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if the error in the measurement of momentum of a particle is 10% and mass is known exactly,the permissible error in the determination of kinetic energy is

if the error in the measurement of momentum of a particle is 10% and mass is known exactly,the permissible error in the determination of kinetic energy is
 

Grade:11

6 Answers

mohit vashishtha
144 Points
8 years ago
2*10= 20%
Bidisha Baruah
20 Points
8 years ago
are u sure??the answer in my book is 21%
Nicho priyatham
625 Points
8 years ago
yes 21%
u can apply calucus only when error is very small or less then 1 
here error is 10% so by calucus u get 2*10 =20 which is wrong 
to do such problems first take P1=100 and calulate KE1
now put p2=110 and calculate KEthen find the error
PLEASE APPROVE
Bidisha Baruah
20 Points
8 years ago
i did not get it i am extemely sorry.i mean why should we put p=110
Ramgopal tiwari
11 Points
6 years ago
Here , K.E- 1/2mv^2Mass is known exactly but there is always a minimum error in any calculation so, Here we let the minimum in mass =1% Now, ∆K/K=∆ M/M +2∆V/V = 1% + 2(10%) = 1% + 20% = 21%
Aan
13 Points
5 years ago
P=mv
Ke = 1/2 mv^2
Ke = P^2/2m
Here error is 10%
Thus
P'= 10% of P
P'=0.1P
Ke'=P'/2m
=1.21×Ke
Percentage error = (Final - Initial)×100/initial
=(1.21Ke - Ke)×100/Ke
= 21%
 

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