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K.E. of a body is increased by 44%. What is the percentage increase in the momentum?
A) $10%$
B) $20%$
C) $30%$
D) $44%$

seo-qna
Last updated date: 16th Jun 2024
Total views: 52.5k
Views today: 0.52k
Answer
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Hint: A body acquires kinetic energy due to the motion of the body. The momentum of a body is a quantity of the motion of the body. Kinetic energy and momentum are related due to their mutual relation to the velocity of the body.

Complete step by step solution:
Kinetic energy is given by:
$K.E.=\dfrac{1}{2}m{{v}^{2}}$………………..equation (1)
where,
$m=$ mass of the body
$v=$ the velocity of the body
A body possesses kinetic energy only when there is a change in its motion due to its work.
Work done= force $\times $ displacement
$\Rightarrow W=f.d$
Here, $d=$ displacement of body
According to equation (3):
$\Rightarrow W=ma.d$
Using the third equation of motion, we can express $a=\dfrac{{{v}^{2}}-{{u}^{2}}}{2d}$ and substituting this value in the above equation:
$\Rightarrow W=m(\dfrac{{{v}^{2}}-{{u}^{2}}}{2d}).d$
$\Rightarrow W=\dfrac{1}{2}m({{v}^{2}}-{{u}^{2}})$
$\Rightarrow W=\dfrac{1}{2}m{{v}^{2}}-\dfrac{1}{2}m{{u}^{2}}$
Hence, work done is equal to the change in kinetic energy of a body.
Momentum is expressed as:
$p=mv$………………….. equation (2)
Now, $f=ma$……………………… equation (3)
where,
$f=$ force on object
$a=$ acceleration of body
The acceleration is the change in velocity of a body per unit time and can be written as $\Rightarrow a=\dfrac{\Delta v}{\Delta t}$
Using this in equation (3), we get
$\Rightarrow f=m\dfrac{\Delta v}{\Delta t}$
$\Rightarrow f=\dfrac{\Delta mv}{\Delta t}$
$\Rightarrow f=\dfrac{\Delta p}{\Delta t}$
Hence, we say that the force applied on a body is the change in the body’s momentum per unit time.
Squaring both sides of equation (2) and substituting in equation (1), we get
$\Rightarrow {{p}^{2}}={{m}^{2}}{{v}^{2}}$
Therefore, the relationship between the kinetic energy and the momentum of a body is given as:
$K.E.=\dfrac{{{p}^{2}}}{2m}$
Let initial K.E. and momentum be: $K.E{{.}_{0}},{{p}_{0}}$ respectively
$\Rightarrow K.E{{.}_{0}}=\dfrac{p_{0}^{2}}{2m}$…………….equation(4)
Change in kinetic energy $=K.E{{.}_{0}}+\dfrac{44}{100}K.E{{.}_{0}}$
$\Rightarrow (1+\dfrac{44}{100})K.E{{.}_{0}}=\dfrac{p_{new}^{2}}{2m}$
From equation (4):
$\Rightarrow (1+\dfrac{44}{100})\dfrac{p_{0}^{2}}{2m}=\dfrac{p_{new}^{2}}{2m}$
$\Rightarrow \dfrac{p_{new}^{2}}{p_{0}^{2}}=(1+\dfrac{44}{100})$
$\Rightarrow \dfrac{{{p}_{new}}}{{{p}_{0}}}=\sqrt{\dfrac{144}{100}}=\dfrac{12}{10}$
$\Rightarrow {{p}_{new}}=1.2{{p}_{0}}$
$\Rightarrow {{p}_{new}}=1.2{{p}_{0}}={{p}_{0}}+0.2{{p}_{0}}$
Multiplying and dividing $0.2{{p}_{0}}$by $100$:
$\Rightarrow {{p}_{new}}={{p}_{0}}+\dfrac{20}{100}{{p}_{0}}$
Percent change in momentum $={{p}_{new}}-{{p}_{0}}$
$\Rightarrow {{p}_{new}}-{{p}_{0}}=\dfrac{20}{100}{{p}_{0}}$

The correct answer is [B], 20%.

Note: The force applied times the displacement of a body is the work done on it, and the change in K.E. is also the work done. Momentum is the force applied on a body per unit time. Also, the S.I. unit of K.E. is Joules, and the S.I. unit of momentum is $Kg.m{{s}^{-1}}$.