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 $Kg.m{{s}^{-1}}$.

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