Question

what will cause greater change in kinetic energy of a body ? changing its mass or changing its velocity . Explain

Answer 1

$\bold{KINETIC \: ENERGY}$

Kinetic eneygy is the energy possed by any body before comming to rest.

Its proved that kinetic energy K is equal to the half of product of mass n and the square of its be velocity v.

$\huge \boxed{K =\frac{1}{2} \times mass \times {velocity}^{2}}$

From which we can convey that kinrtic energy of a body depends on

◼$\bold{Mass \: of \: the \:body}$

Kinetic energy is directly propotional to mass of the body ,so if the mass is increased ,the kinetic energy will also increase by same ratio.

◼$\bold{Velocity }$

Kinetic energy is directly proportional to the square of its velocity.So if the velocity is increased then the kinetic energy will be also increase in same ratio.

$\underline \bold{Example}$

let mass be m

velocity be v

so kinetic energy will be ,

$\huge \boxed{K =\frac{1}{2} \times m \times {v}^{2}}$

◼$\bold{If \: mass\: is \: doubled \: then }$

m= 2m

kinetic energy will be,

$\huge \boxed{K_1 =\frac{1}{2} \times 2m \times {v}^{2}}$

$\huge \boxed{K_1 =\frac{1}{\cancel{2}} \times m \cancel{2} \times {v}^{2}}$

$\huge \boxed{K_1 = m \times {v}^{2}}$

On dividing k and $K_1$

$\frac{K}{K_1} = \frac{ \frac{1}{2} \times 2m \times {v}^{2}}{m \times {v}^{2}}$

$\frac{K}{K_1} = \frac{ \frac{1}{2} \times 2 \cancel{m} \times \cancel{{v}^{2}}}{ \cancel{m} \times \cancel{{v}^{2}}}$

$\frac{K}{K_1} = \frac{1}{2}$

$K_1 = 2K$

So ,the kinetic energy is doubled.

◼$\bold{If \: velocity\: is \: doubled \: then }$

velocity = 2v

mass = m

so kinetic energy will be,

$\huge \boxed{K_2=\frac{1}{2} \times m \times {2v}^{2}}$

On dividing K with $K_2$

$\frac{K}{K_2} = \frac{ \frac{1}{2} \times m \times {2v}^{2}}{\frac{1}{2} \times m \times {v}^{2}}$

$\frac{K}{K_2} = \frac{ \frac{1}{2} \times m \times 4{v}^{2}}{\frac{1}{2} \times m \times {v}^{2}}$

$\frac{K}{K_2} = \frac{\cancel{ \frac{1}{2}} \times \cancel{m} \times \cancel{{v}^{2}}}{\cancel{ \frac{1}{2}} \cancel{m} \times 4 \cancel{{v}^{2}}}$

$\frac{K}{K_2} = \frac{1}{4}$

$K_2 = 4K$

so,the new kinetic energy will be four times the original one.

$\bold{Hence \: ,its \: clear \: changing \: the \: velocity \: will \: increase \: kinetic \: energy \:more}$

Answer 2

Kinetic energy is directly proportional mass as well as velocity. But, you must note the fact that kinetic energy is directly proportional to the square of velocity. Therefore, changing the velocity will cause a greater change in kinetic energy.