Question

A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of \(200 \mathrm{~kg}\) and is traveling east with a velocity of magnitude \(5.00 \mathrm{~m} / \mathrm{s}\). Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks. (a) A \(25.0-\mathrm{kg}\) mass is thrown sideways out of the car with a velocity of magnitude \(2.00 \mathrm{~m} / \mathrm{s}\) relative to the car's initial velocity. (b) A 25.0 -kg mass is thrown back ward out of the car with a velocity of \(5.00 \mathrm{~m} / \mathrm{s}\) relative to the initial motion of the car. (c) A 25.0 -kg mass is thrown into the car with a velocity of \(6.00 \mathrm{~m} / \mathrm{s}\) relative to the ground and opposite in direction to the initial velocity of the car.

Short answer

a) 5.00 m/s east, b) 5.71 m/s east, c) 4.86 m/s east.

Step-by-step solution

Understanding the Initial Conditions

The total mass of the car initially is 200 kg, with an initial velocity of 5.00 m/s east. Since the track is frictionless and there is no air resistance, we'll use the principle of conservation of momentum for this problem.

Case A: Sideways Ejection

Since the mass is thrown sideways, it doesn't affect the forward motion of the car. The velocity in the east-west direction remains unchanged according to the conservation of linear momentum in the east-west direction. Therefore, the final velocity of the car is still 5.00 m/s east.

Case B: Backward Ejection

Calculate the initial momentum of the system: \( p_{initial} = 200 \times 5.00 = 1000 \text{ kg m/s} \). After ejecting a 25.0 kg mass backward with a velocity of 5.00 m/s (velocity of the mass relative to the car): \( v_{mass} = 5.00 - 5.00 = 0 \text{ m/s} \) relative to the ground. The remaining mass is 175 kg. The final momentum is \( p_{car} = 175 \times v_{final} \), so \( 1000 = 175 \times v_{final} \). Solving for \( v_{final} \): \( v_{final} = 1000 / 175 = 5.71 \text{ m/s} \) east.

Case C: Opposite Direction Influx

A 25.0 kg mass is thrown into the car with a velocity of 6.00 m/s opposite to the car's initial velocity. Calculate the relative momentum: \( p_{mass} = 25.0 \times (-6.00) = -150 \text{ kg m/s} \). The initial momentum is 1000 kg m/s. Final momentum is \( p_{total} = (175) \times v_{final} \). Therefore, \( 1000 + (-150) = 175 \times v_{final} \). Solving for \( v_{final} \): \( v_{final} = 850 / 175 = 4.86 \text{ m/s} \) east.

Key concepts

linear momentum

Linear momentum is a fundamental concept in physics, describing the motion of an object. It is a vector quantity, meaning it has both a direction and a magnitude. The formula to calculate linear momentum, represented as \( p \), is: \[ p = m \times v \] where \( m \) is the mass of the object and \( v \) is its velocity. In the context of the problem, the linear momentum of the railroad handcar can be calculated by considering both its total mass and its velocity. In a frictionless environment, like the situation described, momentum is conserved. This means that the total momentum before and after any event must remain constant as long as no external forces act on the system. This principle allows us to solve problems involving changes in mass and velocity, such as objects being thrown off or into the handcar.

railroad handcar

A railroad handcar is a small, manually operated vehicle that moves along train tracks. In this exercise, it is important to consider the handcar along with its contents as a system. The total mass given includes both the handcar and anything inside it, initially totaling 200 kg. This scenario is ideal for studying momentum conservation because the tracks are described as frictionless.

• This means there is no external friction force acting against the handcar's movement, simplifying any calculations.
• Such conditions are theoretical but help to focus on the momentum principles without additional complicating factors.

Understanding these ideal conditions allows us to better apply the momentum conservation law when working through the problem scenarios.

frictionless tracks

Frictionless tracks are a theoretical construct that provide a great way to learn about kinetic principles without additional forces altering the outcome.

• In reality, there is always some degree of friction involved, but in physics problems, it is often set to zero to simplify the calculations.
• This setting ensures that momentum is perfectly conserved, allowing us to focus solely on the effects of internal actions, like mass being thrown from a vehicle.

Because there's no friction, the force normally required to counteract it is not needed, and the handcar only experiences forces from within, such as throwing a mass off or receiving it from another source. This ensures that external factors do not muddy our calculations and makes the track system an isolated system for analysis.

velocity calculation

Velocity calculation is crucial to solving the given problem scenarios. Every time an object is ejected or added to the handcar, we need to calculate the resulting velocity of the car. Let's break it down with the conservation of momentum formula:

• Before an event, the momentum \( p_{initial} \) of the handcar system is computed by multiplying its initial mass (200 kg) by the velocity (5.00 m/s).
• For scenarios involving ejection or addition of mass, the mass and velocity of the thrown object need to be carefully considered.
• The final velocity \( v_{final} \) can be determined by re-arranging the combined momentum equations to solve for the new velocity of the system.

For example, in Case B, when mass is ejected backward, equating the initial and final momentum gives us the new velocity: \[ v_{final} = \frac{p_{initial}}{m_{remaining}} \] This fundamental understanding of velocity and momentum interplay will help in accurately determining the vehicle's speed after each change.