Question

if the percentage error in the measurement of momentum and mass of an object are 2 % and 3 % respectively then maximum percentage error in the calculated value of its kinetic energy is

Answer 1

Hello Dear.

Kinetic Energy(K) = 1/2 × mv²
where, m = mass of the body and v = velocity.
Now, Obtaining the Formula in terms of the Momentum.

∵ p = mv
⇒ v = p/m

∴ K = 1/2 × m (p/m)²
⇒ K = p²/2m
⇒ p² = K × 2m 

Now, Solving the Question.

∵ p² = K × 2m  
∴ Taking Logarithms on both the sides of the above equation.
log p² = log( k 2m)
⇒ 2log p = log K + log 2m 
⇒ log K = 2 log p - log 2m

Now, Differentiating the both sides of the Equation,

∴ dK/K = 2(dp/p) - dm/m

∴ |dK/K|max. = 2|2/100| + |-3/100|
⇒ |dK/K| = 2 × 0.02 + 0.03
⇒ |dK/K|max. = 0.07

∴ Maximum % Error in the Kinetic Energy K is 0.07 × 100 % = 7%.


Hope it helps.

Answer 2

relation between kinetic energy , mass and linear momentum is given by
Kinetic energy = (linear momentum)²/2 × mass
e.g., K = P²/2m
We assume K is kinetic energy , P is linear momentum and m is mass of body .

In case of error,
Use formula,
∆K/K = 2∆P/P + ∆m/m
∴ % error in K = 2 × % error in P + % error in m
[Note :- you should use this formula only when error is less than 10% ]

Given, % error in P = 2 %
% error in m = 3 %
∴ % error in K = 2 × 2 % + 3 % = 4 % + 3 % = 7 %

Hence, maximum percentage error in kinetic energy = 7 %