if the k.E is increase by 800 percentage by what percentage will linear momentum of the body increase?
Answers
Answer:
200%
Explanation:
Hope it will be helpful to you..
Given :
▪ Percentage increase in KE = 800%
To FinD :
▪ Percentage increase in linear momentum of body.
SoluTioN :
↗ Relation b/w kinetic energy of moving body and linear momentum is given by
\bigstar\:\underline{\boxed{\bf{\blue{K=\dfrac{P^2}{2m}}}}}★
K=
2m
P
2
Here, mass remains constant so we can say that
\therefore\bf\:K\propto P^2∴K∝P
2
When the kinetic energy of a body is increased by 800%, its kinetic energy will become
\begin{lgathered}\dashrightarrow\sf\:K'=K+\dfrac{800}{100}K\\ \\ \dashrightarrow\sf\:K'=\dfrac{900}{100}K\\ \\ \dashrightarrow\bf\:\red{K'=9K}\end{lgathered}
⇢K
′
=K+
100
800
K
⇢K
′
=
100
900
K
⇢K
′
=9K
New momentum will become,
\begin{lgathered}\Rightarrow\sf\:\dfrac{{P'}^2}{P^2}=\dfrac{K'}{K}\\ \\ \Rightarrow\sf\:\dfrac{{P'}^2}{P^2}=\dfrac{9K}{K}\\ \\ \Rightarrow\sf\:\dfrac{P'}{P}=\sqrt{9}\\ \\ \Rightarrow\bf\:\green{P'=3P}\end{lgathered}
⇒
P
2
P
′
2
=
K
K
′
⇒
P
2
P
′
2
=
K
9K
⇒
P
P
′
=
9
⇒P
′
=3P
Percentage increase in the momentum of the body is given by
\begin{lgathered}\twoheadrightarrow\sf\:\dfrac{P'-P}{P}\times 100=\dfrac{3P-P}{P}\times 100\\ \\ \twoheadrightarrow\sf\:\%\dfrac{\Delta P}{P}=\dfrac{2P}{P}\times 100\\ \\ \twoheadrightarrow\underline{\boxed{\bf{\gray{\%\dfrac{\Delta P}{P}=200\%}}}}\:\orange{\bigstar}\end{lgathered}
↠
P
P
′
−P
×100=
P
3P−P
×100
↠%
P
ΔP
=
P
2P
×100
↠
%
P
ΔP
=200%
★