Question

Percentage error in mass and momentum are 3% and 2% resp. Maximum possible percentage error in the kinetic energy is​

Answer 1

Answer:

The momentum p = mass × velocity = m v

And kinetic energy E = ½ m v²

E = p²/2m

Taking log of both side we get

log E = log p² - log 2 - log m

=> log E = 2 log p - log 2 - log m

Differentiating we get

dE/E = 2 dp/p - dm/m

But errors only add up, so

dE/E = 2( dp/p) + (dm/m)

Now

dp/p = 3% = 0.03, and

dm/m = 2% = 0.02

Therefore,

dE/E = 2 (dp/p) + (dm/m) = 2 (0.03) + (0.02) = 0.06 + 0.02 = 0.08

Therefore percentage error in calculated value of E from measured value of its momentum and mass = 0.08 = 8%.

hope this helps you and solves your doubts.

Answer 2

Answer:

hope it is helpful..

Plz mark as brainlist