Two bodies of masses $m_{1}$ and $m_{2}$ have same momentum. The ratio of their $K E$ is

\sqrt{\frac{m_{2}}{m_{1}}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{\frac{m_{1}}{m_{2}}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{m_{1}}{m_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{m_{2}}{m_{1}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{m_{2}}{m_{1}}$
$Step 1: Kinetic energy and momentum$
Kinetic energy of a body is given by: $K E = \frac{1}{2} m v^{2}$
Momentum of a body is given by: $P = m v$

$Step 2: Equation Solving$
Kinetic energy can be written as
$K E = \frac{1}{2} m v^{2} \times \frac{m}{m} = \frac{1}{2 m} . (m v)^{2} = \frac{P^{2}}{2 m}$

$\Rightarrow K E \propto \frac{1}{m}$ $[∵ P$ is same for both the bodies $]$
$\Rightarrow \frac{(K E)_{1}}{(K E)_{2}} = \frac{m_{2}}{m_{1}}$

Hence $D$ is the correct option.

Suggest Corrections

Similar questions

Two bodies of masses $m_{1}$ and $m_{2}$ have equal kinetic energies. If $p_{1}$ and $p_{2}$ are their respective momentum, then ratio $p_{1} : p_{2}$ is equal to

Two spheres of masses m1 and m2 (m1>m2) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is:

If the molecular weight of two gases are $M_{1}$ and $M_{2}$ , then at a given tempreture the ratio of their root mean square velocity $v_{1}$ and $v_{2}$ will be

m₁ and m₂ are the slope of two perpendicular lines, if

Q. Same force acts on two bodies of masses

m_{1}

and

m_{2}

(

m_{2}

m_{1}

). After travelling the same distance,

Join BYJU'S Learning Program

Two bodies of masses m1 and m2 have same momentum. The ratio of their KE is

Two bodies of masses $m_{1}$ and $m_{2}$ have same momentum. The ratio of their $K E$ is