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Automobile Accident Analysis. You are called as an expert witness to analyze the following auto accident: Car \(B,\) of mass \(1900 \mathrm{kg},\) was stopped at a red light when it was hit from behind by car \(A,\) of mass 1500 \(\mathrm{kg}\) . The cars locked bumpers during the collision and slid to a stop with brakes locked on all wheels. Measurements of the skid marks left by the tires showed them to be 7.15 \(\mathrm{m}\) long. The coefficient 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 \(A\) speeding, and if so, by how many miles per hour was it exceeding the speed limit?

Short Answer

Expert verified
Car A was traveling at approximately 40.3 mph before the collision and was exceeding the speed limit by about 5.3 mph.

Step by step solution

01

Understand the Formula Needed

In this problem, we need to find the speed of car A just before the collision. The principle of conservation of momentum and the concept of work-energy related to friction will be applied. We must understand:1. Conservation of momentum: \( m_A v_{A\text{before}} = (m_A + m_B) v_{\text{after}} \)2. Work done by friction to stop both cars: \( \text{work} = \text{friction force} \times \text{distance} = \frac{1}{2}(m_A + m_B)v_{\text{after}}^2 \)
02

Calculate the Frictional Force

The cars come to a stop due to the frictional force. To find the frictional force, use:\[ F_{\text{friction}} = \mu_k \times m_{\text{combined}} \times g \]where \( \mu_k = 0.65 \) is the coefficient of kinetic friction, \( m_{\text{combined}} = m_A + m_B = 3400\,\text{kg} \) is the total mass, and \( g = 9.8\,\text{m/s}^2 \) is gravity.
03

Solve for the Velocity after Collision

Using the work-energy principle and the skid distance, \[ F_{\text{friction}} \times 7.15 = \frac{1}{2} \times m_{\text{combined}} \times v_{\text{after}}^2 \]Substitute the values to find \( v_{\text{after}} \) (velocity of the combined mass right after the collision).
04

Use Conservation of Momentum

Now apply the conservation of momentum:\[ m_A v_{A\text{before}} = m_{\text{combined}} v_{\text{after}} \]Solve for car A's velocity before collision \( v_{A\text{before}} \).
05

Compare Speed to Speed Limit

Convert the speed from m/s to mph and compare it to the speed limit of 35 mph to determine how many mph car A was exceeding the speed limit.

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Key Concepts

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

Kinetic Friction in Physics
In the scenario of a car accident where vehicles come to a halt after a collision, kinetic friction plays a crucial role in bringing the cars to a stop. When two vehicles like cars A and B collide and skid together, kinetic friction is what eventually stops them. The force of friction depends on several key factors:
  • The coefficient of kinetic friction (\(\mu_k\)): This dimensionless number represents the ratio of the frictional force between the tires and the road surface to the normal force (weight) acting on the car. In our case, \(\mu_k = 0.65\).
  • The combined mass of the vehicles: After the collision, the total mass \(m_{\text{combined}}\) is the sum of car A's and car B's masses, resulting in \(m_{\text{combined}} = 3400 \text{ kg}\).
  • The gravitational acceleration (\(g\)): Usually taken as \(9.8 \, \text{m/s}^2\), it acts downward, affecting the normal force which equals the weight of the vehicles.
Given these factors, the frictional force \(F_{\text{friction}}\) that resists the motion can be calculated by the formula \(F_{\text{friction}} = \mu_k \times m_{\text{combined}} \times g\). This frictional force does work over the distance of the skid marks to bring the cars to a stop.
Work-Energy Principle
The work-energy principle is instrumental in understanding how the energy associated with an object's motion changes due to forces like friction. In our collision scenario, once car A hits car B, the kinetic energy of the cars is gradually transformed into thermal energy due to friction, effectively halting their motion.
This principle is applied using the formula:
  • Work done by friction: \( \text{work} = F_{\text{friction}} \times \text{distance} \)
  • The change in kinetic energy: \( \frac{1}{2} m_{\text{combined}} v_{\text{after}}^2\), where \(v_{\text{after}}\) is the velocity right after the collision but before stopping.
By equating the work done by friction to the change in kinetic energy, \( F_{\text{friction}} \times 7.15 = \frac{1}{2} \times m_{\text{combined}} \times v_{\text{after}}^2 \), we calculate the velocity of the combined mass immediately after the collision. This approach shows how energy conservation principles help understand and predict post-collision speeds.
Velocity Calculation from Skid Marks
In accident analysis, determining pre-collision speeds is crucial. Here, skid marks provide critical evidence for backtracking to the speed of car A right before the collision. By inspecting the length of these marks, we can apply conservation laws and friction calculations to estimate this speed:
  • Use conservation of momentum initially: \( m_A v_{A\text{before}} = m_{\text{combined}} v_{\text{after}} \)
  • From the work-energy principle, derive \( v_{\text{after}} \).
    • Additionally, comparing this computed speed of car A with the speed limit reveals if car A was speeding precisely.
      The steps involved focus on translating physical evidence (skid marks) into quantitative data like velocity, by bridging different physics concepts – from momentum conservation to energy principles. Ultimately, converting \(\text{m/s}\) to \(\text{mph}\) enables easy comparison with speed limits.

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Most popular questions from this chapter

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?

Suppose you hold a small ball in contact with, and directly over, the center of a large ball. If you then drop the small ball a short time after dropping the large ball, the small ball rebounds with surprising speed. To show the extreme case, ignore air resistance and suppose the large ball makes an elastic collision with the floor and then rebounds to make an elastic collision with the still-descending small. Just before the collision between the two balls, the large ball is moving upward with velocity \(\vec{\boldsymbol{v}}\) and the small ball has velocity \(-\vec{\boldsymbol{v}}\) . (Do you see why? \()\) Assume the large ball has a much greater mass than the small ball. (a) What is the velocity of the small ball immediately after its collision with the large ball? (b) From the answer to part (a), what is the ratio of the small ball's rebound distance to the distance it fell before the collision?

A movie stuntman (mass 80.0 \(\mathrm{kg}\) ) stands on a window ledge 5.0 \(\mathrm{m}\) above the floor (Fig. P8.85). Grabbing a rope attached to a chandelier, he swings down to grapple with the movie's villain (mass 70.0 kg), who is standing directly under the chandelier. (Assume that the stuntman's center of mass moves downward 5.0 \(\mathrm{m} .\) He releases the rope just as he reaches the villain.) (a) With what speed do the entwined foes start to slide across the floor? (b) If the coefficient of kinetic friction of their bodies with the floor is \(\mu_{k}=0.250,\) how far do they slide?

A steel ball with mass 40.0 \(\mathrm{g}\) is dropped from a height of 2.00 \(\mathrm{m}\) onto a horizontal steel slab. The ball rebounds to a height of 1.60 \(\mathrm{m} .\) (a) Calculate the impulse delivered to the ball during impact. (b) If the ball is in contact with the slab for 2.00 \(\mathrm{ms}\) , find the average force on the ball during impact.

A bat strikes a 0.145 -kg baseball. Just before impact, the ball is traveling horizontally to the right at \(50.0 \mathrm{m} / \mathrm{s},\) and it leaves the bat traveling to the left at an angle of \(30^{\circ}\) above horizontal with a speed of 65.0 \(\mathrm{m} / \mathrm{s}\) . If the ball and bat are in contact for 1.75 \(\mathrm{ms}\) find the horizontal and vertical components of the average force on the ball.

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