After traveling all night, you are driving down a narrow country lane. You briefly fall asleep at the wheel, but wake up as the car moves off the shoulder of the road. You wake up but rapidly lose control of the car and it zooms off the road. You are heading towards (1) a stack of bales of hay, or (2) a brick wall. You have seconds to choose, and luckily you choose to hit the hay stack and come to a stop. 

Compare in detail the cases of hitting the brick wall versus hitting the hay stack. In which scenario do you transfer more momentum?  In which scenario is your impulse larger? In which scenario is the force on the car largest? In which scenario do you experience the largest acceleration? Explain why using physical principles you have learned.

In both cases, your car start off traveling at the same initial velocity, \(v_i\). Also, in both cases, you car comes to a stop: \(v_f = 0\). Therefore, in both cases, your car experiences the same velocity change \(\Delta v\) and hence momentum change \(\Delta p\) throughout the collision: \begin{equation} m \Delta v = m \left( v_f - v_i \right) = m v_f - m v_i = p_f - p_i = \Delta p. \end{equation}

According to the impulse-momentum theorem, this momentum change is equal to the impulse \(J\) exerted on your car: \begin{equation} \Delta p = J. \end{equation}

Any impulse is proportional to the average force that causes it \(F_{avg}\) and the time interval \(\Delta t\) during which that force acts: \begin{equation} J = F_{avg} \Delta t. \end{equation}

Therefore, for a given impulse, the average force and time interval are inversely proportional to each other: \begin{equation} F_{avg} = \frac{J}{\Delta t}. \end{equation}

This means that if the time interval is longer, then the force is smaller and vice versa.

Finally, an object’s acceleration is proportional to (\(\propto\)) the force acting upon it (Newton’s 2nd law): \begin{equation} a_{avg} \propto F_{avg}. \end{equation}

This means that if the force acting on your car is greater, then your car’s acceleration is greater, and vice versa.

Combining all these statements into one expression results in \begin{equation} a_{avg} \propto F_{avg} = \frac{J}{\Delta t} = \frac{\Delta p}{\Delta t} = \frac{m \Delta v}{\Delta t}. \end{equation}

Now, imagine an object with mass \(m\) traveling at a velocity \(v\), and imagine two two scenarios in which that object collides with something and comes to a stop. In both cases, the object has the same velocity change \(\Delta v\). Therefore in both cases it has the same momentum change \(\Delta p\) resulting from the same impulse \(J\). Now image that in one case, the collision takes a long time \(\Delta t_L\), while in the collision in the other case takes a short time \(\Delta t_S\). It follows from the result above that: \begin{equation} \begin{aligned} \Delta t_S < \Delta t_L & \implies \frac{J}{\Delta t_L} < \frac{J}{\Delta t_S} \\[1ex] & \implies F_L < F_S \\[1ex] & \implies a_L < a_S. \end{aligned} \end{equation}

The amount of damage you and your car sustain will be proportional to the force exerted on you and your resulting acceleration. And, presumably, a brick wall takes less time to stop a car than does a stack of hay.

So, which one would you rather hit?

—Aaron