Question

If the kinetic energy of particle becomes 16 times, then the magnitude of linear momentum will be

• A. 4 times
• B. 8 times
• C. 2 times
• D. 16 times

Answer 1

The correct option is A 4 times

Kinetic energy of a body of mass m, moving with velocity v, $K.E=\frac{1}{2}m{v}^2\Rightarrow K.E=\frac{1}{2}m{v}^2\times \frac{m}{m}=\frac{(mv{)}^2}{2m}\Rightarrow K.E=\frac{P^2}{2m}(\because Momentum,P=mv)\Rightarrow P=\sqrt{2m(k.E.)}$

Now, new kinetic energy, $K.E\text{'}.=16K.E.$

New linear momentum, $P^{\prime }=\sqrt{2m(K.E^{\prime }.)}=\sqrt{2m(16k.E)}\Rightarrow P^{\prime }=4\sqrt{2m(K..E.)}\Rightarrow P^{\prime }=4P[\because P=\sqrt{2m(K.E.)}]$

Hence, the correct answer is option (a).