Question

If KE is increased by 800% then what will be the % change in P?
1. 100
2. 200
4. 300
4. 400​

Answer 1

P1,P2 momentum before and after let orginal K.E be 100 J

K.E increased by 300%

new KE=400J

P22P12=E2E1=400/100=4

P2P1=2=>P2=2P1

Increase % =P2−P1P1×100

=2P1−P1P1×100

=100%

Hence a is the correct answer

Answer 2

Answer:

Option 2 is the correct answer

Explanation:

Given

Kinetic Energy is increased by 800%

We need

%Change in P = ?????

We know that

$\boxed{ \%P = ( \frac{P_{final} - P_{initial} }{P_{initial}} ) \times 100}$

Now

According to Question

$=>KE_{final} = KE_{initial} + \frac{800}{100} \: KE_{initial} \\ = > KE_{final} = \frac{100KE_{initial} + 800KE_{initial}}{100} \\ = > KE_{final} = \frac{ \cancel900KE_{initial}}{ \cancel100} \\ = > \boxed{KE_{final} = 9KE_{initial}}$

Now

$= > ( { \frac{P_{final}}{P_{initial}} })^{2} = \frac{KE_{final}}{KE_{initial}} \\ = > ( { \frac{P_{final}}{P_{initial}} })^{2} = \frac{9}{1} \\ = > \boxed{\frac{P_{final}}{P_{initial} } = \frac{3}{1} }$

$= > \%P = ( \frac{P_{final}}{P_{initial} } - 1) \times 100 \\ = > \%P = (3 - 1) \times 100 \\ = > \%P = 2 \times 100 \\ = > \huge \boxed{ \%P = 200}$