/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 At illustrates how to model a pr... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 33

At illustrates how to model a problem similar to this one. An automobile has a mass of \(2100 \mathrm{~kg}\) and a velocity of \(+17 \mathrm{~m} / \mathrm{s}\). It makes a rear-end collision with a stationary car whose mass is \(1900 \mathrm{~kg} .\) The cars lock bumpers and skid off together with the wheels locked. (a) What is the velocity of the two cars just after the collision? (b) Find the impulse (magnitude and direction) that acts on the skidding cars from just after the collision until they come to a halt. (c) If the coefficient of kinetic friction between the wheels of the cars and the pavement is \(\mu_{k}=0.68,\) determine how far the cars skid before coming to rest.

Short Answer

Expert verified
(a) 8.925 m/s, (b) 35700 Ns opposite initial direction, (c) 59.53 m.

Step by step solution

01

Conservation of Momentum

To find the velocity of the two cars after the collision, we apply the law of conservation of momentum. Before the collision only the moving car has momentum. The total initial momentum is the momentum of the moving car, calculated as \( p_i = m_1 imes v_1 = 2100 imes 17 \). After the collision, the total momentum is the sum of the momenta of both cars moving together: \( p_f = (m_1 + m_2) imes v_f \). Set \( p_i = p_f \) and solve for \( v_f \).\[ 2100 \times 17 = (2100 + 1900) \times v_f \].
02

Solve for Final Velocity

Solve the equation \[ 2100 \times 17 = 4000 \times v_f \] for \( v_f \). Divide both sides by 4000 to find \( v_f = \frac{2100 \times 17}{4000} \). Calculating gives \( v_f = 8.925 \, \text{m/s} \).
03

Calculate Impulse

Impulse is the change in momentum. The initial momentum of the cars together is \( P_i = 4000 \times 8.925 \). They come to rest, so the final momentum is 0. The impulse \( I \) is \( I = 0 - P_i = - (4000 \times 8.925) \), which simplifies to \( I = -35700 \, \text{Ns} \). The negative sign indicates the impulse direction is opposite to the initial movement.
04

Determine Skid Distance Using Friction

Use the work-energy principle. The work done by friction \( W_f \) equals the initial kinetic energy of the system. Calculate initial kinetic energy: \( KE_i = \frac{1}{2} \times 4000 \times (8.925)^2 \). Solve \( W_f = KE_i \) where \( W_f = \mu_k \times m \times g \times d \). Let \( d \) be the skid distance, \( m = 4000 \), \( g = 9.8 \, \text{m/s}^2 \), and \( \mu_k = 0.68 \). Solve \[ 0.68 \times 4000 \times 9.8 \times d = 159345 \], obtaining \( d \approx 59.53 \, \text{m} \).

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Key Concepts

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

Impulse and Momentum
Impulse and momentum are important concepts in understanding how objects move and react to forces. Momentum is a measure of an object's motion and is calculated by multiplying an object's mass by its velocity. In this context, momentum is expressed as \( p = m \times v \). For an automobile with a mass of \( 2100 \, \text{kg} \) moving at a velocity of \( 17 \, \text{m/s} \), momentum is \( 2100 \times 17 = 35700 \, \text{kg} \cdot \text{m/s} \).

Impulse is the change in momentum caused by a force acting over a period of time. It is equal to the final momentum minus the initial momentum. In the case of a collision, where two cars come to a halt together, their initial combined momentum is \( 4000 \times 8.925 = 35700 \, \text{Ns} \). Since they eventually stop, their final momentum is zero. Thus, the impulse is \( -35700 \, \text{Ns} \), which indicates a force acting opposite to their motion was applied to stop them.

Please keep in mind:
  • Impulse is equal to the change in momentum.
  • The direction is significant; negative impulse often means acting opposite to the initial movement.
  • Momentum is conserved in isolated systems, such as the collision here where no external forces act.
Kinetic Friction
Kinetic friction is the resistive force that acts between moving surfaces. When the cars lock bumpers and skid together, kinetic friction is what slows and eventually stops them.

This type of friction is calculated using the coefficient of kinetic friction \( \mu_k \) and the normal force \( F_n \). The normal force usually equals the weight of the objects involved, in this case the combined weight of the cars \( m \times g \). The formula for kinetic friction is \( F_f = \mu_k \times F_n \).

To determine how far the cars skid, the work done by kinetic friction (\( W_f \)) is analyzed:
  • The frictional work stops the cars by converting their kinetic energy to heat during the slide.
  • \( \mu_k \) is given as 0.68.
  • The normal force is \( 4000 \times 9.8 \) for the weight of the locked cars.


By calculating \( 0.68 \times 4000 \times 9.8 \times d = 159345 \) and solving for \( d \), we find the skid distance is approximately 59.53 meters. Kinetic friction plays the crucial role in determining this stopping distance.
Work-Energy Principle
The work-energy principle connects the kinetic energy of an object with the work done on it. This principle states that the work done by all forces (includiing friction) on a body is equal to its change in kinetic energy.

Kinetic energy (\( KE \)) is given by the formula \( KE = \frac{1}{2} m v^2 \). For the post-collision combined car system:
  • The initial kinetic energy is \( \frac{1}{2} \times 4000 \times (8.925)^2 \).
  • This total, approximately \( 159345 \, \text{J} \), represents the energy available to be dissipated by friction.
  • The work \( W \) done by kinetic friction is equal to the kinetic energy loss as the cars come to rest.


In this exercise, using \( W_f = \mu_k \times m \times g \times d \) connects the energy transformation to the skid distance. Solving this equation ensures understanding of how forces transform energy to stop the moving vehicles. This principle is fundamental in analyzing scenarios with kinetic energy changes due to external work, exemplified here by friction.

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Most popular questions from this chapter

Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. (a) Is it possible that the two-object system has a final total momentum of zero after the collision? (b) Roughly, what is the direction of the final total momentum of the two-object system after the collision? Problem Object A has a mass of \(m_{A}=17.0 \mathrm{~kg}\) and an initial velocity of \(\overrightarrow{\mathbf{v}}_{0 \mathrm{~A}}=8.00 \mathrm{~m} / \mathrm{s},\) due east. Object \(\mathrm{B},\) however, has a mass of \(m_{\mathrm{B}}=29.0 \mathrm{~kg}\) and \(\mathrm{an}\) initial velocity of \(\overrightarrow{\mathbf{v}}_{0 \mathrm{~B}}=5.00 \mathrm{~m} / \mathrm{s},\) due north. Find the magnitude and direction of the total momentum of the two-object system after the collision. Make sure that your answers are consistent with your answers to the Concept Questions.

A \(55-\mathrm{kg}\) swimmer is standing on a stationary \(210-\mathrm{kg}\).oating raft. The swimmer then runs off the raft horizontally with a velocity of \(+4.6 \mathrm{~m} / \mathrm{s}\) relative to the shore. Find the recoil velocity that the raft would have if there were no friction and resistance due to the water.

ssm One average force \(\overrightarrow{\mathrm{F}}_{1}\) has a magnitude that is three times as large as that of another average force \(\overline{\mathrm{F}}_{2} .\) Both forces produce the same impulse. The average force \(\overrightarrow{\mathrm{F}}_{1}\) acts for a time interval of \(3.2 \mathrm{~ms}\). For what time interval does the average force \(\overrightarrow{\mathrm{F}}_{2}\) act?

ssm A golf ball bounces down a flight of steel stairs, striking several steps on the way down, but never hitting the edge of a step. The ball starts at the top step with a vertical velocity component of zero. If all the collisions with the stairs are elastic, and if the vertical height of the staircase is \(3.00 \mathrm{~m}\), determine the bounce height when the ball reaches the bottom of the stairs. Neglect air resistance.

A volleyball is spiked so that its incoming velocity of \(+4.0 \mathrm{~m} / \mathrm{s}\) is changed to an outgoing velocity of \(-21 \mathrm{~m} / \mathrm{s}\). The mass of the volleyball is \(0.35 \mathrm{~kg}\). What impulse does the player apply to the ball?
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