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Chapter 2: Problem 15

Two cars of masses \(1200 \mathrm{kg}\) and \(1400 \mathrm{kg}\) collide head-on and stick to each other. The cars are coming at each other from opposite directions with speeds of \(8.0 \mathrm{m} \mathrm{s}^{-1}\) and \(6 \mathrm{m} \mathrm{s}^{-1}\), respectively. With what velocity does the wreck move away from the scene of the accident?

Short Answer

Expert verified
The wreck moves away with a velocity of approximately \(0.46 \, \text{m/s}\).

Step by step solution

01

- Identify the Principle

The principle to use here is the conservation of momentum, which states that the total momentum of a closed system remains constant before and after a collision.
02

- Write the Momentum Conservation Equation

The equation for the conservation of linear momentum is \[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \]where \(m_1 = 1200 \, \text{kg}\), \(v_1 = 8.0 \, \text{m/s}\), \(m_2 = 1400 \, \text{kg}\), and \(v_2 = -6.0 \, \text{m/s}\). Note that \(v_2\) is negative because the second car is moving in the opposite direction.
03

- Substitute the Values

Substitute the given values into the equation:\[1200 \, \text{kg} \times 8.0 \, \text{m/s} + 1400 \, \text{kg} \times (-6.0 \, \text{m/s}) = (1200 \, \text{kg} + 1400 \, \text{kg}) v_f \].
04

- Simplify and Solve

Calculate the momentum before the collision:\[ (1200 \, \text{kg} \times 8.0 \, \text{m/s}) + (1400 \, \text{kg} \times -6.0 \, \text{m/s}) \].This becomes:\[9600 \, \text{kg} \, \text{m/s} - 8400 \, \text{kg} \, \text{m/s} = 1200 \, \text{kg} \, \text{m/s} \].Now solve for the final velocity:\[ 1200 \, \text{kg} \, \text{m/s} = (1200 \, \text{kg} + 1400 \, \text{kg}) v_f \].\[ 1200 \, \text{kg} \, \text{m/s} = 2600 \, \text{kg} \times v_f \].\[ v_f = \frac{1200 \, \text{kg} \, \text{m/s}}{2600 \, \text{kg}} \].\[ v_f \approx 0.46 \, \text{m/s} \].

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Key Concepts

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

linear momentum
Linear momentum is a fundamental concept in physics that describes the motion of an object. It is defined as the product of an object's mass and its velocity: \( p = m \times v \). Linear momentum is a vector quantity, meaning it has both magnitude and direction.

In a closed system, the total linear momentum before and after an event (like a collision) remains constant. This is the principle of momentum conservation. For example, in the collision of two cars, the sum of their momenta before the collision equals the sum of their momenta after the collision.

Understanding linear momentum helps us predict the outcome of collisions and other interactions between objects.
collision physics
Collision physics studies the interactions between objects when they collide. There are different types of collisions, such as elastic, inelastic, and perfectly inelastic collisions.

An elastic collision is one where the total kinetic energy and momentum are conserved. In an inelastic collision, only the momentum is conserved, while some kinetic energy is converted into other forms of energy, like heat or sound. A perfectly inelastic collision is a special case where the colliding objects stick together after the impact, moving as a single unit.

For instance, in the given exercise, two cars collide head-on and stick together, indicating a perfectly inelastic collision. The principle of momentum conservation is applied to find the final velocity of the combined wreck.
mass and velocity
Mass and velocity are crucial components in calculating an object's momentum.

**Mass**: Mass represents the amount of matter in an object and is measured in kilograms (kg). In the exercise, the masses of the two cars are given as 1200 kg and 1400 kg.

**Velocity**: Velocity is the speed of an object in a specific direction and is measured in meters per second (m/s). It鈥檚 important to consider the direction when dealing with velocity, as it affects the momentum calculation. For instance, if two objects move towards each other, one will have a positive velocity, and the other a negative velocity.

When combined, mass and velocity give us the object's momentum. By understanding these concepts, we can solve complex problems related to motion and collisions efficiently.

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Most popular questions from this chapter

Your brand new convertible Ferrari is parked \(15 \mathrm{m}\) from its garage when it begins to rain. You do not have time to get the keys so you begin to push the car towards the garage. If the maximum acceleration you can give the car is \(2.0 \mathrm{m} \mathrm{s}^{-2}\) by pushing and \(3.0 \mathrm{m} \mathrm{s}^{-2} \mathrm{by}\) pulling back on the car, find the least time it takes to put the car in the garage. (Assume that the car, as well as the garage, are point objects.)

A car of mass 1200 kg starts from rest, accelerates uniformly to a speed of \(4.0 \mathrm{m} \mathrm{s}^{-1}\) in \(2.0 \mathrm{s}\) and continues moving at this constant speed in a horizontal straight line for an additional \(10 \mathrm{s}\). The brakes are then applied and the car is brought to rest in \(4.0 \mathrm{s}\). A constant resistance force of \(500 \mathrm{N}\) is acting on the car during its entire motion. (a) Calculate the force accelerating the car in the first \(2.0 \mathrm{s}\) of the motion. (b) Calculate the average power developed by the engine in the first \(2.0 \mathrm{s}\) of the motion. (c) Calculate the force pushing the car forward in the next \(10 \mathrm{s}\) (d) Calculate the power developed by the engine in those \(10 \mathrm{s}\) (e) Calculate the braking force in the last \(4.0 \mathrm{s}\) of the motion. (f) Describe the energy transformations that have taken place in the 16 s of the motion of this car.

A man of mass \(m\) stands in an elevator. Find the reaction force from the elevator floor on the man when: (a) the elevator is standing still; (b) the elevator moves up at constant speed \(v\) (c) the elevator accelerates down with acceleration a; (d) the elevator accelerates down with acceleration \(a=g\) (e) What happens when \(a>g ?\)

A rocket accelerates vertically upwards from rest with a constant acceleration of \(4.00 \mathrm{m} \mathrm{s}^{-2} .\) The fuel lasts for \(5.00 \mathrm{s}\) (a) What is the maximum height achieved by this rocket? (b) When does the rocket reach the ground again? (c) Sketch a graph to show the variation of the velocity of the rocket with time from the time of launch to the time it falls to the ground. (Take the acceleration due to gravity to be \(10.0 \mathrm{m} \mathrm{s}^{-2} .\)

A mass swings at the end of a string like a pendulum. Draw the forces on the mass at: (a) its lowest position; (b) its highest position.
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