Question

If kinetic energy of a body increase by 400% ,by what percentage will linear momentum of the body increase

Answer 1

Answer:

Given:

Kinetic energy is increased by 400%.

To find:

% change of linear momentum.

Concept:

Let Momentum be p, mass be m and kinetic energy be ke.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \red{ke = \dfrac{ {p}^{2} }{2m}}}$

$= > {p}^{2} = 2m(ke)$

$= > p = \sqrt{2m(ke)}$

Calculation:

Initial momentum be p1 and final momentum be p2.

$\therefore p1 = \sqrt{2m(ke)}$

Now final momentum :

$\therefore p2 = \sqrt{2m(ke \times 500\%)} \\ = > p2 = \sqrt{5} \times \sqrt{2m(ke) } \\ = > p2 = \sqrt{5} \times p1$

percentage is considered 500% , because percentage change is given as 400%. Hence total new percentage = 500%.

So change in Momentum :

$\Delta \: p = p2 - p1 \\ = > \Delta \: p = (\sqrt{5} - 1)p1$

So percentage change :

$= \dfrac{\Delta p}{p1} \times 100\%$

$= \dfrac{ \sqrt{5} - 1 }{1} \times 100\%$

$= 1.23606 \times 100\%$

$= 123.6\% \: increase$

So final answer:

$\boxed{ \red{increase = 123.6\%}}$