Question

- If the error in the measurement of momentum of a particle
is (+100%), then the error in the measurement of kinetic
energy is
​

Answer 1

Explanation:

k=1/2mv^2

k=p^2/2m

here change in error is more than 5

so

k2/k1=p2^2/p1^2

p2=p1+(100/100)p1

p2=2p1

k2=(4p1^2/p1^2)k1

k2=4k1

now

deltk/k×100%=(k2-k1)/k1×100

=4k1-k1)÷k1×100

=3×100=300%