Wed May 25, 2011 By: Shreyash Rai Rai

If the percentage error in the mesurement of momentum and mass of an body are 2% and 3% respectively , then maximun percentage error in the calculated value of its kinetic energy is??....but how is this possible because the % error is always added , i got this quetion in some book but m confused???

Expert Reply
Wed May 25, 2011
Let the initial momentum be 200kg-meter per second and initial mass be 100kg then from the formula
P=MV, the velocity will be 2meter per second.
By question there was an error of 2% in momentum there fore the momentum will be
196kg-meter per second.
By question there was an error of 3% in the mass so the final mass after the error will be 97kg., now the velocity will be 2.02 meter per second.
Kinetic energy before the error is
Kinetic energy after the error is
Error in K.E is 200-197.9=2.1
And the required error percent is
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