If the linear momentum is increased by 5%,the | vedclass

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

Question

(C) The relationship between kinetic energy $E$ and linear momentum $p$ is given by $E = \frac{p^2}{2m}$.For small percentage changes,we can use the d

Vedclass · Accepted Answer

$10.25$

Vedclass · Answer

(C) The relationship between kinetic energy $E$ and linear momentum $p$ is given by $E = \frac{p^2}{2m}$.For small percentage changes,we can use the differential method: $\frac{\Delta E}{E} = 2 \left( \frac{\Delta p}{p} ight)$.Given $\frac{\Delta p}{p} = 5\% = 0.05$,the change in kinetic energy is $\frac{\Delta E}{E} = 2 	imes 5\% = 10\%$.Alternatively,using the exact formula: $E' = \frac{(1.05p)^2}{2m} = 1.1025 E$.The percentage increase is $(1.1025 - 1) 	imes 100 = 10.25\%$. Since $10.25\%$ is the precise value,we select the closest standard approximation or the exact value if provided.

If the linear momentum is increased by $5\%$,the kinetic energy will increase by.......$\%$

Similar Questions

Two bodies of unequal masses have the same momentum. Which one will have more kinetic energy?

$A$ 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:

If the increase in the kinetic energy of a body is $22\%$,then the increase in the momentum will be ........... $\%$

If $p$ is the linear momentum of an object and $v$ is its velocity,write the formula for its kinetic energy.

Define kinetic energy. Give its unit and dimensional formula and mention works by using it.

Vedclass Test Series

Exam Paper Generator

Online Exam Module