The-kinetic-energy-of-a-body-increases-by-100-Find-the-increase-in-its-momentum-please-solve-with-explanations-where-necessary-Thanks-

Question Number 21490 by NECx last updated on 24/Sep/17

The kinetic energy of a body

increases by 100 % . Find the %

increase in its momentum .

please solve with explanations

where necessary . Thanks .

Answered by ajfour last updated on 25/Sep/17

\frac{p_{1}^{2}}{2 m} = K_{1}

\frac{p_{2}^{2}}{2 m} = K_{2}

\frac{K_{2} - K_{1}}{K_{1}} \times 100 = 100

\Rightarrow \frac{p_{2}^{2} - p_{1}^{2}}{p_{1}^{2}} \times 100 = 100

{(\frac{p_{2}}{p_{1}})}^{2} - 1 = 1 \Rightarrow {(\frac{p_{2}}{p_{1}})}^{2} = 2

% c h a n g e i n m o m e n t u m i s

\frac{p_{2} - p_{1}}{p_{1}} \times 100 = (\sqrt{2} - 1) \times 100

= 41 . 4 % .

u n l e s s t h e c h a n g e i s d i f f e r e n t i a l

c h a n g e y o u c a n n o t d i f f e r e n t i a t e

t o g e t t h e c o r r e c t a n s w e r .

Commented by NECx last updated on 25/Sep/17

Mr Ajfour please this solution

isnt explanatory to me . help

simplify .

Commented by NECx last updated on 25/Sep/17

i did it this way so i want to see

whats wrong;

let k = kinetic energy, m = mass

v = velocity, ρ = momentum

k = \frac{1}{2} {mv}^{2} \dots \dots (1)

ρ = mv

hence; v = ρ / m \dots \dots (2)

put (2) into (1)

k = \frac{ρ^{2}}{2 m}

Δ k = \frac{ρ}{m} Δ ρ

since Δ k = k

Δ k = k = \frac{ρ}{m} Δ ρ

\frac{ρ^{2}}{2 m} = \frac{ρ}{m} Δ ρ

∴ \frac{Δ ρ}{ρ} = \frac{1}{2}

∴ % Δ ρ = \frac{Δ ρ}{ρ} \times 100 = 50 %

whats wrong here ? ? ? ? ? ? ?

Commented by Joel577 last updated on 25/Sep/17

K = \frac{p^{2}}{2 m}

Let K_{2} = 2 K_{1}

\frac{K_{1}}{K_{2}} = \frac{\frac{p_{1}^{2}}{2 m}}{\frac{p_{2}^{2}}{2 m}} \to \frac{K_{1}}{2 K_{1}} = \frac{p_{1}^{2}}{p_{2}^{2}} \to p_{2}^{2} = 2 p_{1}^{2}

p_{2} = (\sqrt{2}) p_{1}

% increase = \frac{Δ p}{p_{1}} . 100 % = 41.4 %

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