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Chapter 8: Problem 38

Accident Analysis. Two cars collide at an intersection. Car \(A,\) with a mass of 2000 \(\mathrm{kg}\) , is going from west to east, while car \(B,\) of mass \(1500 \mathrm{kg},\) is going from north to south at 15 \(\mathrm{m} / \mathrm{s}\) . As a result of this collision, the two cars become enmeshed and move as one afterward. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of \(65^{\circ}\) south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car \(A\) going just before the collision?

Short Answer

Expert verified
The enmeshed cars moved at 7.67 m/s. Car A was going at 11.65 m/s before the collision.

Step by step solution

01

Understand the Problem

We have two cars with a given mass colliding and moving together after the collision. We are required to find the velocity of the combined mass after the collision and the velocity of car A before the collision. We will be using the principles of conservation of momentum in two directions: east and south.
02

Establish Conservation of Momentum in the East Direction

Since car B is moving from north to south, it does not contribute to the eastward momentum before the collision. Thus, the total momentum in the east direction before and after the collision is due to car A only. Let the velocity of car A before collision be \( v_A \). The momentum equation is:\[ m_A v_A = (m_A + m_B) v \, \cos(65^\circ) \]where \( m_A = 2000 \, \text{kg} \), \( m_B = 1500 \, \text{kg} \), and \( v \, \cos(65^\circ) \) is the eastward component of the velocity after collision.
03

Establish Conservation of Momentum in the South Direction

Car A does not contribute to the southward momentum before the collision. Initially, only car B has momentum in this direction:\[ m_B v_B = (m_A + m_B) v \, \sin(65^\circ) \]where \( v_B = 15 \, \text{m/s} \) is the velocity of car B before the collision and \( v \, \sin(65^\circ) \) is the southward component of the velocity after collision.
04

Solve for the Velocity After Collision, \( v \)

Using the southward momentum equation:\[ 1500 \, \text{kg} \times 15 \, \text{m/s} = (2000 \, \text{kg} + 1500 \, \text{kg}) v \, \sin(65^\circ) \]Solve for \( v \):\[ v = \frac{1500 \times 15}{3500 \times \sin(65^\circ)} \approx 7.67 \, \text{m/s} \]
05

Solve for the Velocity of Car A Before Collision, \( v_A \)

Using the eastward momentum equation:\[ 2000 \, \text{kg} \times v_A = 3500 \, \text{kg} \times 7.67 \, \text{m/s} \, \cos(65^\circ) \]Solve for \( v_A \):\[ v_A = \frac{3500 \times 7.67 \times \cos(65^\circ)}{2000} \approx 11.65 \, \text{m/s} \]

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Key Concepts

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

Collision Analysis
Collision analysis is a process used to understand the dynamics of two or more objects colliding. In a collision, two key aspects are typically examined: momentum and energy. This analysis begins by identifying the types of objects involved and their movements before and after the collision. For the scenario presented here, we are dealing with two vehicles colliding and then moving together as one unit. This suggests it is an inelastic collision, where the total kinetic energy is not conserved, but the total momentum is preserved. Understanding these dynamics requires breaking down movements and applying the law of conservation of momentum.
Momentum in Two Directions
Momentum, a vector quantity, depends on both the mass and velocity of an object. To fully understand collisions, especially when involving more than one direction, momentum must be considered in each direction separately. In the exercise given, momentum is analyzed in the east and south directions:
  • The eastward momentum is only affected by Car A, as Car B was moving north to south.
  • The southward momentum is only contributed by Car B.
Using components simplifies solving such problems, with calculations involving trigonometric functions (like sine and cosine) to find these components. By applying this principle in both east and south directions, we ensure we account for all aspects of movement in this collision scenario.
Velocity Calculation
Velocity calculation in this problem involves determining the speed of objects before and after the collision. It's vital to apply the conservation of momentum to find unknown velocities. The problem involves determining:
  • The speed of the enmeshed cars just after the collision.
  • The initial speed of Car A before the collision occurred.
Using the southward momentum formula, we calculated the speed of the enmeshed cars after collision: \[ v = \frac{1500 \times 15}{3500 \times \sin(65^\circ)} \approx 7.67 \, \text{m/s} \]From the eastward momentum, we solved for Car A's initial velocity: \[ v_A = \frac{3500 \times 7.67 \times \cos(65^\circ)}{2000} \approx 11.65 \, \text{m/s} \]
Angle of Impact
The angle of impact provides insight into the post-collision trajectory of the combined mass. It helps determine the resultant path when momentum components in different directions are combined.In our scenario, the cars moved at an angle of \(65^\circ\) south of east after the collision. This angle tells us how the momentum components (east and south) affect the post-collision path. By decomposing the total momentum into directional components, we use trigonometric functions:
  • \( \cos(65^\circ) \) relates to the eastward momentum.
  • \( \sin(65^\circ) \) corresponds to the southward component.
Understanding the angle helps with identifying how momentum interactions between colliding objects influence their final direction.

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Most popular questions from this chapter

A 15.0 -kg fish swimming at 1.10 \(\mathrm{m} / \mathrm{s}\) suddenly gobbles up a 4.50 -kg fish that is initially stationary. Neglect any drag effects of the water. (a) Find the speed of the large fish just after it eats the small one. (b) How much mechanical energy was dissipated during this meal?

Three identical pucks on a horizontal air table have repelling magnets. They are held together and then released simultaneously. Each has the same speed at any instant. One puck moves due west. What is the direction of the velocity of each of the other two pucks?

An 8.00 -kg block of wood sits at the edge of a frictionless table, 2.20 \(\mathrm{m}\) above the floor. A \(0.500-\mathrm{kg}\) blob of clay slides along the length of the table with a speed of 24.0 \(\mathrm{m} / \mathrm{s}\), strikes the block of wood, and sticks to it. The combined object leaves the edge of the table and travels to the floor. What horizontal distance has the combined object traveled when it reaches the floor?

A \(1050\) -kg sports car is moving westbound at 15.0 \(\mathrm{m} / \mathrm{s}\) on a level road when it collides with a 6320 -kg truck driving east on the same road at 10.0 \(\mathrm{m} / \mathrm{s}\) . The two vehicles remain locked together after the collision. (a) What is the velocity (magnitude and direction) of the two vehicles just after the collision? (b) At what speed should the truck have been moving so that it and the car are both stopped in the collision? (c) Find the change in kinetic energy of the system of two vehicles for the situations of part (a) and part (b). For which situation is the change in kinetic energy greater in magnitude?

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a 0.400 -kg ball that is traveling horizontally at 10.0 \(\mathrm{m} / \mathrm{s}\) . Your mass is 70.0 \(\mathrm{kg}\) . (a) If you catch the ball, with what speed do you and the ball move afterward? (b) If the ball hits you and bounces off your chest, so afterward it is moving horizontally at 8.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction, what is your speed after the collision?
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