If the error in the measurement of momentum of a particle is 10% and mass is known exactly,the peermissible error in the determination of kinetic energy is
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kinetic energy=1/2mv^2
p=mv
increase in p=p +p'=p+1p/10=11p/10
increase in percentage=11%
therefore increase in k=10+11=21%
or
We know,
K= 1/2 × MV²
We know that 1/2 & M are constants and the only variable thing here is its velocity
So the error only depends on velocity
Using the equation
ΔK / K =ΔV / V + 1
so that,
∴ ( 2 × 10 ) + 1
= 21 %
p=mv
increase in p=p +p'=p+1p/10=11p/10
increase in percentage=11%
therefore increase in k=10+11=21%
or
We know,
K= 1/2 × MV²
We know that 1/2 & M are constants and the only variable thing here is its velocity
So the error only depends on velocity
Using the equation
ΔK / K =ΔV / V + 1
so that,
∴ ( 2 × 10 ) + 1
= 21 %
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