Question

Two bodies $P$ and $Q$ having masses in the ratio $1:4$ and having kinetic energies in the ratio of $4:1$ , then the ratio of linear momentum $P$ and $Q$ is
(A) $1:4$
(B) $1:2$
(C) $1:16$
(D) $1:1$

Answer 1

Hint Use the general equation for kinetic energy and introduce the momentum variable into the equation by necessary substitutions. Use this relation to find two similar equations for the bodies $P$ and $Q$ . Divide one equation by the other and substitute the given value of ratios into the equation and simplify to get the answer.

Complete Step by step solution
Kinetic energy and momentum are related by the equation
$KE=\dfrac{{{p}^{2}}}{2m}$
Where $KE$ is the kinetic energy of the body,
$p$ is the linear momentum of the body,
$m$ is the mass of the body.
Let the kinetic energy of the body $P$ be $K{{E}_{P}}$ and let the kinetic energy of the body $Q$ be $K{{E}_{Q}}$ . Let ${{p}_{P}}$ and ${{p}_{Q}}$ be the linear momentum of bodies the $P$ and $Q$ respectively. And let ${{m}_{P}}$ and ${{m}_{Q}}$ be the mass of the bodies $P$ and $Q$ respectively.
For the body $P$ , the relation will be
$K{{E}_{P}}=\dfrac{{{p}_{P}}^{2}}{2{{m}_{P}}}$
And the relation for the body $Q$ will be
$K{{E}_{Q}}=\dfrac{{{p}_{Q}}^{2}}{2{{m}_{Q}}}$
Now dividing the kinetic energies of bodies $P$ and $Q$ gives us
$\dfrac{K{{E}_{P}}}{K{{E}_{Q}}}=\dfrac{{{p}_{P}}^{2}}{2{{m}_{P}}}\times \dfrac{2{{m}_{Q}}}{{{p}_{Q}}^{2}}$
Now, it is given in the question that the ratio of kinetic energies of the bodies $P$ and $Q$ is $4:1$ . Also, the ratio of masses of the bodies $P$ and $Q$ is $1:4$ . So by using these values in the above equation gives us
$\dfrac{4}{1}=\dfrac{{{p}_{P}}^{2}}{2\times 1}\times \dfrac{2\times 4}{{{p}_{Q}}^{2}}$
By canceling the common terms from the equation and by rearranging, we get
$\dfrac{{{p}_{P}}^{2}}{{{p}_{Q}}^{2}}=\dfrac{1}{1}$
Therefore, by taking square roots on both sides, we get
$\dfrac{{{p}_{P}}}{{{p}_{Q}}}=1$
Or
${{p}_{P}}:{{p}_{Q}}=1:1$

$\Rightarrow$ Option (D) is the correct option.

Note
To get the relation between kinetic energy and linear momentum:
The equation for kinetic energy:
$KE=\dfrac{1}{2}m{{v}^{2}}$
The above equation can be written as
$KE=\dfrac{1}{2}mv\cdot v$
Now multiply and divide by mass $m$
$\Rightarrow KE=\dfrac{1}{2}\times \dfrac{{{\left( mv \right)}^{2}}}{m}$
But as $mv$ is nothing but the equation of momentum, we get
$KE=\dfrac{{{p}^{2}}}{2m}$