A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms^-1. The magnitude of

Q: Why do we care about significant figures at all? Can't we just use the exact calculator value?

Significant figures represent the precision of our physical measuring instruments. A calculator assumes infinite precision, but in reality, we cannot claim our result is more precise than our least precise measurement.

Q: How many significant figures are in $5.00$?

There are three significant figures. Trailing zeros after a decimal point are always considered significant because they indicate the precision of the measurement.

Q: Why did we round up to $17.6$ instead of keeping it $17.5$?

When rounding to three significant figures, we look at the fourth digit. In $17.565$, the fourth digit is $6$. Since $6$ is greater than $5$, we round the previous digit ($5$) up by one, giving us $17.6$.

Q: What if the operation was addition instead of multiplication?

The rule changes! For addition and subtraction, the final answer must have the same number of decimal places as the measurement with the least number of decimal places, regardless of the total significant figures.

Q: Do the units affect the number of significant figures?

No, significant figures only apply to the numerical value of the measurement. The units simply tell us what physical quantity is being measured.

Question

A body of mass $m = 3.513 	ext{ kg}$ is moving along the x-axis with a speed of $5.00 	ext{ ms}^{-1}$. The magnitude of its momentum is recorded as

Options:

(A) $17.6 	ext{ kg ms}^{-1}$

(B) $17.565 	ext{ kg ms}^{-1}$

(C) $17.56 	ext{ kg ms}^{-1}$

(D) $17.57 	ext{ kg ms}^{-1}$

Sigma Percentile · Accepted Answer

### The Illusion of Precision

Imagine you are in a high-tech laboratory, measuring the speed of a moving object. You have a highly precise scale that tells you the mass is exactly $3.513 	ext{ kg}$. That's four significant figures of pure confidence!

But then, you measure the velocity. Your radar gun gives you a reading of $5.00 	ext{ ms}^{-1}$. It's a good reading, but it only has three significant figures.

### The Master Equation

We know that momentum is the product of mass and velocity.

$$p = m \cdot v$$

When we plug our numbers into this equation, our calculator eagerly spits out a highly detailed answer:

$$p = 3.513 	imes 5.00 = 17.565 	ext{ kg ms}^{-1}$$

### The Catch: Significant Figures

Here is where the trap lies. Your calculator doesn't know about the physical limitations of your measuring instruments. It assumes absolute mathematical perfection. But in physics, a chain is only as strong as its weakest link.

The rule for multiplication and division is strict: **The final result must have the same number of significant figures as the measurement with the least number of significant figures.**

Our mass has $4$ significant figures. Our velocity has $3$ significant figures. Therefore, our final answer is restricted to just $3$ significant figures.

### The Final Polish

We must round $17.565$ to three significant figures. We look at the first three digits: $17.5$. The next digit is $6$, which is greater than $5$.

Following the standard rules of rounding, we round up the $5$ to a $6$.

$$p = 17.6 	ext{ kg ms}^{-1}$$

And there we have it! A physically accurate, perfectly precise answer that respects the limits of our tools.

Animated Solution for Physics - Physics and Measurement: A body of mass $m = 3.513 kg$ is moving along the x-axis with a speed of $5.00 ms^{- 1}$ . The magnitude of its momentum is recorded as

Select Answer:

Visualized Solution

Visualizing the Setup

Formula for Momentum

Substituting Values

Raw Calculation

Applying Significant Figures Rule

Final Answer

The Way Forward

The Sigma Insight: Significant Figures

The Illusion of Precision

The Master Equation

The Catch

The Final Polish

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