The relation between momentum (P), mass (m), kinetic energy (KE) and velocity (V) of two bodies is

P_{1} V_{1} = P_{2} V_{2}

where KE is constant

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

K E_{1} m_{1} = {K E}_{2} m_{2}

Where P is constant

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

K E_{1} P_{2} = {K E}_{2} P_{1}

Where m is constant

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

All the above

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

Open in App

Solution

The correct option is D All the above

Suggest Corrections

Similar questions

Q. Two bodies of masses

m_{1}

and

m_{2}

are moving with same kinetic energy. If

P_{1}

and

p_{2}

are their respective momentum, the ratio

\frac{P_{1}}{P_{2}}

is equal to:

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

In the relation Kinetic Energy = p²/2m (where p is momentum, m is mass ) . If Momentum is raised by 0.1 % what will be the percentage increase in Kinetic Energy

Q. Two bodies having kinetic energies