Question

If Velocity is increased by 100%
then what is percentage
increase in Kinetic energy
a) 100%
b) 200%
. C) 300%
d) 400%​

Answer 1

Kinetic energy = p^2/2m

If the momentum increases by 100% , then the p increases to 2p.

Substitute p =1

K before increase in momentum = p^2/2m = 1/2m - (1)

K' after increase in momentum by 100% = 2p^2/2m= 4/2m - (2)

Let p = 1

Divide (2) by (1) and substitute p=1

K'/K = (4/2m) / (1/2m)

= 4 ( 2m gets divided by 2m to give 1)

K'= 4K ( cross-multiplication)

Percentage increase = (K'-K / K) * 100

= (4K-K/K) *100

= (3K /K) *100

= 3*100

= 300%

When momentum of a body increases by 100% then its kinetic energy will increase by 300%.

Answer 2

The correct option regarding the percentage increase in Kinetic energy is C) 300%.

Given:

Velocity is increased by 100%.

To Find:

The percentage increase in Kinetic energy.

Solution:

To find the percentage increase in Kinetic energy, we will follow the following steps:

As we know,

Velocity is increased by 100% which means velocity is doubled.

Now,

Initial kinetic energy (k1) =

$\frac{1}{2} m {v}^{2}$

Here, m is the mass and v is the velocity.

Now,

Initial kinetic energy when velocity is double of initial velocity (k2) =

$\frac{1}{2} m \times {(2v)}^{2} = \frac{1}{2} m {v}^{2} \times 4 = 4k1$

Now,

Percentage change in kinetic energy =

$\frac{k2 - k1}{k1} \times 100 = \frac{4k1 - k1}{k1} \times 100 = 300\%$

Henceforth, The correct option regarding the percentage increase in Kinetic energy is C) 300%.

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