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Question

Question: The kinetic energy of a body is increased by 100% then the percentage change in momentum of the bo...

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

Explanation

Solution

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=12mv2K=\dfrac{1}{2}m{{v}^{2}}
P=mvP=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=12mv2K=\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=mvP=mv … (ii)

From equation (ii) we get that v=Pmv=\dfrac{P}{m},

Substitute this value of v in equation (i).
K=12m(Pm)2\Rightarrow K=\dfrac{1}{2}m{{\left( \dfrac{P}{m} \right)}^{2}}
K=P22m\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 2K=P22m\Rightarrow 2K=\dfrac{P{{'}^{2}}}{2m} …. (iv).

Now, divide (iv) by (iii).
2KK=P22mP22m\Rightarrow \dfrac{2K}{K}=\dfrac{\dfrac{P{{'}^{2}}}{2m}}{\dfrac{{{P}^{2}}}{2m}}
2=P2P2\Rightarrow 2=\dfrac{P{{'}^{2}}}{{{P}^{2}}}
P2=2P2\Rightarrow P{{'}^{2}}=2{{P}^{2}}
P=±2P\Rightarrow P'=\pm \sqrt{2}P

We shall not consider the negative value because we need the magnitude of the momentum.

P=2P\Rightarrow P'=\sqrt{2}P.

The percentage increase in the momentum of the body is PPP×100=2PPP×100=(21)×100=41.1\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.