Question

if p is inclined by 100 % what %
of increase will be there in K.E.​

Answer 1

Correct Question :

The momentum of a body is increased by 100%. The kinetic energy is increased by about

Solution :

❖ Kinetic energy of body of mass m moving at a speed of v is given by

$\bf:\implies\:K=\dfrac{1}{2}\:mv^2\:\dots\:(I)$

❖ Momentum is measured as the product of mass and velocity.

$\bf:\implies\:p=m\:v\:\dots\:(II)$

From the (I) and (II) equation,

$\dag\:\underline{\boxed{\bf{\purple{K=\dfrac{p^2}{2m}}}}}$

Since m remains constant, K ∝ p²

$\sf:\implies\:\dfrac{K_1}{K_2}=\left(\dfrac{p_1}{p_2}\right)^2$

$\sf:\implies\:\dfrac{K_1}{K_2}=\left(\dfrac{p_1}{2p_1}\right)^2$

$\sf:\implies\:\dfrac{K_1}{K_2}=\dfrac{1}{4}$

$\bf:\implies\:K_2=4K_1$

Percentage increase in KE;

$\sf\:=\dfrac{K_2-K_1}{K_1}\times100$

$\sf\:=\dfrac{4K_1-K_1}{K_1}\times100$

$\bf=300\%$

Answer 2

$\large\underline{ \underline{ \sf{ \red{given:} }}}$

• Momentum(p) is increased by 100%

• Mass is constant.

$\\ \\ \large\underline{ \underline{ \sf{ \red{to \: find:} }}}$

• % increase in Kinetic Energy (K.E)

$\\ \\ \large\underline{ \underline{ \sf{ \red{solution:} }}}$

We know ,

$\\ \boxed{ \boxed{\sf{K.E = \frac{1}{2} m {v}^{2} }}}$

$\\ \boxed{ \boxed { \sf{p = mv}}}$

Here ,

• K.E is Kinetic energy.

• p is momentum.

• m is mass.

• v is velocity.

So,

$\\ \boxed{ \sf{ \pink{K.E = \frac{ {p}^{2} }{2m} }}}$

p is increased by 100%.

Therefore , it is increased by 'p'.

Final momentum = p + p = 2p

Since mass is constant,

ㅤㅤㅤㅤK.E ∝ p²

$\\ \sf{ \frac{ K.E_{(1)} }{ K.E_{(2)} } } = \frac{ {p}^{2} \: _ {(1)} }{ {p}^{2} \: _{(2)} } \\ \\ \\ \sf{ \frac{ K.E_{(1)} }{K.E _{(2)} } } = \frac{ {p}^{2} }{ (2 {p)}^{2} } \\ \\ \\ \sf{ \frac{ K.E_{(1)}}{ K.E_{(2)} } = \frac{ \cancel{{p}^{2}} }{4 \cancel{{p}^{2} } }} \\ \\ \\ \bf{ K.E_{(2)} = 4K.E _{(1)}} \:$

$\\ \\ \sf{\% \: increased = \frac{ K.E_{(2)} - K.E _{(1)} }{K.E _{(1)}} } \times 100 \\ \\ \\ \implies \sf{ \frac{4K.E _{(1)} - K.E _{(1)}}{K.E _{(1)}} \times 100 } \\ \\ \\ \sf{ \implies \frac{3K.E _{(1)}}{ K.E _{(1)}}} \times 100 \\ \\ \\ \sf{ \implies \orange{300\%}}$

Hence , K.E is increased by 300% .