Question

Can a body have energy without having momentum and have momentum without having energy? Explain.
The kinetic energy of a body is increased by 21%. What is the percentage increase in the line momentum of the body?

Answer 1

Hint: When a body of mass m is in motion, it possesses an energy called kinetic energy given as $\dfrac{1}{2}m{{v}^{2}}$, where v is the velocity of the body. The momentum of the body is the product of its mass and its velocity i.e. p=mv.
Formula used:
$E=\dfrac{1}{2}m{{v}^{2}}$
p=mv

Complete step-by-step answer:

Momentum of a body is the product of its mass and its velocity. Suppose a body of mass m is moving with a velocity of v. Then its momentum (p) is given as p=mv. Momentum is a vector quantity. Its magnitude is equal to mv and its direction is along the direction of velocity of the body.
Energy of a body is the amount of work done on the body. The most general type of energy is mechanical energy. Mechanical energy consists of kinetic energy and potential energy.
Kinetic energy is the energy possessed by a body when it is in motion. It depends on the mass of the body and square of its speed. Hence, kinetic energy of a body is given as $\dfrac{1}{2}m{{v}^{2}}$, where m is mass of the body and v is velocity of the body.
Potential energy of a body is the amount of work done by an external force to bring a body from infinity into a field where the body is affected by another force. The expression for potential energy depends on the field force acting on it.
Now, if we consider the kinetic energy of a body.
Now we can write the equation for kinetic energy as $E=\dfrac{1}{2}m{{v}^{2}}$…….(i)
Multiply both sides of the equation by m.
$\Rightarrow mE=\dfrac{1}{2}{{m}^{2}}{{v}^{2}}$
We know that mv is the momentum (p) of the body.
Therefore, $mE=\dfrac{1}{2}{{p}^{2}}$.
$\Rightarrow E=\dfrac{{{p}^{2}}}{2m}$.
This implies that if a body has a momentum that it will possess kinetic energy.
However, the converse is not true. A body having some energy may or may not have a momentum. A body at rest may have a potential energy.
Let us solve the problem given in the question. For this, we can use the relation between the momentum and kinetic energy of a body i.e. $E=\dfrac{{{p}^{2}}}{2m}$ ……(ii)
Let the initial energy of the body be E and its initial momentum be p. It is given that kinetic energy increases by 21%. Let the final energy be $E'=\dfrac{121}{100}E$ and final momentum be p’.
Therefore, $E'=\dfrac{{{(p')}^{2}}}{2m}$
Therefore, $\dfrac{E'}{E}=\dfrac{\dfrac{{{(p')}^{2}}}{2m}}{\dfrac{{{p}^{2}}}{2m}}=\dfrac{121}{100}$
$\Rightarrow \dfrac{{{(p')}^{2}}}{{{p}^{2}}}=\dfrac{121}{100}$
$\Rightarrow {{\left( \dfrac{p'}{p} \right)}^{2}}=\dfrac{121}{100}$
$\Rightarrow \left( \dfrac{p'}{p} \right)=\sqrt{\dfrac{121}{100}}=\dfrac{11}{10}$
Therefore, the percentage increase in the momentum of the body is $\dfrac{1}{10}\times 100=10\%$.

Note: Momentum tells us about the force required to stop a body moving with velocity v, in a given amount of time. More the value of the momentum of the body, more the amount of force needed to stop the body.
Larger momentum also indicates more energy possessed by the body. Therefore, more work is to be done on the body to bring it to rest.