# What happens to kinetic energy when momentum is doubled?

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**Hint:**Kinetic energy is defined as the energy possessed by a particle due to its motion. Momentum is defined as the product of mass and velocity. We have to derive the formula of Kinetic energy in terms of momentum.

**Formula used:**The energy related to the motion of any body, known as kinetic energy is of formula,

$K.E. = \dfrac{1}{2}m{v^2}$ where,

$K.E. = $ kinetic energy

$m = $ mass of the particle

$v = $ velocity of the particle

And momentum, the product of mass of velocity has the formula

$p = mv$ where,

$p = $ momentum of the particle

$m = $ mass of the particle

$v = $ velocity of the particle

**Complete step by step answer:**

Now, kinetic energy in terms of momentum is,

$K.E. = \dfrac{1}{2}m{v^2}$

Multiplying both sides by $m$,

$m \times K.E. = \dfrac{1}{2}{m^2}{v^2}$

We know, $p = mv$ so,

$m \times K.E. = \dfrac{1}{2}{p^2}$

$ \Rightarrow \sqrt {2mK.E.} = p - - - - \left( 1 \right)$

Now, when momentum is doubled $p' = 2p$ then let new kinetic energy be $(K.E.)'$

$p' = \sqrt {2m(K.E.)'} - - - - - \left( 2 \right)$

Comparing equation $\left( 1 \right)$ and $\left( 2 \right)$ we get,

$\dfrac{{p'}}{p} = \dfrac{{\sqrt {2m(K.E.)'} }}{{\sqrt {2m(K.E.)} }}$

Substituting $p' = 2p$ we get the equation as,

$\dfrac{{2p}}{p} = \dfrac{{\sqrt {2m(K.E.)'} }}{{\sqrt {2m(K.E.)} }}$

Eliminating and squaring both sides we get,

$

2 \times \sqrt {2m(K.E.)} = \sqrt {2m(K.E.)'} \\

\Rightarrow 4 \times K.E. = K.E.' \\

$

So, if the momentum is doubled then the kinetic energy increases by $4$times.

**Note:**As the question does not mention that the mass is unchanged. But we have to consider the mass to be unchanged. If the mass is not unchanged then it will be two different bodies and not the same body. We must derive kinetic energy in terms of momentum not momentum in terms of kinetic energy as the question provided a change in momentum and not in kinetic energy.

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