# If the kinetic energy of a particle is increased by 44%, the increase in its linear momentum will be

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If the kinetic energy of a particle is increased by 44%, the increase in its linear momentum will be
1. 20%
2. 40%
3. 10%
4. 15%

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Correct Answer - Option 1 : 20%

The correct answer is option 1) i.e. 20%

CONCEPT:

• ​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

CALCULATION:

Let the initial and final kinetic energy of the particle be KE1 and KE2. Similarly, the linear momentum will be p1 and p2.

When the KE of the particle is increased by 44%, KE2 = 1.44KE1

We know,

$⇒ KE =\frac{1}{2} mv^2$

Multiplying and dividing by m we get,

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

$⇒ p =\sqrt{2mKE}$

⇒ p ∝ √KE

• The percentage increase in linear momentum

$\Rightarrow\Delta p\%=\frac{p_2 - p_1}{p_1} \times 100 = \frac{\sqrt{KE_2} - \sqrt{KE_1}}{\sqrt{KE_1}} \times 100$

$\Rightarrow \Delta p\%=\frac{\sqrt{1.44KE_1} - \sqrt{KE_1}}{\sqrt{KE_1}} \times 100 =\frac{\sqrt{1.44} - \sqrt{1}}{\sqrt{1}} \times 100 = \frac{1.2 -1}{1} \times 100 = 20\%$