Question

Kinetic energy of a particle is increased by 300%. Find the percentage increase in its momentum.

Answer 1

Hint: When we say that the kinetic energy of the particle is increased by 300%, it means that the new kinetic energy is equal to 4 times the initial kinetic energy. Use the formulas of momentum (p=mv) and kinetic energy ($K=\dfrac{1}{2}m{{v}^{2}}$) and find relation between momentum and kinetic energy. Then using this relation, find the relation (ratio) between the new momentum and the initial momentum.

Formula used:
p=mv
$K=\dfrac{1}{2}m{{v}^{2}}$

Complete step by step answer:
When a particle of mass m is in motion with a velocity v then we say that it has some momentum. Momentum of a particle is defined as the product of its mass and its velocity.
i.e. p = mv … (i)

We also say that the moving body possesses some amount of energy called kinetic energy. The kinetic energy of the body is given as $K=\dfrac{1}{2}m{{v}^{2}}$ ….. (ii).

Now, divide equation (ii) by equation (i).
Hence, we get
$\dfrac{K}{p}=\dfrac{\dfrac{1}{2}m{{v}^{2}}}{mv}$
This implies, $\dfrac{K}{p}=\dfrac{v}{2}$
Therefore,
$K=\dfrac{vp}{2}$ …. (iii).

The velocity or speed of the particle may change with time. However, the mass of the particle will remain constant. Therefore, we have to write v in terms of m in equation (iii).

For this we will use equation (i). We can write equation (i) as:

$v=\dfrac{p}{m}$ …. (iv).
Substitute the value of v from equation (iii).
$\Rightarrow K=\left( \dfrac{p}{m} \right)\dfrac{p}{2}$
$\Rightarrow K=\dfrac{{{p}^{2}}}{2m}$ …. (v).
Hence, we got a relation between kinetic energy and momentum of a particle.
It is given that the kinetic energy (K) of the particle increases by 300%. This means it has become 4 times the initial kinetic energy (K). Let the new kinetic energy be K’.

Therefore, K’=4K.

Let the new momentum of the particle be p’.
Therefore, according to equation,

 $K'=\dfrac{p{{'}^{2}}}{2m}$.
$\Rightarrow K'=4K=\dfrac{p{{'}^{2}}}{2m}$
And by substituting the value of K from equation (v) we get,

 $\Rightarrow 4\left( \dfrac{{{p}^{2}}}{2m} \right)=\dfrac{p{{'}^{2}}}{2m}$.
$\Rightarrow 4{{p}^{2}}=p{{'}^{2}}$
$\Rightarrow p'=2p$

This means that the new momentum of the particle is two times the initial momentum, which means that the momentum of the particle is increased by 100%.

Note:
Note that momentum is a vector quantity and therefore v in the equation of momentum is velocity of the particle. However, energy is a scalar quantity. Hence, v in the equation of kinetic energy is speed i.e. magnitude of the velocity of the particle.