Question

Kinetic
energy of
a particle is increased by 300%. Find the percentage increase in its momentum.
​

Answer 1

$\dag\:\underline{\sf AnsWer :}$

• Before going to solve this numerical we should know the relationship between momentum (p) and Kinetic energy (K). So first we will see the relationship between Kinetic energy and momentum :

$:\implies \sf K = \dfrac{p^2}{2m}$

$:\implies \sf {p}^{2} = 2mK$

$:\implies \underline{ \boxed{\sf p = \sqrt{2 mK} }}$

• Now we are given that Kinetic energy of a particle is increased by 300%. Let new Kinetic energy as K'

$\dashrightarrow\:\:\sf K'= K + \dfrac{300}{100} K$

$\dashrightarrow\:\:\sf K' = K + 3 K$

$\dashrightarrow\:\: \underline{ \boxed{\sf K'= 4K }}$

• Now we will find new momentum of the particle. Let new momentum of the particle be p' :

$\longrightarrow\:\:\sf p' = \sqrt{2 mK}$

$\longrightarrow\:\:\sf p' = \sqrt{2 mK'}$

$\longrightarrow\:\:\sf p' = \sqrt{2 m(4K)}$

$\longrightarrow\:\:\sf p' = \sqrt{8 mK}$

$\longrightarrow\:\:\sf p' = \sqrt{2 \times 2 \times 2 \times mK}$

$\longrightarrow\:\:\sf p' = 2\sqrt{ 2 mK}$

$\longrightarrow\:\: \underline{ \boxed{\sf p' = 2p}}$

Now, let's find the percentage increase in the momentum :

$\leadsto\sf \dfrac{p' - p}{p} \times 100$

$\leadsto\sf \dfrac{2p- p}{p} \times 100$

$\leadsto\sf \dfrac{ p}{p} \times 100$

$\leadsto \underline{ \boxed{\sf 100 \: \%}}$