Question

If the kinetic energy of body becomes 9
times of its initial value, then new momentum will

Answer 1

We know the kinetic energy of a body of mass $\sf{m}$ moving with velocity $\sf{v}$ is,

$\longrightarrow\sf{K=\dfrac{1}{2}\,mv^2\quad\quad\dots(1)}$

Usually the mass of the body remains constant.

From (1),

$\longrightarrow\sf{K=\dfrac{1}{2}\cdot\,\dfrac{m^2v^2}{m}}$

$\longrightarrow\sf{K=\dfrac{m^2v^2}{2m}}$

$\longrightarrow\sf{K=\dfrac{(mv)^2}{2m}\quad\quad\dots(2)}$

But we know the momentum of the body is,

$\longrightarrow\sf{p=mv}$

Then (2) becomes,

$\longrightarrow\sf{K=\dfrac{p^2}{2m}}$

This implies,

$\longrightarrow\sf{K\propto p^2}$

Or,

$\longrightarrow\sf{\underline{\underline{p\propto\sqrt K}}}$

Thus, the linear momentum of a body is directly proportional to the square root of its kinetic energy.

So if the kinetic energy of a body becomes $\sf{n}$ times then the linear momentum of the body will become $\sf{\sqrt n}$ times.

Here the kinetic energy becomes 9 times. Hence the linear momentum will be $\bf{\sqrt9 = 3}$ times.

We see that,

$\longrightarrow\sf{\dfrac{p_2}{p_1}=\sqrt{\dfrac{K_2}{K_1}}}$

Here, $\sf{K_2=9K_1.}$ Then,

$\longrightarrow\sf{\dfrac{p_2}{p_1}=\sqrt{9}}$

$\longrightarrow\sf{\dfrac{p_2}{p_1}=3}$

$\longrightarrow\sf{\underline{\underline{p_2=3p_1}}}$