Question

KE of a body is increased by 300%. Find the % incerase in momentum

Answer 1

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Answer 2

Answer:  To solve typical questions like this

We have to get 2 equations for (initial momentum) and for (final momentum)

Explanation:

momentum equation we using here is

     $p=\sqrt{2mE}$

P = momentum

m = mass

e = kinetic energy

First, let's take the equation for initial momentum

so the initial is

$p_{1} =\sqrt{2mE_{1} }$

Then let us take an equation for final momentum

special point; if they increase by 30% you have to add a 100 therefore it's become 130%

more example;   increase by 50% =  150%

                             increase by 160%= 260%

[so here they had given a percentage of 300% but we add an additional 100% and make it 400%]

so the final momentum is     $p2=\sqrt{2mE_{2} }\\ \\E_{2} = \frac{400E1}{100} \\ \\E_{2} = 4E1\\\\p2=\sqrt{2mE_{2} } \\ p2=\sqrt{2m4E1 }\\p2=\sqrt{8ME1 }$

THE DIVIDE FINAL MOMENTUM BY INITIAL

$\frac{P2}{P1} = \frac{\sqrt{8ME1} }{\sqrt{2ME1} } = 2$

AND FINALLY TO FIND THE PERCENTAGE

We subtract the final momentum and initial momentum and whole thing into 100%

(p2 - p1)×100

$(\frac{p2}{p1} - 1 )$ ×100

(2 - 1)×100

100%