A body of mass m moving with velocity v has k | vedclass

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

Question

(B) The expression for kinetic energy,$K = \frac{1}{2} mv^2$,is derived from classical mechanics (Newtonian mechanics).In classical mechanics,the mass

Vedclass · Accepted Answer

The body moves with a velocity negligible compared to the speed of light.

Vedclass · Answer

(B) The expression for kinetic energy,$K = \frac{1}{2} mv^2$,is derived from classical mechanics (Newtonian mechanics).In classical mechanics,the mass $m$ of an object is considered a constant,independent of its velocity.However,according to Einstein's theory of special relativity,as an object's velocity $v$ approaches the speed of light $c$,its relativistic mass increases according to the formula $m = \frac{m_0}{\sqrt{1 - v^2/c^2}}$.Therefore,the classical formula $K = \frac{1}{2} mv^2$ is only valid when the velocity $v$ is much smaller than the speed of light $(v \ll c)$,meaning the velocity is negligible compared to the speed of light.

$A$ body of mass $m$ moving with velocity $v$ has kinetic energy equal to $\frac{1}{2} mv^2$. This statement is valid when:

Similar Questions

$A$ running man has half the kinetic energy of a boy who has half the man's mass. The man speeds up by $1 \, m/s$ so that his kinetic energy becomes equal to that of the boy. The original speed of the man is:

Two particles have masses in the ratio $1 : 2$. If they have the same kinetic energy, what is the ratio of their momenta?

If the kinetic energy of an object is increased by $300 \%$,find the percentage $(\%)$ increase in its linear momentum.

In the Olympic games,an athlete covers a distance of $100 \ m$ in $10 \ s$. The estimated range of their kinetic energy is:

$A$ body initially at rest is acted upon by a constant force $F$ for time $t$. The kinetic energy at time $t$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module