Question

If the momentum of a body is increased by 25%, then the kinetic energy is increased approximately by

Answer 1

The relation between kinetic energy, K and momentum p is the following.

K=p^2/2m…………(1)

Since, body is the same, mass remains constant.

If momentum increases by 25%, then new momentum will be

p’=1.25 p. Hence, new kinetic energy will be K’=(1.25)^2p^2/2m……….(2)

Then, (K’-K)/K=[p^2/2m(1.25)^2–1)]/(p^2/2m)=[(1.25)^2–1]=1.56–1.=0.56

Therefore % increase in kinetic energy=0.56x100=56%.



Hope this helps....

Answer 2

Answer:

The kinetic energy is increased approximately by 56.25%.

Explanation:

The kinetic energy (K) of any body is given by:

$K=\frac{\big P^2}{\big 2\big m}$

Where m is the mass and P is the momentum of the body.

From this we can say that, the the kinetic energy is directly proportional to the square of the momentum of the body.

K ∝ P²

The momentum of body (P) is increased by 25%, then new momentum  (P') will be:

$P'=P+\frac{25}{100}P$

$P'=\frac{125}{100}P$

$P'=\frac{5}{4}P$

For the momentum P', the kinetic energy of the body becomes K':

$\frac{K'}{K}=\frac{P'^2}{P^2}$

$\frac{K'}{K}=\frac{(5P/4)^2}{P^2}$

$\frac{K'}{K}=\frac{25}{16}$

$K'=\frac{25}{16}K$

% increase in the kinetic energy of the body will be given by:

$\frac{K'-K}{K}*100=\frac{(25/16)K-K}{K}*100$

                  $=\frac{9}{16}*100$

                  $=56.25\%$

Therefore, percentage increase in the kinetic energy of the body will be approximately 56%.