Answer
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Hint: Both kinetic energy and momentum are related to an object's velocity (or speed) and mass. Momentum is a vector quantity that defines how much mass is moving. Kinetic energy is a scalar representation of an object's energy from motion. We will compare both of these quantities and find a relationship among them.
Complete step by step solution:
It is given in the question that the body's kinetic energy increases to four times its original value.
Now, Let us assume \[m\] to be the body's mass and let $v$ reflect the speed at which the body moves. The body's kinetic energy is then calculated as follows:
$K.E = \dfrac{1}{2}m{v^2} ------ (i)$
The formula to calculate momentum is:
$
P = mv \\
v = \dfrac{P}{m} ------- (ii) \\
$
Now, substituting (ii) equation to (i) equation we will get,
$
K.E = \dfrac{1}{2}m{\left( {\dfrac{P}{m}} \right)^2} \\
\Rightarrow K.E = \dfrac{1}{2}{m}\left( {\dfrac{{{P^2}}}{{{m^{{2}}}}}} \right) \\
\Rightarrow K.E = \dfrac{1}{2}.\dfrac{{{P^2}}}{m} \\
$
Hence, the relation between Kinetic energy and momentum is
$
{P^2} = 2m.K.E \\
\Rightarrow P = \sqrt {2m.K.E} \\
$
So, the momentum is directly proportional to the square root of the kinetic energy.
If the kinetic energy of a body becomes $4$ times, then the momentum will be the square root of it , and we know that the square root of $4$ is $2$. Hence, the percentage increase in its momentum is 2 times.
Note:
Momentum is not the same as energy. Momentum and kinetic energy are terms related to object motion and there will be a shift in kinetic energy if there is a change in momentum but energy is a scalar quantity, while momentum is a vector quantity.
Complete step by step solution:
It is given in the question that the body's kinetic energy increases to four times its original value.
Now, Let us assume \[m\] to be the body's mass and let $v$ reflect the speed at which the body moves. The body's kinetic energy is then calculated as follows:
$K.E = \dfrac{1}{2}m{v^2} ------ (i)$
The formula to calculate momentum is:
$
P = mv \\
v = \dfrac{P}{m} ------- (ii) \\
$
Now, substituting (ii) equation to (i) equation we will get,
$
K.E = \dfrac{1}{2}m{\left( {\dfrac{P}{m}} \right)^2} \\
\Rightarrow K.E = \dfrac{1}{2}{m}\left( {\dfrac{{{P^2}}}{{{m^{{2}}}}}} \right) \\
\Rightarrow K.E = \dfrac{1}{2}.\dfrac{{{P^2}}}{m} \\
$
Hence, the relation between Kinetic energy and momentum is
$
{P^2} = 2m.K.E \\
\Rightarrow P = \sqrt {2m.K.E} \\
$
So, the momentum is directly proportional to the square root of the kinetic energy.
If the kinetic energy of a body becomes $4$ times, then the momentum will be the square root of it , and we know that the square root of $4$ is $2$. Hence, the percentage increase in its momentum is 2 times.
Note:
Momentum is not the same as energy. Momentum and kinetic energy are terms related to object motion and there will be a shift in kinetic energy if there is a change in momentum but energy is a scalar quantity, while momentum is a vector quantity.
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