54 Analyzing Collisions

Analyzing a Car Crash

In the previous chapters we saw that the combination of Newton’s three laws of motion is actually equivalent to the combination of conservation of momentum and the impulse momentum theorem. Using Newton’s Laws of motion can help us to break down complex problems into smaller steps related to each law and in some cases we strategically apply the laws of motion and momentum conservation to solve a complex problem more efficiently.

Everyday Examples: Collisions

Let’s analyze the car accident in the previous video. The initial momentum of the stopped car (car #1) was zero: \bold{p_{1,i}} = 0.  The momentum of the jeep (car #2) 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 perfectly inelastic collisions, and for such collisions we can treat the objects moving together as a single object that has their combined total mass. The final momentumis 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 perfectly inelastic collision 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 that avoids dealing with the frame-rate is to use the slo-mo feature on a smartphone to film the video playing at full speed next to a running stopwatch and then play back the slow motion video and record times displayed by the stop watch).  We see that before the collision 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 of the jeep-car combination:

    \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 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.

We start from Newton’s Second Law:

(1)   \begin{equation*} \bold{F_{ave}} =m\bold{a} = \end{equation*}

Then write out the acceleration as change in velocity over change in time:

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

We have the final velocity of 12 m/s from our previous work using momentum conservation, so we can enter all our values:

(3)   \begin{equation*} \bold{F_{ave}} =\frac{m \left(\bold{v_f}-\bold{v_i}\right)}{\Delta t} = \frac{1500\,\bold{kg}\left(12\,\bold{m/s}-19\,\bold{m/s})\right)}{0.5\,\bold{s}} = -23,000 \,\bold{N} \end{equation*}

hat’s over 5,000 lbs of average force, and we know the peak force would be greater!

We expected the force on the Jeep to be negative because we originally chose  backward as the negative direction. We also known from Newton's Third Law that the car feels an equal and opposite 23, 000 N force from the Jeep in the forward direction.



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