Question

the kinetic energy of a body is increased by 100% then by what percent its Momentum will increase​

Answer 1

Answer:

Given:

KE is increased by 100%

To find:

Percentage change in Momentum.

Formulas used:

$KE = \frac{ {p}^{2} }{2m}$

$= > p = \sqrt{2m(KE)}$

"p" is momentum , KE is Kinetic energy.

Calculation:

Let initial momentum be p1

$= > p1 = \sqrt{2m(KE)}$

Now, New Kinetic energy is {KE(200/100)}

as it is increased by 100%.

Let new momentum be p2.

$= > p2 = \sqrt{2m(KE \times \frac{200}{100} )}$

$= > p2 = \sqrt{2m(KE \times 2)}$

$= > p2 = \sqrt{2 } \times \sqrt{2m(KE )}$

$= > p2 = \sqrt{2 } \times p1$

Hence :

Change in Momentum :

∴ ∆p = p2 - p1

=> ∆p = (√2) p1 - p1

=> ∆p = (√2 - 1)p1

=> ∆p =( 0.41)p1

% change in Momentum:

∴ % change = ∆p/p × 100%

=> % change = (0.41)p/p × 100%

=> % change = 41 %

So final answer is 41%

Answer 2

Hey!

good question ☺

☆Refer to the attachment please