Question

If the linear momentum is increased by 50% , then kinetic energy will be increased by... *

Answer 1

Answer:

K.E.=P²/2M,P being increased 50% =>it becomes 1.5P=>new K.E.=(1.5P)²/2M=2.25P²/2M=>2.25K.E. =>the KE is changed by (2.25–1)×100%=125%.

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Answer 2

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✦ Required Answer:

♦️ GiveN:

• Linear momentum increased by 50%

♦️ To FinD:

• Increase in Kinetic energy...?

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✦ Expalnation of Concept:

The question is asking about Kinetic energy and linear momentum is already given. So, Let's discuss the concept and formula of Linear Momentum and Kinetic energy.

Linear Momentum:

Product of mass and velocity

$\large{\boxed{\rm{\purple{p=mv}}}}$

Kinetic Energy:

The energy possessed by a body on virtue of its motion.

$\large{\boxed{\rm{\purple{K.E = \frac{1}{2}m{v}^{2}}}}}$

✏ Note:

Symbols have their usual meanings...

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✦ Solution:

The linear momentum is the product of mass and velocity

$\large{\rm{\purple{p = mv}}}$

So, it is given that linear momentum increased by 50%, it means velocity increases by 50%, because mass remains constant for a particular body.

$\large{\rm{\purple{v' = 1.5 v}}}$

where, v' = final velocity/new velocity

v = initial velocity/ original velocity

The Kinetic energy is given as,

$\large{\rm{\purple{K.E = \frac{1}{2} m{v}^{2}}}}$

So, New Kinetic energy is,

$\large{\rm{\purple{K.E' = \frac{1}{2}m{v'}^{2}}}}$

So, Relation between original and new kinetic energy,

$\large{ \rm{ \longrightarrow \: \frac{KE'}{KE} = \frac{ \cancel{\frac{1}{2}m}v' {}^{2} }{ \cancel{\frac{1}{2}m} {v}^{2} }}} \\ \\ \large{ \rm{ \longrightarrow \: \frac{KE'}{KE} = \frac{ {v'}^{2} }{ {v}^{2} } }} \\ \\ \large{ \rm{ \longrightarrow \: \frac{KE'}{KE} = \frac{(1.5 {v})^{2} }{ {v}^{2} }}} \\ \\ \large{ \rm{ \longrightarrow \: \frac{KE'}{KE} = \frac{2.25 \: \cancel{{v}^{2}} }{ \cancel{{v}^{2}} } }} \\ \\ \large{ \rm{ \longrightarrow \: KE'= 2.25 \: KE}}$

Then, % change in the new kinetic energy,

$\large{ \rm{ \longrightarrow \: \% \: change = \frac{KE' - KE}{KE} \times 100}} \\ \\ \large{ \rm{ \longrightarrow \: \% \: change = \frac{2.25 \: KE - KE}{KE} \times 100}} \\ \\ \large{ \rm{ \longrightarrow \: \% \: change = \frac{1.25 \: \cancel{KE}}{ \cancel{ke} } \times 100}} \\ \\ \large{ \rm{ \longrightarrow \: \% \: change = \boxed{ \rm{ \red{125\%}}}}}$

Thus, the Kinetic energy increased by 125%

$\large{ \therefore{ \underline{ \rm{ \purple{Hence,\: solved \: \dag}}}}}$

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