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A ball of mass $0.05$ kg moves horizontally at $8$ m/s. It is struck with a racquet causing it to travel at $18$ m/s in the opposite direction. Find the impulse imparted to the ball considering uncertainties in the given data.

My understanding is that $0.05$ means $0.050 \pm 0.005$, $8$ means $8.0 \pm 0.5$, and $18$ means $18.0 \pm 0.5$, taking uncertainties into consideration.

Mass: $m=0.050\pm0.005$ kg

Initial velocity: $\vec u = +8.0\pm 0.5 \vec i$ m/s

Final velocity: $\vec v = -18.0\pm 0.5 \vec i$ m/s

Q1: Is that correct?

My formula for Impulse is then $$\vec I = m(\vec v - \vec u)$$

I get max and min values for impulse of: $\vec I_{max}=-1.485 \vec i$ Ns, and $\vec I_{min}=-1.125 \vec i$ Ns.

Taking the mean and the range of these two values I get an answer of: $$\vec I = -1.305 \pm 0.18 \vec i Ns$$

Q2: I learned that a calculation can only be as accurate as the data used in the calculation. Some of the data here has only one significant figure. So how should I report/present my result? Should I round it up to something like $\vec I = -1.3 \pm 0.2$ Ns ? But why should I ? I mean the result above gives the entire range of possible values given the data. Even though the data may have only 1 significant figure the result seemingly can have many more. What is the standard procedure here?

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It all depends on the level of sophistication that you need but if you want a rule of thumb the quote your answer to the number of significant figures given or significant figures given plus one when the result is "slightly" more than $ \rm 1 \, or\, 10 \, or\, 1000 .... \, or \, 0.1 \, or \, 0.01\, or \, 0.001 ....$
So in this case you could quote a value of $1.3$.
If you had found the value to be $8.3$ the $8$ would probably be reasonable given that the mass was only given to one significant figure.

Using your data values what you have found is the maximum (*see below) range of values that the impulse can have and if you were to quote the value as $ -1.3 \pm 0.2$ you are really saying that there is a $\pm 15\%$ range.
Now what does quoting $ -1.30 \pm 0.18$ do for you?
The range drops a little to $\pm 13\%$ but do you feel that the $0$ in $-1.30$ tells you any more about the value which has been found?.
So I think that $ -1.3 \pm 0.2$ is fine remembering that there will be an error in your error as well.

(*) To get the minimum value you have assumed that all three data values are a minimum at the same time eg $(7.5+15.5)\times 0.45 = 1.125$.
The chances of this happening are probably not that great so a better method of estimating the error is as follows.

The error in $18+8=16$ is $\pm \sqrt{0.5^2+0.5^2} = \pm 0.7(1)$
This represents a percentage error in $26$ of $\pm \dfrac {0.7}{26} \times 100 = 0.02(7)$
The percentage error in the mass is $\pm \dfrac {0.005}{0.05} \times 100 = 10$ and this immediately tells you that the best way you can improve the accuracy of the experiment is to measure the mass more accurately.

The percentage error in $0.05 \times 26 = 1.3$ is $\pm \sqrt{10^2+0.03^2} = \pm 10$

This percentage error represents an actual error of $\pm 0.13$ and so you result is $1.30 \pm 0.13$ which I think is better quoted as $1.3\pm 0.1$.
This result was obtained on the assumption that the all the readings were independent of one another and that on average all the data value are not going to be skewed to a maximum or minimum value.

Error analysis is a very important part of any experiment and there are many Internet sites and textbooks which are available for you to look at.
Here is one that may be of use?

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Q1 : See Farcher's answer

Q2. In fact, the significant figure method is less accurate than error calculation. That is why it is frustrating.

The rule is that when multiplying or dividing, one keeps the lower number of significant figure. For example, 1.3*2 = 2.6 = 3

However, when adding, one keeps the lowest accurate significant figure. For example, 1000.0 + 0.123 = 1000.1.

So you should write I = 1 Ns. Which is less accurate than $I = 1.30 \pm 0.18 Ns$. Error calculation is a way to cope with significant figures' inaccuracy.

Comment : You should never have more figures in the mean value than in the error, because in this case the last figures would be useless.

So you shouldn't write $$\vec I = -1.305 \pm 0.18 \vec i Ns$$ But $$\vec I = -1.31 \pm 0.18 \vec i Ns$$

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