Question

the linear momentum of a body is increased by 10% what is the percentage increase in its kinetic energy?​

Answer 1

Answer:If p is momentum and corresponding kinetic energy is K, we have ,for particle of mass m, the relation,

K=p^2/2m

When p is increased by 10%, the new kinetic energy will be,

K’=(1.1p)^2/2m=1.21(p^2/2m)

% increase=(change/original value)x100.=[(1.21–1)(p^2/2m/(p^2/2m)]x100=21%

Explanation:

Answer 2

Answer:

$\large\bold\red{21\%}$

Explanation:

We know that,

Kinetic Energy, K.E of a particle having linear momentum ,p and mass, m is given by,

$\large\bold{K.E = \frac{ {p}^{2} }{2m}}$

Now,

It is Given,

The linear momentum, p of a body is increased by 10%.

Therefore,

the new linear momentum ,

$p' = p + \frac{10}{100} p \\ \\ = >p' = \frac{(100 + 10)p}{100} \\ \\ = > p' = \frac{110p}{100} \\ \\ = > p' = \frac{11p}{10}$

Therefore,

the new Kinetic Enery,

$K.E' = \frac{ {p' }^{2} }{2m} \\ \\ = >K.E' = \frac{ {( \frac{11p}{10}) }^{2} }{2m} \\ \\ = > K.E' = \frac{121 {p}^{2} }{200m}$

Therefore,

Increase in Kinetic Enery is,

$= K.E' - K.E \\ \\ = \frac{121 {p}^{2} }{200m} - \frac{ {p}^{2} }{2m} \\ \\ = \frac{(121 - 100) {p}^{2} }{2m} \\ \\ = \frac{21 {p}^{2} }{2m}$

Therefore,

Percentage increase in Kinetic energy is,

$= (\frac{increased\:K.E}{original\:K.E} \times 100)\% \\ \\ = (\frac{ \frac{21 {p}^{2} }{200m} }{ \frac{ {p}^{2} }{2m} } \times 100 )\% \\ \\ = 21\%$

Hence,

21% Kinetic Energy increased.