An object has momentum p and kinetic energy E | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The relationship between kinetic energy $E$ and momentum $p$ for an object of mass $m$ is given by $E = \frac{p^2}{2m}$.When the momentum becomes

Vedclass · Accepted Answer

$4E$

Vedclass · Answer

(D) The relationship between kinetic energy $E$ and momentum $p$ for an object of mass $m$ is given by $E = \frac{p^2}{2m}$.When the momentum becomes $p' = 2p$,the new kinetic energy $E'$ is given by:$E' = \frac{(p')^2}{2m} = \frac{(2p)^2}{2m} = \frac{4p^2}{2m}$.Substituting $E = \frac{p^2}{2m}$ into the equation for $E'$,we get:$E' = 4 	imes \left( \frac{p^2}{2m} ight) = 4E$.Therefore,the new kinetic energy will be $4E$.

An object has momentum $p$ and kinetic energy $E$. If its momentum becomes $2p$,then its kinetic energy will be:

Similar Questions

Two bodies with kinetic energies in the ratio of $4 : 1$ are moving with equal linear momentum. The ratio of their masses is

The kinetic energy of a body moving along a straight line varies with time as shown in the figure. The force acting on the body is:

The kinetic energy acquired by a mass $m$ in travelling a certain distance $d$ starting from rest under the action of a constant force is directly proportional to

If the momentum of a body is increased by $100\%$,then the percentage increase in the kinetic energy is ............ $\%$

The kinetic energy possessed by a body of mass $m$ moving with a velocity $v$ is equal to $\frac{1}{2}mv^2$,provided

Vedclass Test Series

Exam Paper Generator

Online Exam Module