Question

A police officer is investigating the scene of an accident where two cars collided at an intersection. One car with a mass of \(1100 \mathrm{kg}\) moving west had collided with a \(1300-\mathrm{kg}\) car moving north. The two cars, stuck together, skid at an angle of \(30^{\circ}\) north of west for a distance of \(17 \mathrm{m} .\) The coefficient of kinetic friction between the tires and the road is \(0.80 .\) The speed limit for each car was $70 \mathrm{km} / \mathrm{h} .$ Was either car speeding?

Short answer

To answer this question, first, compute the initial velocities of the cars using the final velocity found in step 3, and then compare them to the converted speed limit: - Calculate the initial velocities of the two cars: \(v_{1i_{West}} = (1100 + 1300) * u * \cos(30^\circ) /1100\) \(v_{2i_{North}} = (1100 + 1300) * u * \sin(30^\circ) /1300\) - Compare the initial velocities to the converted speed limit: \((v_{1i_{West}}, v_{2i_{North}}) > 70 * (1000/3600) \,\text{m/s}\) If either of the initial velocities is greater than the speed limit, the corresponding car was speeding.

Step-by-step solution

Calculate the force of friction

To find the force of friction acting on the two cars, we'll use the formula: $$F_{friction} = μ * F_{normal} $$ Since the normal force is equal to the total weight of the two cars, and because weight = mass × gravity, we can find the normal force. Let's calculate the force of friction: \(F_{normal} = (1100 + 1300) * 9.81\) Now, calculate the force of friction using the coefficient of kinetic friction: \(F_{friction} = 0.80 * F_{normal}\).

Calculate the acceleration

Using Newton's second law of motion, we can calculate the acceleration of the system: $$F = m * a$$ The force acting on the two cars is the force of friction in the negative direction. The total mass of the cars is \((1100+1300)\,\text{kg}\). Rearrange the equation as: $$a = \frac{-F_{friction}}{m_{total}}$$ Calculate the acceleration using the force of friction found in step 1: \(a = \frac{-F_{friction}}{(1100+1300)}\)

Determine the final velocity

Now we need to determine the final combined velocity of the two cars to analyze the collision. We can use the formula: $$v^2 = u^2 + 2 * a * s$$ Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the distance covered. Since the cars came to a stop after skidding, \(v=0\) Rearrange the equation to find the initial velocity: $$u=\sqrt{- 2 * a * s}$$ We'll now calculate the initial velocity using the distance \(s=17\,\text{m}\) and acceleration found in step 2: \(u = \sqrt{- 2 * a * 17}\) The direction of the velocity is \(30^\circ\) north of west.

Analyze the collision using conservation of momentum

Now we will use conservation of momentum to determine the initial velocities of the individual cars: $$m_1\vec{v}_1 + m_2\vec{v}_2 = (m_1 + m_2)\vec{v_f}$$ Let the initial velocities of the two cars be \(\vec{v}_{1i}\) and \(\vec{v}_{2i}\). We know the final velocity \(\vec{v}_f\) and their masses \(m_1\) and \(m_2\). Separate the equation into two components (north and west): $$\begin{cases} 1100 * v_{1i_{West}} + 0 = (1100 + 1300) * v_f * \cos(30^\circ) \\ 0 + 1300 * v_{2i_{North}} = (1100 + 1300) * v_f * \sin(30^\circ) \end{cases}$$ Solve the system of equations for \(v_{1i_{West}}\) and \(v_{2i_{North}}\): $$\begin{cases} v_{1i_{West}} = (1100 + 1300) * v_f * \cos(30^\circ) /1100 \\ v_{2i_{North}} = (1100 + 1300) * v_f * \sin(30^\circ) /1300 \end{cases}$$ Calculate the initial velocities using the final velocity found in step 3: \(v_{1i_{West}} = (1100 + 1300) * u * \cos(30^\circ) /1100\) \(v_{2i_{North}} = (1100 + 1300) * u * \sin(30^\circ) /1300\)

Compare initial velocities to the speed limit

Convert the speed limit from km/h to m/s: \(70\,\text{km/h} = 70 * (1000/3600) \,\text{m/s}\) Now compare the initial velocities \(v_{1i_{West}}\) and \(v_{2i_{North}}\) with the speed limit. If either of the initial velocities is greater than the speed limit, the respective car was speeding.