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If the linear momentum of a body is increased by 50%, the increase in kinetic energy will be-
1. 50%
2. 75%
3. 100%
4. 125%

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Correct Answer - Option 4 : 125%

The correct answer is option 4) i.e. 125%


  • Kinetic energy is the energy possessed by a moving object. Kinetic energy (KE) is expressed as:

\(⇒ KE =\frac{1}{2} mv^2\)
Where m is the mass of the object and v is the velocity of the object.

  • Momentum: Momentum is the impact due to a moving object of mass m and velocity v.
  • The momentum (p) of an object is expressed as:
    ⇒ p = mv


Let the initial and final linear momentum be p1 and p2.

Let the initial and final kinetic energy be KE1 and KE2

We know, \(KE =\frac{1}{2} mv^2\)

Multiplying and dividing by m we get, 

\(⇒ KE =\frac{1}{2} mv^2 \times \frac{m}{m} = \frac{p^2}{2m}\)

⇒ KE ∝ p2

Let p1 = p

For 50% increase in linear momentum, p2 = 1.5 p1 = 1.5p


KE1 ∝ p2

KE2 ∝ (1.5p)2

Percentage change in kinetic energy \(= \frac{KE_2 -KE_1}{KE_1} \times 100 \)

\(⇒ \frac{(1.5p)^2 -p^2}{p^2} \times 100 \)

\(⇒ \frac{2.25 - 1}{1} \times 100 = 125\%\)

Thus, the change in kinetic energy is 125%.

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