If the percentage error in the measurement of momentum and mass of an

1.	If the percentage error in the measurement of momentum and mass of an object are
Answer» <P>"MAXIMUM percentage error in the KINETIC Energy K is $0.07%\quad \times \quad 100%\quad =\quad 7%$ Solution: We know that, Kinetic Energy, $({ K.E })\quad =\quad \frac { 1 }{ 2 } \quad \times \quad m{ v }^{ 2 }$ Where, m = mass of the BODY and v = velocity Now, Obtaining the Formula in terms of the Momentum. Hence, p = mv ${ v }\quad =\quad \frac{ p }{ m }$ Therefore, $K\quad =\quad \frac { 1 }{ 2 } \quad \times \quad m\left( \frac { p }{ m } \right)^{ 2 }$ $K\quad =\quad \frac {p^{ 2 }}{ 2m }$ $p^{ 2 }\quad =\quad K\quad \times\quad 2m$ Hence, TAKING Logarithms on both the sides of the above equation. $\log p^{2}\quad =\quad\log(K.2m)$ $2\log p\quad =\quad\log K\quad +\quad \log2m$ $\log{K}\quad =\quad 2\log p\quad -\quad \log2{ m }$ Now, differentiating the both sides of the equation, we get $\frac { dK }{ K }\quad =\quad 2\left( \frac {dp }{p} \right)\quad -\quad \frac { dm }{ m }$ $\left\| \frac { dK }{ K } \right\| (\max)\quad =\quad 2\left\| \frac { 2 }{ 100 } \right\| \quad +\quad \left\| -\frac { 3 }{ 100 } \right$ $\left\| \frac { dK }{ K } \right\| \quad =\quad 2\quad \times \quad 0.02\quad +\quad 0.03$ $\left\| \frac { dK }{ K } \right\| \quad =\quad 0.07(\max)$ "

If the percentage error in the measurement of momentum and mass of an object are

Answer»

<P>"MAXIMUM percentage error in the KINETIC Energy K is $0.07%\quad \times \quad 100%\quad =\quad 7%$

Solution:

We know that,

Kinetic Energy, $({ K.E })\quad =\quad \frac { 1 }{ 2 } \quad \times \quad m{ v }^{ 2 }$

Where, m = mass of the BODY and v = velocity

Now, Obtaining the Formula in terms of the Momentum.

Hence, p = mv

${ v }\quad =\quad \frac{ p }{ m }$

Therefore, $K\quad =\quad \frac { 1 }{ 2 } \quad \times \quad m\left( \frac { p }{ m } \right)^{ 2 }$

$K\quad =\quad \frac {p^{ 2 }}{ 2m }$

$p^{ 2 }\quad =\quad K\quad \times\quad 2m$

Hence, TAKING Logarithms on both the sides of the above equation.

$\log p^{2}\quad =\quad\log(K.2m)$

$2\log p\quad =\quad\log K\quad +\quad \log2m$

$\log{K}\quad =\quad 2\log p\quad -\quad \log2{ m }$

Now, differentiating the both sides of the equation, we get

$\frac { dK }{ K }\quad =\quad 2\left( \frac {dp }{p} \right)\quad -\quad \frac { dm }{ m }$

$\left| \frac { dK }{ K } \right| (\max)\quad =\quad 2\left| \frac { 2 }{ 100 } \right| \quad +\quad \left| -\frac { 3 }{ 100 } \right$

$\left| \frac { dK }{ K } \right| \quad =\quad 2\quad \times \quad 0.02\quad +\quad 0.03$

$\left| \frac { dK }{ K } \right| \quad =\quad 0.07(\max)$ "