Problem 5 - Applied Physics

Question

A vehicle with a mass of \(1000 \mathrm{~kg}\) is going east at a velocity of \(30.0 \mathrm{~m} / \mathrm{s}\). It collides with a stationary vehicle of the same mass and is deflected \(35.0^{\circ}\) north of its original course. The second vehicle's final path is \(90^{\circ}\) to the right of the final path of the first vehicle. (a) What is the momentum of the first vehicle after the collision? (b) What is the momentum of the second vehicle after the collision? (c) What is the velocity of the first vehicle after the collision? (d) What is the velocity of the second vehicle after the collision?

Step-by-Step Solution

Answer

The momentum of Vehicle 1 after the collision is 24,570 kg m/s. The momentum of Vehicle 2 is 13,390 kg m/s. Vehicle 1's velocity is approximately 24.57 m/s; Vehicle 2's is about 13.39 m/s.

Step 1: Diagram and Known Values

We start by visualizing the collision. Initially, Vehicle 1 travels east. After the collision, it moves 35 degrees north of east. Vehicle 2 moves 90 degrees to the right of Vehicle 1's final path, meaning it heads directly south of east, or 55 degrees east of south. The mass of each vehicle is known to be 1000 kg.

Step 2: Conservation of Momentum

The conservation of momentum must hold for this collision, as no external forces are acting on the system. The total system momentum before the collision is equal to the total system momentum after:\[m_{1}v_{1i} = m_{1}v_{1f} + m_{2}v_{2f}\]where \(v_{1i} = 30 \text{ m/s}\) east. Since \(v_{2i} = 0\), the initial momentum is entirely from Vehicle 1.

Step 3: Resolve into Components

We decompose the momentum into x (east-west) and y (north-south) components. Initially, all momentum is eastward:\[p_{xi} = m_{1}v_{1i} = 30,000 \text{ kg m/s}, \quad p_{yi} = 0 \text{ kg m/s}\] For Vehicle 1 after the collision:\[p_{x1f} = m_{1}v_{1f}\cos(35^{\circ}), \ p_{y1f} = m_{1}v_{1f}\sin(35^{\circ})\]For Vehicle 2:\[p_{x2f} = m_{2}v_{2f}\cos(55^{\circ}), \ p_{y2f} = -m_{2}v_{2f}\sin(55^{\circ})\]

Step 4: Solve for Final Momenta

We use the conservation of momentum equations for each axis:\[p_{xi} = p_{x1f} + p_{x2f}, \ p_{yi} = p_{y1f} + p_{y2f}\]Substituting in values and solving gives:For x-direction:\[30,000 = 1000v_{1f}\cos(35^{\circ}) + 1000v_{2f}\cos(55^{\circ})\]For y-direction:\[0 = 1000v_{1f}\sin(35^{\circ}) - 1000v_{2f}\sin(55^{\circ})\]

Step 5: Calculate Velocities

By solving the system of equations from Step 4, we find\[v_{1f} \approx 24.57 \text{ m/s}\] east 35 degrees north, and \[v_{2f} \approx 13.39 \text{ m/s}\] east 55 degrees south.Vehicle 1's momentum after the collision is:\[m_1v_{1f} = 1000 \times 24.57 = 24,570 \text{ kg m/s}\]Vehicle 2's momentum after the collision is:\[m_2v_{2f} = 1000 \times 13.39 = 13,390 \text{ kg m/s}\]

Key Concepts

Collision Physics

In collision physics, two or more objects exert forces on each other for a relatively short time, often resulting in significant changes in their velocities and momenta. When considering a collision, we categorize it as elastic or inelastic.
A key principle in collision physics is the conservation of momentum. This principle states that in a closed system, the total momentum before the collision equals the total momentum after the collision.
Momentum is a vector quantity, meaning it has both magnitude and direction. Therefore, when we analyze collisions, we not only consider the speed of the objects but also the direction in which they are moving. This directional aspect becomes critical, especially in scenarios involving angular deflection post-collision, as seen in our given exercise.

Momentum Components

When dealing with momentum in vector form, breaking it down into components is crucial. This process simplifies the mathematics involved, particularly when dealing with collisions at angles.
For our scenario, we need to analyze the motion in two dimensions: the x-axis (east-west direction) and the y-axis (north-south direction). Initially, all momentum is in the x-direction since the first vehicle travels east.
After the collision, the momentum of each vehicle will have components in both the x and y directions.
To find each component, we use trigonometric functions:

• The x-component is found using: \(p_{x} = p \cos(\theta)\) where \(\theta\) is the angle to the x-axis.
• The y-component is found using: \(p_{y} = p \sin(\theta)\)

This breakdown allows us to write separate equations for each direction, aiding in the resolution of the system.

Velocity Calculation

Velocity is a critical aspect of analyzing collisions, being a vector itself, meaning it has both speed and direction. In our problem, calculating post-collision velocities is quintessential for understanding how the vehicles move after impact.
To calculate these velocities, we rely on the conservation of momentum principle and resolve the momentum into components.
For the calculations:

• First, we apply conservation of momentum in the x-direction: \[30,000 = 1000v_{1f}\cos(35^{\circ}) + 1000v_{2f}\cos(55^{\circ})\]
• Next, we apply it in the y-direction: \[0 = 1000v_{1f}\sin(35^{\circ}) - 1000v_{2f}\sin(55^{\circ})\]

These equations reflect how momentum is distributed between the two vehicles post-collision. Solving this system of equations yields the final velocities:
\(v_{1f} \approx 24.57 \text{ m/s}\) and \(v_{2f} \approx 13.39 \text{ m/s}\).

Vector Resolution

Vector resolution involves breaking a vector into its constituent components, usually in the x and y directions for two-dimensional motion.
This process is essential for analyzing collisions, especially in our problem where vehicles move in non-linear paths.
Understanding vector resolution includes knowing:

• The angle at which an object moves after an interaction (like a collision).
• How to decompose the momentum or velocity vectors into x and y components using trigonometric identities.

Let's see this in our exercise. After the collision, the first vehicle moves at a deflection of 35 degrees north of east. By resolving its velocity vector, we find:

• \(v_{x1f} = v_{1f}\cos(35^{\circ})\)
• \(v_{y1f} = v_{1f}\sin(35^{\circ})\)

Such decomposition allows accurate calculations of forces and movements in each direction, ensuring precise application of conservation laws.