Question

if the kinetic energy of bodu increased by 300% its momentum is increased by​

Answer 1

Answer:

41.42%

Explanation:

Originally-

K.E = $\frac{1}{2} mv^{2}$

Momentum (P)= mv

Now K.E. is increased by 300%

We know that mass of a body will always remain constant.

So only the velocity of the body will change.

So, new K.E. (KE') = K.E. + 300% of K.E.= KE + $\frac{300}{100}$* KE= 4KE

Therefore KE' = 4KE

so, $\frac{1}{2} mv'^{2}$ = 4* $\frac{1}{2} mv^{2}$

There fore, $v'^{2}$ = 2$v^{2}$

v' = $\sqrt{2}$v

So New momentum (P')=  mv'

P' =$\sqrt{2}$mv

So increase in momentum = P'-P

= $\sqrt{2}$mv- mv= ($\sqrt{2}$-1) mv

Increase % ={ ($\sqrt{2}$-1)mv/mv}* 100

Increase %= 41.42%

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Answer 2

$KE = \frac{1}{2} MV^{2}$  .................. 1

Momentum (P) = Mass × Velocity ....... 2

so if KE increase its due to Velocity so

if KE + 3KE then it will be 4 times of its original value

So in equation ( 1 ) Velocity will become 2V

So Momentum will increase by 100%

using equation (2)