Question

If momentum of an object is increased by 10% then its kinetic energy will increase by
(1) 20%
(3) 40%
(2) 21%
(4) 19%
TL​

Answer 1

Answer:

Let the initial mass of the body be 'm' and it's velocity be 'v'. 

Initial momentum of the body = mv 

$\boxed{\sf{Initial\:KE = \frac{ {mv}^{2} }{2}}}$

Now:

If the momentum increases by 10%, new momentum:

$\implies$ $\boxed{\sf{mv + 10 \%\:of \:mv}}$

So:

$\implies$ $\sf{mv + \frac{mv}{10}}$

$\implies$ $\sf{\frac{11mv}{10}}$

$\implies$ $\sf{m( \frac{11v}{10} )}$

Here:

New velocity = $\sf{\frac{11v}{10}}$

Increase in velocity = $\sf{\frac{v}{10}}$

And:

New kinetic energy:

$\implies$ $\sf{ \frac{m( \frac{11v}{10})^{2} }{2}}$

$\implies$ $\sf{ \frac{ \frac{121}{100mv ^{2} } }{2} }$

Increase in KE = 21%

Therefore:

Correct option: (2) 21%

Answer 2

Answer:-

21 %

Option → 3

Given :-

If the momentum of an object is increased by 10 %.

To find :-

The change in kinetic energy.

Solution:-

Let the body of mass m moving with velocity v.

The momentum of the body is :-

$P = mv$

The kinetic energy of the body is :-

$K_e = \dfrac{1}{2}mv^2$

• After increasing its momentum up to 10%.

It's new momentum will be :-

→$P' = mv + 10\% of mv$

→$P' = mv + \dfrac{10}{100}mv$

→$P' = mv + \dfrac{mv}{10}$

→$P' = \dfrac{10mv+mv}{10}$

→$P' = \dfrac{11mv}{10}$

→$P' = m \dfrac{11v}{10}$

• After increasing the momentum of the body . It's velocity is increased by v/10.

Now,

New kinetic energy will be :-

→$K_e' = \dfrac{1}{2}m (\dfrac{v}{10}) ^2$

→$K_e' = \dfrac{1}{2}m \dfrac{v^2}{100}$

→$K_e' = \dfrac{121}{100}mv^2$

Increase in kinetic energy :-

→$K_e' - K_e$

→$\dfrac{121mv^2}{100}-\dfrac{mv^2}{2}$

→$\dfrac{121mv^2 -100mv^2}{100}$

→$\dfrac{21mv^2}{100}$

→$21 \%$

hence,

The new kinetic energy is increased by 21% times than previous kinetic energy.