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{excerpt:hidden=true}Compare the forces on the occupants of two cars in a 1-D totally inelastic collision.{excerpt}
h2. Part A

A 4460 lb Ford Explorer traveling 35 mph has a head on collision with a 2750 lb Toyota Corolla, also traveling 35 mph.  Assuming that the automobiles become locked together during the collision, what is the speed of the combined mass immediately after the collision?

System:  Explorer plus Corolla as [point particles|point particle].  Impulse resulting from external influences will be neglected, as we assume that the collision is instantaneous.

Model:  [Momentum and Impulse].

Approach:  We begin by sketching the situation and defining a coordinate system.


Since we assume that external forces are negligible during the collision, we set the external impulse to zero which gives:

{latex}\begin{large}\[ p^{TC}_{x,i} + p^{FE}_{x,i} = p^{system}_{x,f} \]\end{large}{latex}

or, in terms of the masses:

{latex}\begin{large}\[ m^{TC}v^{TC}_{x,i} + m^{FE}v^{FE}_{x,i} = (m^{TC}+m^{FE})v_{x,f} \]\end{large}{latex}

which gives:

{latex}\begin{large}\[ v_{x,f} = \frac{m^{TC}v^{TC}_{x,i} + m^{FE}v^{FE}_{x,i}}{m^{TC}+m^{FE}} = \mbox{3.71 m/s} = \mbox{8.3 mph}\]\end{large}{latex}

{warning}Remember that in our coordinate system, the Corolla has a negative x-velocity before the collision.{warning}

h2. Part B

Find the impulse that acted on each of the vehicles during the collision.

Systems:  Corolla and Explorer as *separate* [point particle] systems.

Model: [Momentum and Impulse].

Approach: With the results of Part A it is straightforward to calculate the impulse on the Explorer due to the collision.  To be specific, we label the impulse "EC" to remind ourselves the impulse on the Explorer is provided by the Corolla.

{latex}\begin{large}\[ I^{EC}_{x} = p^{E}_{x,f}-p^{E}_{x,i} = m^{E}(v^{E}_{x,f}-v^{E}_{x,i}) = (\mbox{2030 kg})(\mbox{3.71 m/s}-\mbox{15.6 m/s}) = -\mbox{24000 kg m/s} \] \end{large}{latex}

Similarly, for the Corolla:

{latex}\begin{large}\[ I^{CE}_{x} = p^{C}_{x,f}-p^{C}_{x,i} = m^{C}(v^{C}_{x,f}-v^{C}_{x,i}) = (\mbox{1250 kg})(\mbox{3.71 m/s}+\mbox{15.6 m/s}) = +\mbox{24000 kg m/s} \] \end{large}{latex}

{note}Again, it is important to note that the Corolla's initial x-velocity is negative in our chosen coordinate system.{note}

It is no accident that _I_^EC^ = -- _I_^CE^.  The relationship is guaranteed by [Newton's 3rd Law|Newton's Third Law].

h2. Part C

Assuming the collision lasted for 0.060 seconds, find the average force exerted on each vehicle.

Systems:  Corolla and Explorer as *separate* [point particle] systems.  We will assume the collsion force dominates.

Model: [Momentum and Impulse].

Approach:  From the Law of Interaction, we know:

{latex}\begin{large}\[ I^{EC}_{x} = \int F^{EC}_{x}\:dt \equiv \bar{F}^{EC}_{x} \Delta t\]\end{large}{latex}

so the average force exerted on the Explorer is:

{latex}\begin{large}\[\bar{F}^{EC}_{x} = \frac{I^{CE}_{x}}{\Delta t} = -\mbox{400,000 N} \]\end{large}{latex}

Similarly, the average force exerted on the Corolla is:

{latex}\begin{large}\[\bar{F}^{CE}_{x} = \frac{I^{EC}_{x}}{\Delta t} = \mbox{400,000 N}\]\end{large}{latex}

{note}Again, the relationship is guaranteed by Newton's 3rd Law.{note}

h2. Part D

Suppose a 75 kg person in each vehicle underwent the same change in velocity as their automobile in the same amount of time.  Find the average force exerted on these people.

Systems: First, the Corolla and Explorer as separate [point particle] systems, then the passengers as *separate* [point particle] systems subject to some decelerating force, possibly a seatbelt or airbag.

Model: [Point Particle Dynamics].

Approach:  The passengers clearly are not subject to the same force as their vehicles.  Rather, they are subject to the same accelerations.  They are (hopefully) strapped into their automobiles, so that whatever happens to their vehicle happens to them as well.  Thus, our first goal is to determine the vehicles' accelerations.  This is easily accomplished using the results of Part C in [Newton's 2nd Law|Newton's Second Law].

{latex}\begin{large}\[ a^{E}_{x}= \frac{F^{EC}_{x}}{m^{E}} = -\mbox{200 m/s}^{2}\]
\[ a^{C}_{x} = \frac{F^{CE}_{x}}{m^{C}} = \mbox{320 m/s}^{2} \]\end{large}{latex}

{tip}The Corolla's acceleration should clearly be larger, even though the beginning and ending speeds of the Explorer and Corolla are the same.  The reason is that the Corolla has _changed direction_.  Thus, it effectively dropped all the way to zero from 35 mph and then _accelerated_ back up to 8.3 mph the other way.  The Explorer simply dropped its forward speed from 35 mph to 8.3 mph.{tip}

{info}Note (discounting safety measures like airbags) the SUV driver's acceleration would be about 20 g's, near the limit of human endurance.  The Corolla's driver would experience almost 33 g's.{info}

With these accelerations, we can find the force on each driver:

{latex}\begin{large}\[ F^{E,driver}_{x} = m^{driver}a^{E}_{x} = -\mbox{15000 N} = -\mbox{3,300 lbs}\]
\[ F^{C,driver}_{x} = m^{driver}a^{C}_{x} = \mbox{24000 N} = \mbox{5,400 lbs}\]\end{large}{latex}