Question

By how much will the speed of a body, of fixed mass, increase if its kinetic energy becomes four times its initial kinetic energy ?

Answer 1

Given :

• Kinetic energy of a body is increased by 4 times the initial kinetic energy of fixed mass

To Find :

• Change in speed of the body due to the increase in kinetic energy

Solution :

Kinetic energy of a body is given by ,

$\\ \star \: {\boxed{\purple{\sf{KE = \dfrac{1}{2}m {v}^{2} }}}}$

Where ,

• m is mass of the body
• v is speed of the body

Let the mass of the body be "m" and velocity be "v". Then initial kinetic energy of the body is ,

$\\ : \implies \sf \:KE = \dfrac{1}{2} m {v_1}^{2}$

$\\ : \implies \sf \: 2KE = m {v_1}^{2}$

$\\ : \implies \sf \: \sqrt{\dfrac{2KE}{m}} = {v_1}^{2} \: .......(1)$

Now , According to the second condition ;

$\\ : \implies \sf \: 4KE= \dfrac{1}{2}m {v_2}^{2}$

$\\ : \implies \sf \: 8KE = m {v_2}^{2}$

$\\ : \implies \sf \sqrt{ \dfrac{8KE}{m} } = v_2$

$\\ : \implies \sf \: 2 \sqrt{ \dfrac{2KE}{m} } = v_2$

$\\ : \implies{\underline{\boxed{\pink{\mathfrak{v_2 = 2v_1}}}}} \: \bigstar$

Hence ,

• The speed of a body becomes 2 times the initial one when the kinetic energy becomes 4 times the initial kinetic energy by fixed mass.

Answer 2

Given :-

• Mass is constant.

• Initial K.E = K.E

• Final K.E = 4K.E

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To find :-

• Increase in speed.

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Solution :-

K.E = (1/2)mv²

Here ,

• m is mass.

• v is velocity.

• K.E is Kinetic Energy.

According to question,

Initally ,

$\\ \sf \: K.E = \frac{1}{2} m {v}^{2} \: \: \: \: \: \: \: \: \: - - - (1)$

Finally,

$\\ \sf \: 4K.E = \frac{1}{2}m {( v_{2} )}^{2}\: \: \: \: \: - - - (2)$

Dividing eqⁿ (1) by eqⁿ (2) ,

$\\ \sf \: \frac{4 \: \cancel{K.E}}{ \cancel{K.E}} = \frac{ \frac{1}{2} \: m {v}^{ 2} _{ \tiny2} }{ \frac{1}{2}m {v}^{2} } \\ \\ \\ \sf 4 = \frac{ { v_{ \tiny2}}^{2} }{ {v}^{2} } \\ \\ \\ \sf \: 4 {v}^{2} = { v_{ \tiny2} }^{2} \\ \\ \\ \sf \: \boxed{ \bf \: v_{2} = 2v}$

Increase in speed = v(2) - v

Increase in speed = 2v - v

Increase in speed = v

$\\ \therefore \: \bf \: speed \: is \: increased \: by\: v.$