# When the linear momentum of a particle is increased by 1% its kinetic energy increases by x%. When the kinetic energy of the particle is increased by

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When the linear momentum of a particle is increased by 1% its kinetic energy increases by x%. When the kinetic energy of the particle is increased by 300%, its linear momentum increases by y%. The ratio of y to x is:
1. 300
2. 150
3. 50
4. None of these

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Correct Answer - Option 3 : 50

CONCEPT:

Kinetic Energy

• The energy possessed by a particle by the virtue of its motion is called kinetic energy. It is given by,

$⇒ KE=\frac{1}{2}mv^{2}$

Linear Momentum:

• It is defined as the product of the mass of an object, m, and its velocity, v. It, is a vector quantity.​

⇒ P = mv

Where KE = kinetic energy, P = linear momentum, m = mass and v = velocity

CALCULATION:

Given when %ΔP = 1%, %ΔKE = x% and when %ΔKE = 300%, %ΔP = y%

• We know that the relation between the kinetic energy and the momentum is given as,

$⇒ KE=\frac{P^{2}}{2m}$     -----(1)

Let P1 = P and KE1 = KE

For first case when %ΔP = 1% and %ΔKE = x%,

$⇒ P_{2}=P+\frac{1}{100}P$

$⇒ P_{2}=\frac{101}{100}P$     -----(2)

$⇒ KE_{2}=KE+\frac{x}{100}KE$

$⇒ KE_{2}=\frac{(100+x)}{100}KE$     -----(3)

By equation 1, equation 2, and equation 3,

$⇒ KE_{2}=\frac{P_{2}^{2}}{2m}$

$⇒ \frac{(100+x)}{100}KE=\frac{101^{2}}{100^2}\frac{P^2}{2m}$

$⇒(100+x)KE=\frac{101^{2}}{100}KE$

⇒ 100 + x = 102.01

⇒ x = 2.01     -----(4)

For second case when %ΔP = y% and %ΔKE = 300%,

$⇒ P'_{2}=P+\frac{y}{100}P$

$⇒ P'_{2}=\frac{100+y}{100}P$     -----(5)

$⇒ KE'_{2}=KE+\frac{300}{100}KE$

$⇒ KE'_{2}=4KE$     -----(6)

By equation 1, equation 5, and equation 6,

$⇒ KE'_{2}=\frac{(P_{2}')^{2}}{2m}$

$⇒ 4KE=\frac{\left ( 100+y \right )^{2}}{100^{2}}\frac{P^{2}}{2m}$

$⇒ 4KE=\frac{\left ( 100+y \right )^{2}}{100^{2}}KE$

⇒ 100 + y = 200

⇒ y = 100     -----(7)

By dividing equation 7 and equation 4,

$\Rightarrow \frac{y}{x}=\frac{100}{2.01}$

$\Rightarrow \frac{y}{x}=50$

• Hence, option 3 is correct.