91Ó°ÊÓ

At a classic auto show, a \(840-\mathrm{kg}\) 1955 Nash Metropolitan motors by at 9.0 \(\mathrm{m} / \mathrm{s}\) , followed by a \(1620-\mathrm{kg} 1957\) Packard Clipper purring past at 5.0 \(\mathrm{m} / \mathrm{s}\) , (a) Which car has the greater kinetic energy? What is the ratio of the kinetic energy of the Nash to that of the Packard? (b) Which car has the greater magnitude of momentum? What is the ratio of the magnitude of momentum of the Nash to that of the Packard? (c) Let \(F_{N}\) be the net force required to stop the Nash in time \(t\) , and let \(F_{\mathrm{p}}\) be the net force required to stop the Packard in the same time. Which is larger: \(F_{\mathrm{N}}\) or \(F_{\mathrm{P}}\) ? What is the ratio \(F_{\mathrm{N}} / F_{\mathrm{p}}\) of these two forces? (d) Now let \(F_{\mathrm{N}}\) be the net force required to stop the Nash in a distance \(d\) , and let \(F_{\mathrm{p}}\) be the net force required to stop the Packard in the same distance. Which is larger. \(F_{\mathrm{N}}\) or \(F_{\mathrm{P}}\) ? What is the ratio \(F_{\mathrm{N}} / F_{\mathrm{P}} ?\)

Step by step solution

01

Calculating Kinetic Energy

Kinetic energy is given by the formula \( KE = \frac{1}{2}mv^2 \). Let's calculate the kinetic energy for both cars: - Nash: \( m = 840\,\text{kg}, v = 9.0\,\text{m/s} \) \[ KE_{\text{Nash}} = \frac{1}{2} \times 840 \times (9.0)^2 = 34020\,\text{J} \] - Packard: \( m = 1620\,\text{kg}, v = 5.0\,\text{m/s} \) \[ KE_{\text{Packard}} = \frac{1}{2} \times 1620 \times (5.0)^2 = 20250\,\text{J} \] The Nash has greater kinetic energy. The ratio of the KE of the Nash to the Packard is: \[ \text{Ratio} = \frac{34020}{20250} \approx 1.68 \]

02

Calculating Momentum

Momentum is given by the formula \( p = mv \). Calculate the momentum for both cars: - Nash: \( m = 840\,\text{kg}, v = 9.0\,\text{m/s} \) \[ p_{\text{Nash}} = 840 \times 9.0 = 7560\,\text{kg m/s} \] - Packard: \( m = 1620\,\text{kg}, v = 5.0\,\text{m/s} \) \[ p_{\text{Packard}} = 1620 \times 5.0 = 8100\,\text{kg m/s} \] The Packard has a greater momentum. The ratio of the momentum of the Nash to the Packard is: \[ \text{Ratio} = \frac{7560}{8100} \approx 0.93 \]

03

Comparing Forces to Stop in Same Time

The force to stop an object is related to its momentum over time: \( F = \frac{\Delta p}{t} \). Both cars are stopped in time \( t \), so we compare their momentum: - Since the Packard has greater momentum, it requires a larger force to stop in the same time. The ratio of the forces is: \[ \text{Ratio} = \frac{F_{\text{Nash}}}{F_{\text{Packard}}} = \frac{7560}{8100} \approx 0.93 \]

04

Comparing Forces to Stop in Same Distance

For stopping over a distance \( d \), the work-energy principle \( W = \Delta KE = Fd \) applies: - The Nash has greater kinetic energy, requiring a larger force to stop in the same distance. The ratio of the forces is: \[ \text{Ratio} = \frac{F_{\text{Nash}}}{F_{\text{Packard}}} = \frac{34020}{20250} \approx 1.68 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum

Momentum is a measure of the motion of an object and depends on its mass and velocity. It's a key concept in physics, representing the quantity of motion that a body possesses. The formula to calculate momentum (\( p \)) is \( p = mv \), where \( m \) is the mass and \( v \) is the velocity of the object.

• The 1955 Nash Metropolitan, with a mass of 840 kg and velocity of 9.0 m/s, has a momentum of 7560 kg m/s.
• The 1957 Packard Clipper, heavier at 1620 kg but slower at 5.0 m/s, results in a higher momentum of 8100 kg m/s.

Even though the Nash is faster, the Packard packs more mass, giving it greater momentum. Understanding these values helps in determining the forces needed for stopping, important in safety analysis.

Force Stopping Distance

The stopping force needed for a moving car depends on how quickly or how suddenly the car needs to stop. Here, we focus on stopping over the same distance. Considerations of stopping force are crucial for safe vehicle operation. When stopping distance \( d \), the force is related to the change in kinetic energy. According to the work-energy principle, the work needed to stop a car is equal to the car's kinetic energy.

• The Nash has a kinetic energy of 34020 J, meaning it needs more force to stop over the same distance compared to the Packard, which has 20250 J.

This difference shows how speed and mass influence stopping. Practically, this is why heavier and faster cars require more powerful brakes.

Work-Energy Principle

The work-energy principle is a powerful tool in physics which states that the work done on an object is equal to the change in its kinetic energy. Applied to vehicle stopping, the principle helps explain how a change in speed or mass affects the force needed to come to a stop. For a car stopping over a distance \( d \), its kinetic energy (\( KE \)) \( W = Fd = \Delta KE \), where \( F \) is the force applied.
In our example, the Nash requires more force to stop because its higher kinetic energy (34020 J) needs to be reduced to zero. Understanding this principle is essential for engineers designing safer vehicles or devices that manage forces and energy effectively.

Formula Application

Applying formulas correctly is crucial to solving physics problems effectively. Let's break down how the formulas were used:

• Kinetic Energy (\( KE \)): Calculated for Nash as \( KE = \frac{1}{2}mv^2 = 34020 J \) and for Packard as \( KE = 20250 J \)
• Momentum (\( p \)): Nash calculated as \( p = mv = 7560 kg \cdot m/s \), Packard as \( 8100 kg \cdot m/s \)

Through these calculations, it's clear how mass and speed affect both momentum and kinetic energy. This problem-solving process highlights how knowing the right formula can unravel complex relationships between different physical quantities in everyday scenarios.