
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
507.5k+ 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.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

