Question

The kinetic energy of a body increased by $21\% $ .What is the percentage increase in the linear momentum of the body?
$\left( A \right)20\% $
$\left( B \right)10\% $
$\left( C \right)30\% $
$\left( D \right)5\% $

Answer 1

Hint: We can calculate the percentage increase by using the kinetic energy and linear momentum inter relation. First calculate the velocity of the body with the help of kinetic energy then using that velocity values we can find the linear momentum after this we can calculate the percentage increase.

Formula used:
$KE = \dfrac{1}{2}m{v^2}$
$p = mv$
Where,
${\text{KE = Kinetic}}\,\,{\text{energy}}$
${\text{m = Mass}}\,{\text{of}}\,{\text{the}}\,{\text{body}}$
${\text{v = Velocity}}$
${\text{p = Momentum}}$

Complete step-by-step answer:
Kinetic Energy is defined as the multiplication of half of the mass of the body with the square of the velocity of the body whereas momentum of a body is defined as the product of mass and velocity of the body
As per the given problem
The kinetic energy of the body is increased by $21\% $ .
Now we can write the above statement in mathematical form as,
$K{E_1} = \dfrac{1}{2}m{v^2}$
Now the new kinetic energy after $21\% $ increment we get,
$K{E_2} = \left( {1 + 21\% } \right)K{E_1}$
Here,
$K{E_2} = New\,kinetic\,energy\,after\,increment$
Thus,
$K{E_{.2}} = \dfrac{{121}}{{100}}K{E_1}$
As there is a change in the kinetic energy the velocity of the body must changes up to some extend to meet this increased kinetic energy.
Now by using kinetic energy formula with changed velocity we will get,
$\dfrac{1}{2}m{v_2}^2 = \dfrac{{121}}{{100}} \times \dfrac{1}{2}m{v_1}^2$
By cancelling the common terms we will get,
${v_2}^2 = \dfrac{{121}}{{100}}{v_1}^2$
By taking the square to other side we will get,
${v_2} = \sqrt {\dfrac{{121}}{{100}}{v_1}^2} $
$ \Rightarrow {v_2} = \dfrac{{11}}{{10}}{v_1}$
Now putting this ${v_2}\,,{v_1}$ relation in linear momentum formula we will get,
 $m{v_2} = \dfrac{{11}}{{10}}m{v_1}$
$ \Rightarrow {p_2} = \dfrac{{11}}{{10}}{p_1}$
Taking ${p_1}$ to other side we will get,
$\dfrac{{{p_2}}}{{{p_1}}} = \dfrac{{11}}{{10}}$
We know that
Change in linear momentum divided by initial linear momentum whole multiplied by $100$ gives the increased percentage of the linear momentum.
Apply this above statement in the above momentum ratio we will get,
$\dfrac{{{p_2}}}{{{p_1}}} = \dfrac{{11}}{{10}}$
Subtracting both the sides with one we will get,
$\dfrac{{{p_2}}}{{{p_1}}} - 1 = \dfrac{{11}}{{10}} - 1$
$ \Rightarrow \dfrac{{{p_2} - {p_1}}}{{{p_1}}} = \dfrac{{11 - 10}}{{10}}$
Now multiply both side with $100$ we will get,
$\dfrac{{{p_2} - {p_1}}}{{{p_1}}} \times 100 = \dfrac{1}{{10}} \times 100$
$ \Rightarrow Percentage\,\,Increase\,in\,Linear\,Momentum = 10\% $
So the percentage increase in the magnitude of linear momentum is $10\% $ with the kinetic energy of a body increased by $21\% $ .

So, the correct answer is “Option B”.

Note: Apply the proper formula of kinetic energy and linear momentum or else you will not be able to find a relationship between kinetic energy and linear momentum. And also keep in mind that after the change in kinetic energy the velocity of the body must be changed but the mass will remain as the change is produced in the same body.