Question

You are called as an expert witness to analyze the following auto accident: Car \(B\), of mass 1900 kg, was stopped at a red light when it was hit from behind by car \(A\), of mass 1500 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 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 mph, was car \(A\) speeding, and if so, by how many miles per hour was it \(exceeding\) the speed limit?

Short answer

Car A was traveling at 48.14 mph, exceeding the speed limit by 13 mph.

Step-by-step solution

Identify Known Variables and Physical Concepts

First, identify the known values and the concepts involved. We know:- Mass of car A: \( m_A = 1500 \, \text{kg} \)- Mass of car B:\( m_B = 1900 \, \text{kg} \)- Coefficient of kinetic friction:\( \mu_k = 0.65 \)- Length of skid marks:\( d = 7.15 \, \text{m} \)- Total mass after collision due to locked bumpers:\( m_{\text{total}} = m_A + m_B \)- We use the principle of conservation of momentum and kinematic equations to analyze the collision and subsequent skid.

Calculate the Deceleration

Use the formula for force of friction to find the deceleration:\[ f_k = \mu_k \times m_{\text{total}} \times g \]Where:- \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.The frictional force provides the deceleration:\[ a = \frac{f_k}{m_{\text{total}}} = \mu_k \times g \]Substitute the values:\[ a = 0.65 \times 9.81 = 6.3765 \, \text{m/s}^2 \]

Use Kinematics to Determine Skid Speed After Collision

Apply the kinematic equation to determine the speed of the cars just after the collision when they began to skid:\[ v^2 = u^2 + 2as \]Where:- \( v = 0 \) (since the cars come to a stop),- \( s = 7.15 \, \text{m} \) (skid distance), - \( a = -6.3765 \, \text{m/s}^2 \).Rearrange to solve for initial speed \( u \):\[ 0 = u^2 - 2 \times 6.3765 \times 7.15 \]\[ u = \sqrt{2 \times 6.3765 \times 7.15} \approx 9.5 \, \text{m/s} \]

Apply Conservation of Momentum for Initial Speed of Car A

Before the collision, car B was stationary. Apply conservation of momentum:\[ m_A \times v_A = m_{\text{total}} \times u \]Where:- \( v_A \) is the initial speed of car A,- \( m_{\text{total}} = 3400 \, \text{kg} \).Substitute the known values:\[ 1500 \times v_A = 3400 \times 9.5 \]Solve for \( v_A \):\[ v_A = \frac{3400 \times 9.5}{1500} \approx 21.53 \, \text{m/s} \]

Convert Speed to Miles per Hour and Check Against Speed Limit

Convert the initial speed of car A to miles per hour:\[ 1 \, \text{m/s} = 2.237 \, \text{mph} \]So:\[ 21.53 \, \text{m/s} \times 2.237 = 48.14 \, \text{mph} \]Compare to speed limit (35 mph):\( 48.14 \, \text{mph} - 35 \, \text{mph} = 13.14 \, \text{mph} \)Car A was exceeding the speed limit by approximately 13 mph.

Key concepts

Conservation of momentum

The principle of conservation of momentum is a fundamental concept in physics. It states that the total momentum of a closed system does not change unless it is acted upon by an external force. In the context of an accident such as the one described, we apply this principle to understand the velocities of the vehicles involved both before and after the collision.
Momentum, usually represented by the symbol \( p \), is the product of an object's mass \( m \) and its velocity \( v \): \[ p = mv \]When two cars collide and lock bumpers, as described in the accident, they can be considered as a single system post-collision. Before the collision, the momentum of the entire system (Car A + Car B) is predominantly due to Car A since Car B is stationary.
Post-collision, the shared momentum of the combined cars slides to a stop due to friction. By setting up the momentum equations before and after the collision, we can solve for unknown quantities like the initial speed of Car A.

Kinematic equations

Kinematic equations are essential tools in physics for analyzing motion. They describe the relationships between velocity, acceleration, and displacement. When applied to the scenario of cars sliding to a stop, they allow us to determine the vehicles' velocities and deceleration.
A key kinematic equation used in the problem is:\[ v^2 = u^2 + 2as \]where:

• \( v \) is the final velocity (0 m/s since the cars stop)
• \( u \) is the initial velocity just after the collision
• \( a \) is the acceleration (or deceleration in this case)
• \( s \) is the distance slid (7.15 m)

In this scenario, since the cars come to rest, we only need to rearrange this equation to find the initial velocity \( u \) just after the collision. Solving this gives us an insight into how fast they were moving immediately after impact.

Coefficient of kinetic friction

The coefficient of kinetic friction, \( \mu_k \), is a dimensionless constant that describes the friction between two moving surfaces. It significantly affects how an object slows down when sliding. In this problem, it helps us determine how quickly the two collided cars stop moving.
Frictional force can be expressed using the formula:\[ f_k = \mu_k \times m_{\text{total}} \times g \]where:

• \( \mu_k \) is the coefficient of kinetic friction (0.65)
• \( m_{\text{total}} \) is the total mass of the locked cars
• \( g \) is the acceleration due to gravity (9.81 m/s²)

The frictional force causes deceleration, which is crucial for finding the velocity just before the car reaches a complete stop. Understanding friction allows us to solve for deceleration and subsequent velocities using the kinematic equations.

Acceleration due to gravity

The acceleration due to gravity, denoted as \( g \), is a constant that represents the gravitational pull on objects near the Earth's surface. It has a standard value of 9.81 m/s². This constant plays a critical role in calculating forces, including the frictional force in the context of sliding cars.
When linked with friction, gravity helps determine the force exerted on sliding cars through:\[ f_k = \mu_k \times m_{\text{total}} \times g \]Gravity is crucial because it affects how much friction slows down the car. Without it, we wouldn't be able to quantify the resistance the road offers as the cars skid. In this accident scenario, gravity assists us in understanding the deceleration of the cars due to friction.