the momentum of the body is increased by 50%.

Therefore, the new momentum (P') of the body is:

P' = P + (50/100) * P = 1.5P

The kinetic energy of a body is given by:

K = (1/2) * m * v^2

where m is the mass of the body and v is its velocity.

Since the mass of the body remains constant, the kinetic energy is directly proportional to the square of its velocity.

Let the initial velocity (v) of the body be v and its new velocity (v') be v'.

Since momentum is given by:

P = m * v

and

P' = m * v'

we can write:

m * v' = 1.5 * m * v

or

v' = (3/2) * v

Therefore, the new kinetic energy (K') of the body is:

K' = (1/2) * m * v'^2

Substituting the value of v' in the above equation, we get:

K' = (1/2) * m * (3/2)^2 * v^2

K' = (9/8) * K

Therefore, the percentage increase in kinetic energy is:

(K' - K)/K * 100%

= [(9/8)K - K]/K * 100%

= (1/8) * 100%

= 12.5%

Hence, the correct option is (c) 125%.

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