
The kinetic energy of a body is increased by 100% then the percentage change in
momentum of the body is?
A. 4.14%
B. 41.4%
C. 141.4%
D. none of these
Answer
418.7k+ views
Hint:Kinetic energy is the energy possessed by a moving body and the product of its mass and velocity is defined to be its momentum. Write the expressions for the kinetic energy and momentum of a body and find a relation between the two quantities. The use the given data and find the percentage change in momentum.
Formula used:
$K=\dfrac{1}{2}m{{v}^{2}}$
$P=mv$
Complete step by step answer:
When a body is in motion we say that it possesses some energy. This energy of the body is called the kinetic energy of the body.
If the body of mass m is moving with a speed v then it possesses a kinetic energy equal to
$K=\dfrac{1}{2}m{{v}^{2}}$ …. (i)
We also define another term for a moving body and that is momentum of the body. When the body of mass m is moving with a velocity whose magnitude is v, the magnitude of its momentum is given
as $P=mv$ … (ii)
From equation (ii) we get that $v=\dfrac{P}{m}$,
Substitute this value of v in equation (i).
$\Rightarrow K=\dfrac{1}{2}m{{\left( \dfrac{P}{m} \right)}^{2}}$
$\Rightarrow K=\dfrac{{{P}^{2}}}{2m}$ …. (iii)
Now, it is given that the kinetic energy of the body is increased by 100%. This means that the new kinetic energy of the body is 2K.
Let the new momentum of the body be P’.
Then $\Rightarrow 2K=\dfrac{P{{'}^{2}}}{2m}$ …. (iv).
Now, divide (iv) by (iii).
$\Rightarrow \dfrac{2K}{K}=\dfrac{\dfrac{P{{'}^{2}}}{2m}}{\dfrac{{{P}^{2}}}{2m}}$
$\Rightarrow 2=\dfrac{P{{'}^{2}}}{{{P}^{2}}}$
$\Rightarrow P{{'}^{2}}=2{{P}^{2}}$
$\Rightarrow P'=\pm \sqrt{2}P$
We shall not consider the negative value because we need the magnitude of the momentum.
$\Rightarrow P'=\sqrt{2}P$.
The percentage increase in the momentum of the body is $\dfrac{P'-P}{P}\times
100=\dfrac{\sqrt{2}P-P}{P}\times 100=\left( \sqrt{2}-1 \right)\times 100=41.1$%.
Hence, the correct option is B.
Note:Note that kinetic energy is scalar quantity and momentum is a vector quantity. However, in this solution we understand that the magnitude of momentum and kinetic energy are related.
We can say that momentum of a body tells us how much kinetic energy the body possesses.
Formula used:
$K=\dfrac{1}{2}m{{v}^{2}}$
$P=mv$
Complete step by step answer:
When a body is in motion we say that it possesses some energy. This energy of the body is called the kinetic energy of the body.
If the body of mass m is moving with a speed v then it possesses a kinetic energy equal to
$K=\dfrac{1}{2}m{{v}^{2}}$ …. (i)
We also define another term for a moving body and that is momentum of the body. When the body of mass m is moving with a velocity whose magnitude is v, the magnitude of its momentum is given
as $P=mv$ … (ii)
From equation (ii) we get that $v=\dfrac{P}{m}$,
Substitute this value of v in equation (i).
$\Rightarrow K=\dfrac{1}{2}m{{\left( \dfrac{P}{m} \right)}^{2}}$
$\Rightarrow K=\dfrac{{{P}^{2}}}{2m}$ …. (iii)
Now, it is given that the kinetic energy of the body is increased by 100%. This means that the new kinetic energy of the body is 2K.
Let the new momentum of the body be P’.
Then $\Rightarrow 2K=\dfrac{P{{'}^{2}}}{2m}$ …. (iv).
Now, divide (iv) by (iii).
$\Rightarrow \dfrac{2K}{K}=\dfrac{\dfrac{P{{'}^{2}}}{2m}}{\dfrac{{{P}^{2}}}{2m}}$
$\Rightarrow 2=\dfrac{P{{'}^{2}}}{{{P}^{2}}}$
$\Rightarrow P{{'}^{2}}=2{{P}^{2}}$
$\Rightarrow P'=\pm \sqrt{2}P$
We shall not consider the negative value because we need the magnitude of the momentum.
$\Rightarrow P'=\sqrt{2}P$.
The percentage increase in the momentum of the body is $\dfrac{P'-P}{P}\times
100=\dfrac{\sqrt{2}P-P}{P}\times 100=\left( \sqrt{2}-1 \right)\times 100=41.1$%.
Hence, the correct option is B.
Note:Note that kinetic energy is scalar quantity and momentum is a vector quantity. However, in this solution we understand that the magnitude of momentum and kinetic energy are related.
We can say that momentum of a body tells us how much kinetic energy the body possesses.
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