Question

if the kinetic energy of the body is doubled by what percent does its linear momentum will change​

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

Kinetic Energy

$\huge{ \boxed{ \boxed{\sf{k = \frac{1}{2}mv {}^{2} }}}}$

Linear Momentum

$\huge{ \boxed{ \boxed{ \sf{p = mv}}}}$

Now,

$\sf{p = mv} \\ \\ \large{\implies \sf{v = \frac{p}{m} }}$

Here,

$\sf{k = \frac{1}{2}m( \frac{p}{m}) {}^{2} } \\ \\ \huge{\implies \: \sf{k = \frac{p {}^{2} }{2m} }}$

Here,

• Kinetic Energy is directly proportional to the square of linear momentum

When Kinetic Energy is doubled, the rate of linear momentum also doubles

Answer 2

$\huge{\star}{\underline{\boxed{\red{\sf{Answer :}}}}}{\star}$

We know that,

$\huge{\boxed{\boxed{\sf{K.E \: = \: \frac{1}{2} \: mv^2}}}--(1)}$

Where,

K.E is kinetic energy

m is mass of body

v is velocity

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And formula for linear momentum is :-

$\huge{\boxed{\boxed{\sf{p \: = \: mv}}}}$

$\LARGE{\implies}{\boxed{\boxed{\sf{v \: = \: \frac{p}{m}}}}}$

Put value of v in (1)

We, get.

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$\Large{\sf{K.E \: = \: \frac{1}{2} \: m {\frac{p}{m}}^{2}}}$

Now solving it

$\Large{\sf{K.E \: = \: \frac{1}{2} \: m \: \frac{(p)^2}{(m)^2}}}$

$\Large{\sf{K.E \: = \: \frac{1}{2} \: \cancel{m} \: \frac{(p) ^2}{(m)^{\cancel{2}}}}}$

$\huge{\implies}{\boxed{\boxed{\green{\sf{K.E \: = \: \frac{p^2}{2m}}}}}}$

$\rule{200}{2}$

But, here 2m is constant.

then ,

$\huge{\boxed{\boxed{\sf{K.E \: \propto \: p^2}}}}$

∴ If kinetic energy is doubled then liner momentum will also increase twice.