Correct Answer - Option 2 : 50%
Concept:
Kinetic Energy:
The energy possessed by a particle by the virtue of its motion is called kinetic energy. It is given by,
\(⇒ KE=\frac{1}{2}m× v^{2}\)
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:
%ΔKE = 125%
Let KE1 = KE
\(⇒ KE_{2}=KE+\frac{125}{100}KE\)
⇒ KE2 = 2.25 KE
The relation between the kinetic energy and momentum is given by
\(⇒ P=\sqrt{2\times KE\times m}\)
In initial momentum,
\(⇒ P_{1}=\sqrt{2\times KE\times m}\) -----(1)
Final momentum,
\(⇒ P_{2}=\sqrt{2\times KE_{2}\times m}\)
\(⇒ P_{2}=\sqrt{2\times 2.25KE\times m}\)
\(⇒ P_{2}=1.5\sqrt{2\times KE\times m}\) -----(2)
From equation 1 and equation 2,
⇒ P2 = 1.5 P1
Percentage increase in momentum,
\(⇒ \%\Delta P = \frac{P_{2}-P_{1}}{P_{1}}\times100\)
\(⇒ \%\Delta P = \frac{1.5P_{1}-P_{1}}{P_{1}}\times100=50\%\)
Hence, option 2 is correct.
