Question

if the kinetic energy of a moving body of linear momentum p is doubled the final momentum of the body will be? answer with explanation​

Answer 1

Explanation:

K.E = 1/2  mv^2

      = 1/2  (mv)^2 /m

      = 1/2   p^2/m    { mv = p}

K.E = p^2 /2m ⇒ √2mK.E = p

 

     when K E is doubled.

√2m * 2K.E = √2*2mKE

                   = √2*√(2mKE)

                   = √2  * p

momentum is now √2 times of the initial momentum.

Answer 2

Given:

• Linear momentum= p
• Kinetic Energy is doubled

Need to Find:

• The final momentum= ?

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We know,

K.E

$= \dfrac{1}{2}m {v}^{2}$

$= \dfrac{1}{2} \times m {v}^{2} \times \dfrac{m}{m}$

$= \dfrac{1}{2} \dfrac{{m}^{2} \times {v}^{2}}{m}$

     

$= \dfrac{1}{2} \dfrac{ \times {p}^{2}}{m}$

Thus,

K.E= $\dfrac{1}{2} \dfrac{ \times {p}^{2}}{m}$

==>$p = \sqrt{2 \times m \times ke}$

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When K E is doubled p2:

$= \sqrt{2m\times 2ke}$

$= \sqrt{2} \times \sqrt{2mke}$

$= \sqrt{2} \times p$

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The Final Momentum is $\red{ \sqrt{2} }$times of initial momentum