Question

by how much will the speed of a body of fixed mass increase if its kinetic energy becomes four times its initial kinetic energy​

Answer 1

Explanation:

If the kinetic energy increases four times then the moment increases two times

P2=2m×Kinetic energy

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{v_{1}=2v}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies Increased \: kinetic \: energy = 4 \: K.E \\ \\ \red{\underline{\bold{To \: Find:}}} \\ \tt: \implies Increased \: speed =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies Initial \: kinetic \: energy = \frac{1}{2} {mv}^{2} \\ \\ \tt: \implies K.E = \frac{1}{2} m {v}^{2} - - - - - (1) \\ \\ \bold{Similarly : } \\ \tt: \implies Increased \: kinetic \: energy = \frac{1}{2} m { (v_{1})}^{2} \\ \\ \tt: \implies 4 \: K.E = \frac{1}{2} m {( v_{1}) }^{2} \\ \\ \tt: \implies K.E= \frac{1}{8} {m (v_{1} })^{2} - - - - - (2)\\ \\ \text{From \: (1) \: and \: (2)} \\ \\ \tt: \implies \frac{1}{2} m {v}^{2} = { \frac{1}{8} } m { (v_{1} )}^{2} \\ \\ \tt: \implies 4 {v}^{2} = (v_{1})^{2} \\ \\ \tt: \implies {(2v)}^{2} = { (v_{1}) }^{2} \\ \\ \green{\tt: \implies v_{1} = 2v} \\ \\ \green{\tt \therefore Speed \: becomes \: 2 \: times \: in \: increased \: kinetic \: energy}$