/*! 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 76 Automobile Accident Analysis. Yo... [FREE SOLUTION] | 91Ó°ÊÓ

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Automobile Accident Analysis. You are called as an expent witness to analyre the following auto accident: Car \(B\), of mass \(1900 \mathrm{~kg}\), was stopped at a red light when it was hit from bchind by car \(A\), of mass \(1500 \mathrm{~kg}\). The cars locked bumpers during the collision and slid to a stop with hrakes locked on all wheels. Measurements of the skid marks left by the tires showed them to be \(7.15 \mathrm{~m}\) long. The coeflicient of kinetic friction between the tires and the road was 0.65 . (a) What was the speed of car \(A\) just before the collision? (b) If the speed limit was \(35 \mathrm{mph}\), was car \(A\) spceding, and if so, by how many milcs per hour was it exceeding the speed limit?

Short Answer

Expert verified
The speed of Car A just before collision is calculated to define if it was speeding. By comparing this speed with the speed limit, we can determine if Car A was over the speed limit and, if so, by how much.

Step by step solution

01

Calculate the combined momentum before and after the collision

The momentum is conserved during the collision. Hence the total momentum before the collision will be equal to the momentum after the collision. Before the collision, the momentum is only from car A as car B was stationary. And after the collision, the momentum is from both the cars as they moved together. Let \(v\) be the velocity of both cars after collision. Therefore, \( Momentum_{before} = Momentum_{after} \Rightarrow m_A * v_A = (m_A + m_B) *v \) (equation 1).Here, \(m_A\) is the mass of car A, \(m_B\) is the mass of car B and \(v_A\) is the velocity of car A.
02

Calculate the initial speed of car A using the friction and length of skid marks

After the collision, kinetic friction acted on the cars until they came to a stop. The work done by the friction is equal to the change in kinetic energy of the cars. Using the work-energy theorem:\(0.5 * (m_A + m_B) * v^2 = Friction * Distance => 0.5 * (m_A + m_B) * v^2 = \mu_k * (m_A + m_B) * g * d\) (equation 2).Here, \( \mu_k \) is the kinetic friction coefficient, g is the acceleration due to gravity and d is the length of the skid marks.
03

Solve for \(v\)

Solving equation 2 for \(v\) to determine the speed just after the collision (\(v = \sqrt{\mu_k * g * d * 2}\)), and substitute this into equation 1 to determine the speed before collision, \(v_A= \frac{(m_A + m_B) * v }{m_A}\).
04

Compare to the speed limit

At last, transform \(v_A\) into miles per hour and compare with the speed limit to see if car A was speeding and by how much.

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Key Concepts

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

Momentum Conservation
In collision physics, momentum conservation is a fundamental concept. Momentum, a measure of the motion of an object, is the product of its mass and velocity. During a collision, the total momentum before the event is equal to the total momentum after, assuming no external forces are acting.

In the provided car collision scenario, only car A had initial momentum, as car B was stationary. Once the collision occurs, the combined mass of both cars moves together, so their combined momentum post-collision must be equivalent to car A’s initial momentum. This is mathematically expressed as:
  • Before collision: Momentum = Mass of A * Velocity of A ( m_A * v_A )
  • After collision: Momentum = Combined Mass * Velocity after impact ( (m_A + m_B) * v )
By equating these two momentum expressions, we can solve for the unknown velocity just after the collision. This step is vital for understanding the events leading to the skid marks observed.
Work-Energy Theorem
The work-energy theorem connects the forces acting on an object to its kinetic energy. In the context of the car collision, this theorem helps to explain how the kinetic energy of the cars changes due to friction while they skid to a stop.

After the cars lock together post-collision, kinetic friction opposes their motion. The work done by friction is equal to the change in kinetic energy—essentially, the initial kinetic energy is fully converted into work done by friction until they stop. This is described by:
  • Work done by friction = Kinetic energy change
  • Friction force * Distance = 0.5 * Combined Mass * Velocity2
By integrating the coefficient of kinetic friction ( \( \mu_k \) and skid distance, we derive the velocity immediately post-collision. Subsequently, this helps pinpoint the speed of car A prior to impact when combined with momentum conservation.
Kinetic Friction
Kinetic friction plays a pivotal role in one's understanding of why and how objects come to a stop after collisions. It is the force that opposes the relative motion of sliding surfaces in contact. In the car accident analysis, kinetic friction helps bring the combined vehicle mass to rest.

This force can be calculated using the coefficient of kinetic friction ( \( \mu_k \) which is a measure of how resistive the surfaces are. The kinetic friction force is expressed as:
  • Kinetic Friction Force = \( \mu_k * \text{Normal Force} \)
  • Normal Force = mass * gravity ( (m_A + m_B) * g )
This frictional force then performs work over the skid distance, converting the kinetic energy of the moving mass into thermal energy, thus fully stopping the cars. Understanding how this energy conversion details the mechanics of stopping distances allows for accurate speed estimates post-collision.

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Most popular questions from this chapter

A \(0.160 \mathrm{~kg}\) hockey puck is moving on an icy, frictionless, horizontal surface. At \(t=0\), the puck is moving to the right at \(3.00 \mathrm{~m} / \mathrm{s}\). (a) Calculate the velocity of the puck (magnitude and direction) after a force of \(25.0 \mathrm{~N}\) directed to the right has been applied for \(0.050 \mathrm{~s}\). (b) II. instead, a forec of \(12.0 \mathrm{~N}\) dirccted to the left is applied from \(t=0\) to \(t=0.050 \mathrm{~s},\) what is the final velocity of the puck?

Just before it is struck by a racket, a tennis ball weighing \(0.560 \mathrm{~N}\) has a velocity of \((20.0 \mathrm{~m} / \mathrm{s}) \hat{\imath}-(4.0 \mathrm{~m} / \mathrm{s}) \hat{\jmath} .\) During the \(3.00 \mathrm{~ms}\) that the racket and ball are in contact, the net external force on the ball is constant and equal to \(-(380 \mathrm{~N}) \hat{\imath}+(110 \mathrm{~N}) \hat{\jmath}\). What are the \(x\) - and \(y\) -components (a) of the impulse of the net external force applied to the ball; (b) of the final velocity of the ball?

When cars are equipped with flexible bumpers, they will bounce off cach other during low-specd collisions, thus causing less damage, In one such accident, a \(1750 \mathrm{~kg}\) car traveling to the right at \(1.50 \mathrm{~m} / \mathrm{s}\) collides with a \(1450 \mathrm{~kg}\) car going to the left at \(1.10 \mathrm{~m} / \mathrm{s}\) Measurements show that the heavier car's speed just after the collision was \(0.250 \mathrm{~m} / \mathrm{s}\) in its original direction. Ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the change in the combined kinetic eneryy of the two-car system during this collision.

A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of \(200 \mathrm{~kg}\) and is traveling east with a velocity of magnitude \(5.00 \mathrm{~m} / \mathrm{s}\). Find the final velocity of the car in cach casc, assuming that the handcar docs not leave the tracks. (a) A \(25.0 \mathrm{~kg}\) mass is thrown sideways out of the car with a velocity of magnitude \(2.00 \mathrm{~m} / \mathrm{s}\) relative to the car's initial velocity, (b) \(\mathrm{A} 25.0 \mathrm{~kg}\) mass is thrown backward out of the car with a velocity of \(5.00 \mathrm{~m} / \mathrm{s}\) relative to the initial motion of the car. (c) A \(25.0 \mathrm{~kg}\) mass is thrown into the car with a velocity of \(6.00 \mathrm{~m} / \mathrm{s}\) relative to the ground and opposite in direction to the initial velocity of the car.

The mass of a regulation tennis ball is 57 g (although it can vary slighaly), and tests have shown that the ball is in conthet with the tennis racket for \(30 \mathrm{~ms}\). (This number can also vary, depending on the racket and swing.) We shall assume a \(30.0 \mathrm{~ms}\) contact time. One of the fastest-known served tennis balls was served by "Big Bill" Tilden in \(1931,\) and its speed was measured to be \(73 \mathrm{~m} / \mathrm{s}\). (a) What impulse and what total force did Big Bill exert on the tennis ball in his record serve? (b) If Big Bill's opponcnt returned his serve with a speed of \(55 \mathrm{~m} / \mathrm{s},\) what total force and what impulse did he exert on the ball, assuming only horizontal motion?

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