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Solutions
Given,
\(KE _{2}=9 KE _{1}\)
We know that the kinetic energy of a body is given as,
\( K E=\frac{1}{2} m v^{2}\)...(1)
The linear momentum of a body is given as,
\( P = mv\)...(2)
By equation (1) and equation (2) the relation between the kinetic energy and the momentum is given as,
\( K E=\frac{P^{2}}{2 m}\)...(3)
By equation (3) for the initial position,
\( K E_{1}=\frac{P_{1}^{2}}{2 m}\)...(4)
By equation (3) for the final position,
\( K E_{2}=\frac{P_{2}^{2}}{2 m} \)
\(\Rightarrow 9 \times K E_{1}=\frac{P_{2}^{2}}{2 m}\)...(5)
By equation (4) and equation (5),
\(\frac{P_{2}^{2}}{2 m} \times \frac{2 m}{P_{1}^{2}}=\frac{9 \times K E_{1}}{K E_{1}}\)
\(\Rightarrow \frac{P_{2}^{2}}{P_{1}^{2}}=9 \)
\(\Rightarrow \frac{P_{2}}{P_{1}}=3 \)
\(\Rightarrow P_{2}=3 P_{1}\)
So, if the kinetic energy of a body is increased nine times then the momentum of the body will be increased by three times.