78 Collisions

We can extend the concept of introduced in the previous chapter to analyze forces and changes in experienced by multiple objects during a collision. For example, we could determine the of two cars after a rear-end collision like this one:

Let’s analyze this specific collision seen in the video. We start with our definition for impulse:

(1)   \begin{equation*} \Delta t \bold{F_{ave}} = m \left(\bold{v_f}-m \bold{v_i}\right) \end{equation*}


times (mv) is known as the (p) , or \bold{p} = m\bold{v}. The change in momentum (Δp) is defined the same way as any other change, final momentum minus the initial momentum: (\bold{\Delta p} = m\bold{ v_f}-m \bold{v_i}) . We can use these definitions to write the .

(2)   \begin{equation*} \Delta t \bold{F_{ave}} = \bold{p_f}-\bold{p_i} \end{equation*}


(3)   \begin{equation*} \Delta t \bold{F_{ave}} = \bold{\Delta p} \end{equation*}

Conservation of Momentum

The indicates that if the average force acting on single object or a system of objects is zero, then the of the object or system of objects is (conserved). The previous statement is known as the .  The related states that the combined total momentum of all objects in a system must be the same immediately before and immediately after a collision. We can treat both cars in our example collision as a single system as long as we account for the initial momentum and final momentum of both cars.

The initial momentum of the first (stopped) car was zero: \bold{p_{1,i}} = 0.  The momentum of the second car was: \bold{p_{2,i}} = m_2 \bold{v_{2,i}}. Therefore the total initial momentum was

    \begin{equation*} \bold{p_i} = 0+ m_2 \bold{v_{2,i}} \end{equation*}

The cars lock together immediately after the collision and only separate later so they have the same final velocity immediately after the collision, we’ll call it v_f. Sticky collisions like this are known as and for such collisions we can treat the objects moving together as a single object that has their combined total . The final is then:

    \begin{equation*} \bold{p_f} =  (m_2+m_1) \bold{v_{f}} \end{equation*}

Conserving momentum during the collision tells us to set the initial and final momenta equal:

    \begin{equation*} m_2 \bold{v_{2,i}} = (m_2+m_1) \bold{v_{f}} \end{equation*}

If we want to solve this equation for the final velocity we divide by the combined mass:

    \begin{equation*} \bold{v{f}} = \frac{m_2 \bold{v_{2,i}}}{m_2+m_1} \end{equation*}

We can look up some data on the cars and find that the length of the Jeep is 3.8 m,  the mass of the jeep is about 1500 kg and a small sports car mass is about 1000 kg[1][2] Now let’s estimate some numbers from the video (a quick method is to use the slo-mo feature on a smartphone to film the video as it plays with a running stopwatch also visible in the frame).  We see the Jeep covers at least two of its own lengths in about 0.4 s so calling its direction the positive, its initial velocity will be:

    \begin{equation*} \bold{v_{2,i}} = \frac{2\times 3.8\,\bold{m}} {0.4\,\bold{s} = 19\,\bold{m/s}}= 42\,\bold{MPH} \end{equation*}

We are ready to calculate the final velocity:

    \begin{equation*} \bold{v_{f}} = \frac{1500\,\bold{kg}\times 19\,\bold{m/s}}{1500\,\bold{kg}+1000\,\bold{kg}} = 12 \,\bold{m/s} = 27 \,\bold{MPH} \end{equation*}

This is an interesting result, but what’s really cool is that if we estimate the collision interval we can reapply the to calculate the average force applied to each car. From the video the collision time appears to be about 0.5 s. Let’s do the Jeep first. Just to think ahead, the force on the jeep should be in the negative direction based on our choice of backward as negative.

(4)   \begin{equation*} \bold{F_{ave}} =\frac{m \left(\bold{v_f}-\bold{v_i}\right)}{\Delta t} = \frac{1500\,\bold{kg}\left(11.5\,\bold{m/s}-19\,\bold{m/s})\right)}{0.5\,\bold{s}} = -23,000 \,\bold{N} \end{equation*}

We known from that the -23,000 N force on the Jeep from the car is paired with a 23,000 force felt by the car from the Jeep. That’s over 5,000 lbs of average force on each vehicle, the peak force would be much greater.

Reinforcement Exercise

A 48 kg soccer player running 5 m/s and watching the ball in the air collides with another player of mass 68 kg running at 2 m/s in the other direction and they tangle together before falling. What is the of the pair immediately after the collision, before hitting the ground?

If the entire collision lasted only 0.30 s, what was the average force on each player?


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