Question

When the linear momentum of a particle is increased by 1% it's kinetic energy increased by x% . when the kinetic energy of the particle is increased by 300%, it's linear momentum increased by y% . the ratio of y to x ​

Answer 1

In 1st case:

• Linear momentum is increased by 1% and KE gets increased x%.

$KE = \dfrac{{P}^{2}}{2m}$

Now, new KE :

$\implies \: KE2 = \dfrac{{(P \times 101\%)}^{2} }{2m}$

$\implies \: KE2 = \dfrac{{P }^{2} }{2m} \times 1.02$

So, % increase in KE is :

$= \dfrac{(1.02 - 1) \times \frac{ {P}^{2} }{2m} }{ \frac{ {P}^{2} }{2m} } \times 100\%$

$= \dfrac{0.2}{ 1} \times 100\%$

$= 20\%$

So, value of x is 20 !

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In 2nd case

• KE increased by 300% and linear momentum increases by y%.

$KE = \dfrac{{P}^{2}}{2m}$

$\implies \: P = \sqrt{2m \: KE}$

New linear momentum will be :

$\implies \: P2 = \sqrt{2m \: KE \times \dfrac{400}{100} }$

$\implies \: P2 = \sqrt{2m \: KE \times 4}$

$\implies \: P2 = 2 \times \sqrt{2m \: KE}$

$\implies \: P2 = 2 \times P$

So, % increase of linear momentum:

$= \dfrac{(2 - 1)P}{P} \times 100\%$

$= 100\%$

So, value of y = 100.

Required ratio = y : x = 100 : 20 = 5 : 1.