Question

the linear momentum of a body is increased by 200% calculate the percentage increase in kinetic energy

Answer 1

$\huge\orange{As-Salamu-Alykum}$

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$\huge\red{Kinetic energy is given by :- }$
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K.E = P^2/2m ---------------(1)

Now new Momentum is :-
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P'= 200% P + P

P'= 200/100P + P

P'= 2P +P

P'= 3P

So new kinetic energy will be :-

K.E '= P' /2m ----------------------(2)

Now ..........

Divide equation 2 by 1

K.E'/K.E = P'^2/2m ÷ P^2/2m

Or

K.E'/K.E = P'^2/2m × 2m/P^2

K.E'/K.E = 3P^2/2m ×2m/P^2

P^2 cancel P^2 , and 2m will cancel 2m

So...........

K.E'/K.E = 3^2

K.E'/K.E = 9

K.E' = 9K.E

So.......................

$\huge\blue{ New Momentum is 9K.E }$

$\huge\pink{Fraction change in Kinetic energy is }$
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Fraction change in kinetic energy = K.E'_K.E /K.E

Fraction change in Kinetic energy = 9K.E _K.E/K.E

Fraction change in Kinetic energy = 8K.E /K.E

$\huge\green{Fraction change in kinetic energy = 8 }$

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$\huge\pink{INSHALLAH it will help U}$

Answer 2

Have a NYC tym.........