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.