Question

Please solve question 1 with solution

Answer 1

Since the errors are not given, we may take the least count as the error.

Here mass is measured as 1.00 kg with a precision of 2 decimal places, so we take 0.01 kg as error in mass.

Similarly, error in velocity is 0.01 ms^(-1).

Now, the percentage error in the calculation of mass is,

δm = (0.01 / 1.00) × 100 = 1%

And that of velocity is,

δv = (0.01 / 2.00) × 100 = 0.5%

We have the expression for kinetic energy,

K = (mv²)/2

So the percentage error in kinetic energy is,

δK = δm + 2δv

δK = 1% + 2 × 0.5%

δK = 1% + 1%

δK = 2%

Hence (2) is the answer.