Question

Two vehicles are approaching an intersection. One is a 2500-kg pickup traveling at 14.0 m/s from east to west (the \(-x\)-direction), and the other is a 1500-kg sedan going from south to north (the \(+y\) direction) at 23.0 m/s. (a) Find the \(x\)- and \(y\)-components of the net momentum of this system. (b) What are the magnitude and direction of the net momentum?

Short answer

(a) Net momentum components: \(-35,000 \text{ kg}\,\text{m/s}\) in x, \(34,500 \text{ kg}\,\text{m/s}\) in y. (b) Magnitude: \(49,155 \text{ kg}\,\text{m/s}\); Direction: \(44.98°\) above \(-x\)-axis.

Step-by-step solution

Identify Given Data

Let's identify the information provided: The pickup has a mass of 2500 kg and is moving at a velocity of 14.0 m/s in the \(-x\)-direction. The sedan has a mass of 1500 kg and moves at a velocity of 23.0 m/s in the \(+y\)-direction.

Calculate Momentum Components of the Pickup

The momentum \((p)\) of an object is given by the product of its mass \((m)\) and its velocity \((v)\). For the pickup:- Mass \(m_1 = 2500 \text{ kg}\)- Velocity \(v_{x1} = 14.0 \text{ m/s}\) (since it's in the -x direction, velocity will be \(-14.0 \, \text{m/s}\)).Thus, the \(x\)-component of the momentum \(p_{x1}\) can be calculated as:\[ p_{x1} = m_1 \times v_{x1} = 2500 \times (-14.0) = -35,000 \, \text{kg} \, \text{m/s}\]The \(y\)-component is zero because the pickup is only moving in the \(x\)-direction.

Calculate Momentum Components of the Sedan

Similarly, we calculate the momentum for the sedan, which is only moving in the \(y\)-direction.- Mass \(m_2 = 1500 \text{ kg}\)- Velocity \(v_{y2} = 23.0 \text{ m/s}\)The \(x\)-component of the sedan's momentum \(p_{x2}\) is zero because it has no motion in the \(x\)-direction.The \(y\)-component \(p_{y2}\) is calculated as:\[ p_{y2} = m_2 \times v_{y2} = 1500 \times 23.0 = 34,500 \, \text{kg} \, \text{m/s}\]

Calculate Net Momentum Components

Combine the momentum components calculated to find the net momentum components. The net \(x\)-component of momentum \(p_{x_{net}}\) is:\[ p_{x_{net}} = p_{x1} + p_{x2} = -35,000 + 0 = -35,000 \, \text{kg} \, \text{m/s}\]The net \(y\)-component of momentum \(p_{y_{net}}\) is:\[ p_{y_{net}} = p_{y1} + p_{y2} = 0 + 34,500 = 34,500 \, \text{kg} \, \text{m/s}\]

Calculate Magnitude of Net Momentum

Use the Pythagorean theorem to find the magnitude of the net momentum \(p_{net}\): \[ p_{net} = \sqrt{(p_{x_{net}})^2 + (p_{y_{net}})^2} = \sqrt{(-35,000)^2 + (34,500)^2}\]Calculate this to get:\[ p_{net} = \sqrt{1225000000 + 1190250000} = \sqrt{2415250000} \approx 49,155 \, \text{kg} \, \text{m/s}\]

Determine Direction of Net Momentum

The direction \(\theta\) of the net momentum vector can be found using the arctangent function:\[ \theta = \tan^{-1} \left( \frac{p_{y_{net}}}{p_{x_{net}}} \right) = \tan^{-1} \left( \frac{34,500}{-35,000} \right) \]Calculate this to find:\[ \theta \approx \tan^{-1}(-0.9857) \approx -44.98°\]Since the angle is measured from the \(-x\)-axis toward the \(+y\)-axis, the net momentum is \(44.98°\) above the \(-x\)-axis.

Key concepts

Understanding Collision and Its Effect on Momentum

Collision refers to the event where two or more objects exert forces on each other for a short duration. In physics, collisions play a crucial role in understanding how objects interact, especially in terms of momentum.
When a collision occurs, the total momentum of the system (combining all involved objects) before the collision is equal to the total momentum after the collision. This principle is known as the conservation of momentum.
For example:

• The pickup truck and the sedan at the intersection each contribute differently to the system's momentum.
• This scenario, though focusing on momentum components, is built on the concept of collision where two different paths meet, theoretically creating a point of impact.

Understanding how these vehicles interact helps us determine the new direction and velocity after a potential collision, crucial in many fields like automotive safety and physics research.

Breaking Down Momentum into Components

Momentum, a vector quantity, can be broken into components. This is especially useful when objects are moving in two or more dimensions. Each component corresponds to motion in a specific direction, like the example of cars traveling along negative x and positive y axes.

• The pickup's x-component of momentum illustrates this directionality as it moves westward.
• For the sedan, we consider the y-component because its motion is northward.

To find the components of momentum, we multiply the mass by velocity in each respective direction:

• Pickup’s x-component: \[ p_{x1} = m_1 \times v_{x1} \]Where \( m_1 = 2500 \text{ kg} \) and \( v_{x1} = -14.0 \text{ m/s}\).
• Sedan’s y-component: \[ p_{y2} = m_2 \times v_{y2} \]Where \( m_2 = 1500 \text{ kg} \) and \( v_{y2} = 23.0 \text{ m/s} \).

This breakdown is essential for solving real-world problems where objects don’t travel in a straight line but along varied paths.

Applying the Pythagorean Theorem to Momentum

When we need to find the magnitude of momentum that has been resolved into perpendicular components, we use the Pythagorean theorem. This ancient principle from geometry helps us combine x and y components into one resultant momentum vector.
Given our vehicles, we have:

• The \( x\) component,\(-35,000 \text{ kg m/s} \),from the pickup.
• The \( y\) component,\(34,500 \text{ kg m/s} \),from the sedan.

To find the net momentum, we calculate:\[ p_{net} = \sqrt{(p_{x_{net}})^2 + (p_{y_{net}})^2} \]Where \( p_{x_{net}} = -35,000 \) and \( p_{y_{net}} = 34,500 \).This results in \( \sqrt{2415250000} \approx 49,155 \text{ kg m/s} \).Applying this theorem confirms the magnitude of momentum considering both directional influences, indispensable in determining the system's overall behavior.

Calculating the Direction of Momentum

The direction of the resultant momentum gives us insightful information about the motion pattern after analyzing components. The focus here is to combine components to establish overall direction using trigonometric functions.
We use the arctangent function (tan⁻¹) to find the angle \( \theta \):

• \[ \theta = \tan^{-1} \left( \frac{p_{y_{net}}}{p_{x_{net}}} \right) \]
• Substituting our components:\[ \theta = \tan^{-1} \left( \frac{34,500}{-35,000} \right) \]results in \(-44.98^{\circ}\).

Since angles are measured counterclockwise from the positive x-axis, \(-44.98^{\circ}\) translates to \(44.98^{\circ}\) above the negative x-axis.
This calculation completes our analysis by indicating not just how fast, but also where collectively the movement tends to go,vital for fully understanding collision dynamics and resultant paths.