Question

when momentum of body increases by 200%, its kinetic energy increases by 1. 200% 2. 300% 3. 400% 4. 800%

Answer 1

Momentum,p = mv
P increases by 200%
P1 = 200/100P + P = 3P
m1 = m
v1 = 3v

Increase in K.E = {1/2m(3v)^2 - 1/2mv^2} / 1/2mv^2 x 100%
= 9 - 1 x 100%
= 800%

So correct option is 4

Answer 2

Kinetic energy will get increased by 800%

Given:

The increase in momentum is 200%

Solution:

The momentum changes kinetic energy also changes.

The kinetic energy is given by the formula given below:

$K . E .=\frac{1}{2}\left(\frac{p^{2}}{m}\right)$

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

After increase in kinetic energy, we get,

$K . E^{\prime}=\frac{(p+200 \% p)^{2}}{2 m}$

$=\frac{p^{2}+4 p^{2}+2\left(2 p^{2}\right)}{2 m}$

$\Rightarrow K \cdot E^{\prime}=\frac{9 p^{2}}{2 m}$

Therefore,

Increase in K.E will be total sum = K.E initial + K.E initial of a percent increase = K.E’  

K.E. initial of a percent increase = K.E’ – K.E

Increase in initial K.E = $K.E \times \frac{a}{100}$

$(K . E) \times \frac{a}{100}=\frac{9 p^{2}}{2 m}-\frac{p^{2}}{2 m}$

$\frac{p^{2}}{2 m} \times \frac{a}{100}=\frac{8 p^{2}}{2 m}$

$\frac{a}{100}=\frac{8 p^{2}}{2 m} \times \frac{2 m}{p^{2}}$

$\frac{a}{100}=8$

$a = 800\%$

Therefore, the kinetic energy will get increased by 800 %