Kinetic energy (K) depends on momentum (P) as | vedclass

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

Question

(C) The relationship between kinetic energy $(K)$ and momentum $(P)$ is given by $K = \frac{P^2}{2m}$.$1$. For the graph between $K$ and $P$,the equat

Vedclass · Accepted Answer

$A$ and $C$

Vedclass · Answer

(C) The relationship between kinetic energy $(K)$ and momentum $(P)$ is given by $K = \frac{P^2}{2m}$.$1$. For the graph between $K$ and $P$,the equation is $K = \frac{1}{2m} P^2$. This is of the form $y = ax^2$,which represents a parabola opening upwards. Thus,graph $(A)$ is correct.$2$. For the graph between $K$ and $P^2$,let $X = P^2$. Then the equation becomes $K = \frac{1}{2m} X$. This is of the form $y = mx$,which represents a straight line passing through the origin. Thus,graph $(C)$ is correct.Therefore,graphs $(A)$ and $(C)$ are correct.

Kinetic energy $(K)$ depends on momentum $(P)$ as $K = \frac{P^2}{2m}$. Which of the following graphs is/are correct?

Similar Questions

$A$ ball of mass $2\,kg$ and another of mass $4\,kg$ are dropped together from a $60\,ft$ tall building. After a fall of $30\,ft$ each towards the earth,their respective kinetic energies will be in the ratio of:

$A$ horizontal force of $5 \text{ N}$ is applied on a body of mass $5 \text{ kg}$, which is initially at rest on a frictionless table. What is the change in kinetic energy of the body after $10 \text{ s}$?

Which of the following graphs represents the graphical relation between momentum $(p)$ and kinetic energy $(K)$ for a body in motion?

When kinetic energy of a body becomes $36$ times of its original value,the percentage increase in the momentum of the body will be : (in $\%$)

$A$ bullet of mass $200\,g$ having initial kinetic energy $90\,J$ is shot inside a long swimming pool as shown in the figure. If its kinetic energy reduces to $40\,J$ within $1\,s$,the minimum length of the pool the bullet has to travel so that it completely comes to rest is $.....m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module