Question

study the problem and proposed solution below. is the final answer correct? then click the no mistakes button. is the final answer wrong? then click the button in front of the line with the first mistake and correct the mistake. note: dont worry about correcting the rest of the solution, after the first mistake. problem: momentum is mass times velocity. suppose the momentum of a 335 g object is measured to be 303 \\(\frac{kg\cdot m}{s}\\). calculate the object’s velocity. be sure your answer has a unit symbol and the correct number of significant digits. solution: \\(p = mv\\) write down the definition of momentum. \\(v = \frac{p}{m}\\) solve for velocity. \\(v = \frac{303}{0.335} \frac{m}{s}\\) put in the data given. \\(v = 904.48... \frac{m}{s}\\) use the calculator. \\(v = 904.48 \frac{m}{s}\\) round to the correct number of significant digits. no mistakes

Explanation:

Step1: Identify the mistake in mass conversion

The mass is given as 335 g, but when using in the momentum formula \( p = mv \), mass should be in kilograms. So we need to convert grams to kilograms: \( 335\ g=\frac{335}{1000}\ kg = 0.335\ kg \). However, in the step "Put in the data given", the mass used is \( 0.335\) (but the units of momentum are \( \frac{kg\cdot m}{s} \), so when we substitute \( p = 301\frac{kg\cdot m}{s} \) and \( m = 0.335\ kg \), the formula should be \( v=\frac{p}{m}=\frac{301\frac{kg\cdot m}{s}}{0.335\ kg} \), not \( \frac{301\frac{m}{s}}{0.335} \). Wait, actually, the mistake is in the substitution step. Let's re - check:

The momentum \( p = 301\frac{kg\cdot m}{s} \), mass \( m = 335\ g=0.335\ kg \). The formula for velocity is \( v=\frac{p}{m} \). So substituting the values, we have \( v=\frac{301\frac{kg\cdot m}{s}}{0.335\ kg} \). The \( kg \) units cancel out, giving \( v=\frac{301\ m/s}{0.335} \). Wait, no, the initial substitution step in the solution has a mistake in the units of momentum. Wait, the momentum is \( 301\frac{kg\cdot m}{s} \), so when we write \( v=\frac{301\ \frac{kg\cdot m}{s}}{0.335\ kg} \), the \( kg \) cancels, and we get \( v=\frac{301\ m/s}{0.335} \). But in the solution's "Put in the data given" step, they wrote \( v = \frac{301\ \frac{m}{s}}{0.335} \), which is incorrect because the momentum has units of \( \frac{kg\cdot m}{s} \), not \( \frac{m}{s} \). But actually, the first mistake is in the mass conversion? Wait, no, the mass is 335 g, which is 0.335 kg, that part is correct. Wait, the mistake is in the "Put in the data given" step. Let's correct it:

The correct substitution should be \( v=\frac{p}{m}=\frac{301\ \frac{kg\cdot m}{s}}{0.335\ kg} \). So the numerator has units \( \frac{kg\cdot m}{s} \) and the denominator has units \( kg \), so when we divide, the \( kg \) cancels, and we get \( \frac{m}{s} \) as the unit for velocity. But in the solution's "Put in the data given" step, they wrote \( v=\frac{301\ \frac{m}{s}}{0.335} \), which is wrong because the momentum is \( 301\ \frac{kg\cdot m}{s} \), not \( 301\ \frac{m}{s} \). So the first mistake is in the "Put in the data given" step. The correct expression should be \( v=\frac{301\ \frac{kg\cdot m}{s}}{0.335\ kg} \) (or simplified as \( v=\frac{301\ m/s}{0.335} \) since \( kg \) cancels, but the initial numerator should have \( \frac{kg\cdot m}{s} \) not \( \frac{m}{s} \)). But actually, the mass is 335 g = 0.335 kg, momentum is 301 \( \frac{kg\cdot m}{s} \). So \( v=\frac{301\ \frac{kg\cdot m}{s}}{0.335\ kg}=\frac{301\ m/s}{0.335} \). Wait, maybe the mistake is in the significant figures. Let's check the significant figures: the mass is 335 g (3 significant figures), momentum is 301 \( \frac{kg\cdot m}{s} \) (3 significant figures). So the answer should have 3 significant figures.

But the first mistake is in the "Put in the data given" step. The correct expression should be \( v=\frac{301\ \frac{kg\cdot m}{s}}{0.335\ kg} \) (or \( v=\frac{301\ m/s}{0.335} \) after unit cancellation). So we need to correct that step.

Step1: Correct the substitution step

The momentum \( p = 301\frac{kg\cdot m}{s} \) and mass \( m = 335\ g=0.335\ kg \). The formula for velocity is \( v=\frac{p}{m} \). Substituting the values, we get \( v=\frac{301\ \frac{kg\cdot m}{s}}{0.335\ kg} \). After canceling the \( kg \) units, this becomes \( v=\frac{301\ m/s}{0.335} \).

Step2: Calculate the velocity

\( v=\frac{301}{0.335}\ m/s\approx900\ m/s \) (rounded to 3 significant figures, since 301 and 0.335 both have 3 significant figures). Wait, \( \frac{301}{0.335}=900. \) (when rounded to 3 significant figures, it's 900. or 9.00×10², but let's do the calculation: 301÷0.335 = 900 (exactly? 0.335×900 = 301.5, which is close to 301. Wait, 0.335×898.5≈301. So maybe my initial thought was wrong. Wait, 301 divided by 0.335: 0.335×900 = 301.5, so 301÷0.335≈898.5 m/s, which rounds to 9.00×10² m/s or 900 m/s (if we consider 3 significant figures). But the main mistake is in the substitution step.

The first mistake is in the "Put in the data given" step. The correct expression should be \( v=\frac{301\ \frac{kg\cdot m}{s}}{0.335\ kg} \) (or \( v=\frac{301\ m/s}{0.335} \) after unit cancellation, but the numerator's units should be \( \frac{kg\cdot m}{s} \) initially). So we correct that step. Then, when we calculate \( \frac{301}{0.335} \), we get approximately 898.5 m/s, which rounds to 900 m/s (3 significant figures) or 899 m/s (more precise). But the key mistake is in the substitution step.

Answer:

The first mistake is in the "Put in the data given" step. The correct expression should be \( v=\frac{301\ \frac{kg\cdot m}{s}}{0.335\ kg} \) (or simplified as \( v=\frac{301\ m/s}{0.335} \) after unit cancellation). After correcting this and calculating, the velocity is approximately \( 9.00\times 10^{2}\ \frac{m}{s} \) (or 900 \( \frac{m}{s} \)) when rounded to 3 significant figures.