Question

9.) If kinetic energy of body decreases by 36%then percentage change in its linear momentum will be
(1) +20% 2) -20%
(3) +18% (4) -18% ​

Answer 1

Answer:

hey mate your answer is

Explanation:

K= p /2m

p= square root.2mK

decrease in Kinetic energy= 36% of K

i.e 36/100 * K = 0.36k

Final K.E , K'= 0.64

Final momentum, p' = square rt. 2mK' = sqr. rt. 2m x 0.64

= 0.64 square root 2mK = 0.8

increase in p

p-p'/p x 100 = p1.2p/p x 100 = 20%.

plz mark the brilliant answer thank you.

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Let the initial Kinetic energy be "K".
• Let the Final Kinetic energy be "K' ".
• Initial momentum be "P".
• Final momentum be "P' ".
• Kinetic energy is decreased by 36%.

$\huge{\bold{\underline{Explanation:-}}}$

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As question said that 36% is decreased,

Then,

$\large{\tt Final \: K.E - Initial \: K.E = - 36 \% K}$

$\large{\tt K' - K = - \dfrac{36}{100}K}$

[Here negative sign indicates that the Kinetic energy decreases]

$\large{\tt K' - K = - 0.36K}$

$\large{\tt K' = K - 0.36K}$

$\large{\boxed{\tt K' = 0.64K}}$

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As we know, the momentum and , Kinetic energy relation,

$\large{\boxed{K = \dfrac{P^2}{2m}}}$

Now,

$\large{\tt P^2 = 2mK}$

$\large{\boxed{\tt P = \sqrt{2mK}}}$

This is the initial momentum of the body,

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Now,Final momentum,

$\large{\tt P' = \sqrt{2mK'}}$

Substituting the values of K',

$\large{\tt P' = \sqrt{2m \times 0.64K}}$

$\large{\tt P' = \sqrt{0.64 \times 2mK}}$

$\large{\tt P' = 0.8 . \sqrt{2mK} \: \: \: \: [\sqrt{0.64} = 0.8]}$

As we know, ${\tt P = \sqrt{2mK}}$

Substituting it

$\large{\tt P' = 0.8 P}$

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Now, Percentage change,

$\large{\boxed{\tt Change \: \% = \dfrac{Final \ Momentum - Initial \: Momentum}{Initial \: momentum} \times100}}$

Substituting the values,

$\large{\tt Change \: \% = \dfrac{P' - P}{P} \times 100}$

$\large{\tt Change \: \% = \dfrac{0.8P - P}{P} \times 100}$

$\large{\tt Change \: \% = \dfrac{-0.2P}{P} \times 100}$

$\large{\tt Change \: \% = \dfrac{-0.2\cancel{P}}{\cancel{P}} \times 100}$

It comes as,

$\large{\tt Change \: \% = - 0.2 \times 100}$

$\huge{\boxed{\boxed{\tt Change\: \% = - 20\: \%}}}$

So, the change in momentum is - 20%[Option - (1)].

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Note:-

• ${\tt K = Initial \: Kinetic energy}$
• ${\tt K' = Final \: Kinetic \: energy}$
• ${\tt P = Initial \: Linear \: momentum}$
• ${\tt P' = Final \: Linear \: momentum}$