Acceleration in One Dimensional Inelastic Collision

[Calvin Pitts]

Homework Statement

Two cars of 540 kg and 1400 kg collide head on while moving 80 km/h in opposite directions. After the collision, the automobiles remain locked together.
Find the velocity of the wreck, the kinetic energy of the two-automobile system before and after the collision, and find the acceleration of the passenger compartment of each vehicle given that the front end of each car crumpled by 0.6 m during the collision.

(I've already found the velocity and energies, I just need help with the acceleration.)

Homework Equations

Kinematic equations?

The Attempt at a Solution

a(x-x₀) = (1/2)(v² - v₀²)

 

[Chestermiller]

They are asking for the acceleration, but what they really want is the average acceleration during the collision. The equation you wrote is the relationship they expect you to use.

 

[Calvin Pitts]

They are asking for the acceleration, but what they really want is the average acceleration during the collision. The equation you wrote is the relationship they expect you to use.

Maybe I'm doing something really wrong with this equation, but I'm not getting anywhere near the right answer (-130 m/s, 850 m/s). I keep getting -330 m/s using the final velocity I calculated in the first part, which is 9.8 m/s in the direction of the 1400 kg car.

 

[Chestermiller]

Maybe I'm doing something really wrong with this equation, but I'm not getting anywhere near the right answer (-130 m/s, 850 m/s). I keep getting -330 m/s using the final velocity I calculated in the first part, which is 9.8 m/s in the direction of the 1400 kg car.

Let's see the details of your calculation for the velocity of the wreck. I get a wreck velocity of <10 m/s.

 

[Calvin Pitts]

Sorry for the lack of LaTeX. The app doesn't support it.

9.8 m/s is the same value the book gives.

 

[Hiero]

As Chestermiller said, they most likely want you to pretend the acceleration is constant.
(I'm curious how much larger the maximum acceleration would be than the average accel. in a real crash.)

You need to be careful though with this equation though:

a(x-x₀) = (1/2)(v² - v₀²)

x-x₀ is the distance(s) that the passenger compartment(s) move(s) during the collision (which is not necessarily just the given 'crumple distance').

Does the acceleration (that we want to find) differ in two frames moving at a constant speed relative to each other?
The answer to this question is a hint as to how I would approach the problem.

 

[Calvin Pitts]

As Chestermiller said, they most likely want you to pretend the acceleration is constant.
(I'm curious how much larger the maximum acceleration would be than the average accel. in a real crash.)

You need to be careful though with this equation though:

x-x₀ is the distance(s) that the passenger compartment(s) move(s) during the collision (which is not necessarily just the given 'crumple distance').

Does the acceleration (that we want to find) differ in two frames moving at a constant speed relative to each other?
The answer to this question is a hint as to how I would approach the problem.

So that means the displacement would be 1.2 meters, because the passenger moves his .6 m and the other .6 as well, right?

I'm trying to wrap my head around your question. Haha

 

[Hiero]

So that means the displacement would be 1.2 meters, because the passenger moves his .6 m and the other .6 as well, right?

Can you appreciate the fact (qualitatively) that the smaller car should accelerate more?
(It slows all the way down from 80km/hr and reverses its direction of motion, whereas the big car just slows down a bit from 80km/hr.)

If you look at your equation,

a(x-x₀) = (1/2)(v² - v₀²)

doesn't the fact that the accelerations should come out different for each car mean that (x-x₀) should be different for each car?
So how can we understand this difference in (x-x₀) of each car?
[Hints: does the contact point between the two cars move during the collision? Does this contribute to the displacement (x-x₀)?]

 

[haruspex]

what they really want is the average acceleration

I would put it differently. Since the duration of impact is not given, there is no easy way to calculate the average acceleration unless we assume it is constant. So just asking for "acceleration" is ok, but they ought to say assume it is constant.
In fact, in a survivable head-on car crash it is more likely to be roughly constant than in many other collisions. That is because car extremities are designed to crumple at approximately constant load so as to minimise the maximum acceleration. That breaks down at unsurvivable speeds.

 

[haruspex]

Can you appreciate the fact (qualitatively) that the smaller car should accelerate more?
(It slows all the way down from 80km/hr and reverses its direction of motion, whereas the big car just slows down a bit from 80km/hr.)

If you look at your equation,

doesn't the fact that the accelerations should come out different for each car mean that (x-x₀) should be different for each car?
So how can we understand this difference in (x-x₀) of each car?
[Hints: does the contact point between the two cars move during the collision? Does this contribute to the displacement (x-x₀)?]

It might be conceptually simpler to calculate the duration of the impact first, using relative velocities and displacements.

 

[Chestermiller]

I would put it differently. Since the duration of impact is not given, there is no eay to calculate the average acceleration unless we assume it is constant. So just asking for "acceleration" is ok, but they ought to say assume it is constant.
In fact, in a survivable head-on car crash it is more likely to be roughly constant than in many other collisions. That is because car extremities are designed to crumple at approximately constant load so as to minimise the maximum acceleration. That breaks down at unsurvivable speeds.

As you indicated, this is seems to be a pretty complicated problem. I have been trying to model this by putting a damper on the front of each of the cars to represent the (dissipating) crumpling region of each. I have started out by assuming that the velocity of the contact point between the two cars is equal to the final velocity. But, when I do this, and, requiring that, by Newton's third law, the contact force is the same for each car (at each instant of time), I find that the two crushing zones are not equal. The crushing region of the lighter car comes out larger than the crushing region of the heavier car. I'll keep working on this an keep you posted on progress.

Chet

 

[Hiero]

As you indicated, this is seems to be a pretty complicated problem. I have been trying to model this by putting a damper on the front of each of the cars to represent the (dissipating) crumpling region of each. I have started out by assuming that the velocity of the contact point between the two cars is equal to the final velocity. But, when I do this, and, requiring that, by Newton's third law, the contact force is the same for each car (at each instant of time), I find that the two crushing zones are not equal. The crushing region of the lighter car comes out larger than the crushing region of the heavier car. I'll keep working on this an keep you posted on progress.

Chet

I made the same assumption, that 'the velocity of the contact point is the final velocity,' but I thought of it as, 'the contact point does not move in the center of mass frame.'

In the CoM frame, if the contact point were stationary, the rate at which each car is crushed will be the same as the speed of the back of the car. Since the lighter car moves faster at all times, it will be crushed more at all times, and so we will not find the 'crumple distances' to be equal.

Conclusion: The assumption that the contact point moves with the center of mass is inconsistent with the fact that they crumple the same amount.When I used that inconsistent assumption, my answer agreed with what the OP said was correct, so I now think his given answer isn't right. (I think @haruspex has the proper idea in post #10.)

 

[Calvin Pitts]

I made the same assumption, that 'the velocity of the contact point is the final velocity,' but I thought of it as, 'the contact point does not move in the center of mass frame.'

In the CoM frame, if the contact point were stationary, the rate at which each car is crushed will be the same as the speed of the back of the car. Since the lighter car moves faster at all times, it will be crushed more at all times, and so we will not find the 'crumple distances' to be equal.

Conclusion: The assumption that the contact point moves with the center of mass is inconsistent with the fact that they crumple the same amount.When I used that inconsistent assumption, my answer agreed with what the OP said was correct, so I now think his given answer isn't right. (I think @haruspex has the proper idea in post #10.)

So how do I go about solving for time in this way? Kinematic equations still?

 

[Calvin Pitts]

EDIT*
The book says to assume the accelerations are constant. I'm not sure if that affects your answers in any way.

 

[haruspex]

So how do I go about solving for time in this way? Kinematic equations still?

Yes. As I wrote, think in terms of the relative speeds and displacements. From the instant they make contact, how much closer do the drivers get, and at what average approach speed?

The book says to assume the accelerations are constant.

Good.

 

[Chestermiller]

So how do I go about solving for time in this way? Kinematic equations still?

In my judgment, the person who formulated this problem tremendously underestimated the complexity of what he was asking. I think he intended for the student to set the change in kinetic energy of each of the cars (as reckoned by an observer in the laboratory frame of reference) equal to the average force times the crumple distance. As pointed out by Haruspex, Hiero, and myself, this is an incorrect approach. It would be better to allow the two crumple distances to be different, and to specify the average of the two crumple distances. Then, at least, the analyses of Hiero and myself would be applicable.

 

[Calvin Pitts]

Yes. As I wrote, think in terms of the relative speeds and displacements. From the instant they make contact, how much closer do the drivers get, and at what average approach speed?

Good.

They each move 80 km/h towards each other, relative velocity of 160 km/h, and the displacement is 1.2 m (.6 m for each car). Is it as simple as 1.2 m divided by 160 km/h to get the time?

 

[Hiero]

They each move 80 km/h towards each other, relative velocity of 160 km/h, and the displacement is 1.2 m (.6 m for each car). Is it as simple as 1.2 m divided by 160 km/h to get the time?

It is very nearly that simple! That would be wrong though, because 160km/h is the relative speed at the initial moment. (At the final moment, for example, the relative speed is zero.) You'd want to use the average relative speed instead of the initial relative speed.

 

[Calvin Pitts]

Looking again, should I use the equation
x-x₀ = (1/2)(v₀+v)t
?

 

[Hiero]

Looking again, should I use the equation
x-x₀ = (1/2)(v₀+v)t
?

It depends what you mean by these variables. It can be correct, yes, if you have the correct meaning for the variables.

 

[Calvin Pitts]

It depends what you mean by these variables. It can be correct, yes, if you have the correct meaning for the variables.

Should the average velocity just be 80 km/h then? I'm trying to figure out what my final velocity is (0, or 9.8 m/s). If it's zero, then the time of the collision is .11 seconds. If it's 9.8 m/s, the time is .075 seconds (assuming that the displacement is 1.2 m).

 

[Hiero]

Should the average velocity just be 80 km/h then? I'm trying to figure out what my final velocity is (0, or 9.8 m/s). If it's zero, then the time of the collision is .11 seconds. If it's 9.8 m/s, the time is .075 seconds (assuming that the displacement is 1.2 m).

Yes, the average relative velocity is 80 km/h, because the initial is 160, and the final is 0, and it varies linearly between so the average is the midpoint.
The reason why the final is zero is because it is the relative velocity of the objects; but they move together at the final moment.

I do not get 1.2m/(80km/hr) = 0.11 seconds though... I would double check your numbers.

 

[Calvin Pitts]

I carried the half from a previous equation...oops :) it should be .054 seconds, correct?

 

[Hiero]

I carried the half from a previous equation...oops :) it should be .054 seconds, correct?

Right, so then the acceleration of each car is as simple as a=Δv/Δt

 

[Calvin Pitts]

Maybe I'm using wrong numbers, but I still can't get answers that match the book. Change in time is .054 seconds, change in velocity is 22.2 m/s? Or 22.2 m/s - (-9.8 m/s)?

 

[Hiero]

Maybe I'm using wrong numbers, but I still can't get answers that match the book. Change in time is .054 seconds, change in velocity is 22.2 m/s? Or 22.2 m/s - (-9.8 m/s)?

The change in velocity is different for each car. Your book said (130, 850) right? I do not agree with your books answers, I believe they are mistaken.

 

[Calvin Pitts]

The change in velocity is different for each car. Your book said (130, 850) right? I do not agree with your books answers, I believe they are mistaken.

They should be different accelerations according to the book. I wouldn't be totally opposed to the thought of it being wrong haha there have been a couple problems that were wrong.

 

[haruspex]

They should be different accelerations according to the book. I wouldn't be totally opposed to the thought of it being wrong haha there have been a couple problems that were wrong.

They will be different accelerations because the two velocity changes are different. (What are they?) But I agree with Hiero that the book answers are wrong.

 

[Calvin Pitts]

They will be different accelerations because the two velocity changes are different. (What are they?) But I agree with Hiero that the book answers are wrong.

The 540 kg car goes from 22.2 m/s (80 km/h) to -9.8 m/s (opposite direction). The 1400 kg car goes from 22.2 m/s (maybe negative due to the direction?) to 9.8 m/s (again, maybe negative). I draw my diagram with the 540 kg car on the left and the 1400 kg car on the right, implying that traveling to the right is positive and to the left is negative.

 

[haruspex]

The 540 kg car goes from 22.2 m/s (80 km/h) to -9.8 m/s (opposite direction).

Right, so what is the magnitude of the change?

The 1400 kg car goes from 22.2 m/s (maybe negative due to the direction?) to 9.8 m/s (again, maybe negative).

For finding the magnitude of the change, it doesn't matter which sign convention you use as long you are consistent. What is the magnitude of the change for this car?

From those magnitudes, what are the magnitudes of the accelerations?

 

[Calvin Pitts]

The 540 kg car has a magnitude of -32 m/s, and the 1400 kg car has a change of -12.4 m/s. Dividing each number by .054 seconds gives accelerations of -600 m/s and -230 m/s (I hope :) )

 

[haruspex]

The 540 kg car has a magnitude of -32 m/s, and the 1400 kg car has a change of -12.4 m/s. Dividing each number by .054 seconds gives accelerations of -600 m/s and -230 m/s (I hope :) )

That looks about right.

 

[Calvin Pitts]

That looks about right.

Works for me! Thank you so much for your help.

 

[Chestermiller]

CAR COLLISION MODEL WITH CRUMPLING

In this development, the energy dissipating bumper crumpling effect is modeled as a viscous damper, representing the front end of each vehicle. The force of a viscous damper is proportional to the velocity difference between the two ends of the damper. The dampers for the two vehicles are taken as identical so that, by Newton's 3rd law, since the forces on the two dampers are the same, the amount of crushing will be the same. Also, the velocity of the interface between the two dampers is always equal to the average of the two vehicle velocities. For this model, the force balances on the two vehicles are given by:
$$m_h\frac{dv_h}{dt}=\frac{k}{2}(v_l-v_h)\tag{1}$$
$$m_l\frac{dv_l}{dt}=-\frac{k}{2}(v_l-v_h)\tag{2}$$
where the subscript l refers to the light vehicle, h refers to the heavy vehicle, v is velocity in the positive x direction, m is mass, and k is the proportionality constant for each damper. The factor of 2 in the denominators takes into account the fact that the overall velocity difference is equal to twice the velocity difference across each individual damper.

The analytic solution to Eqns. 1 and 2 is given by:
$$v_h=v_{\infty}+\frac{m_l}{(m_h+m_l)}(v_{h0}-v_{l0})e^{-\frac{(m_h+m_l)}{2m_hm_l}kt}\tag{3}$$
$$v_l=v_{\infty}+\frac{m_h}{(m_h+m_l)}(v_{l0}-v_{h0})e^{-\frac{(m_h+m_l)}{2m_hm_l}kt}\tag{4}$$
where $v_{\infty}$ is the final wreck velocity at long times, given by:
$$v_{\infty}=\frac{(m_hv_{h0}+m_lv_{l0})}{(m_h+m_l)}\tag{5}$$with $v_{h0}$ and $v_{l0}$ representing the initial velocities before collision.

From these results, it follows that the relative velocity of the two ends of each of the dampers is given by:
$$\Delta v=\frac{ (v_{10}-v_{20})}{2}e^{-\frac{(m_h+m_l)}{2m_hm_l}kt}\tag{6}$$
The crumple distance of each vehicle is then given by:
$$d=\int_0^{\infty}{(\Delta v)dt}=\frac{m_hm_l}{k(m_h+m_l)}(v_{h0}-v_{l0})\tag{7}$$
From this, it follows that the damper constant k is given by:
$$k=\frac{m_hm_l}{d(m_h+m_l)}(v_{h0}-v_{l0})\tag{8}$$
Substituting Eqn. 8 into Eqns. 3 and 4 for the velocities of the two vehicles during the collision yields:
$$v_h=v_{\infty}+\frac{m_l}{(m_h+m_l)}(v_{h0}-v_{l0})e^{-\frac{(v_{h0}-v_{l0})t}{2d}}\tag{9}$$
$$v_l=v_{\infty}+\frac{m_h}{(m_h+m_l)}(v_{l0}-v_{h0})e^{-\frac{(v_{h0}-v_{l0})t}{2d}}\tag{10}$$
Taking the derivatives of these velocities with respect to time, we obtain the accelerations of the two vehicles:
$$a_h=-\frac{m_l}{(m_h+m_l)}\frac{(v_{h0}-v_{l0})^2}{2d}e^{-\frac{(v_{h0}-v_{l0})t}{2d}}\tag{11}$$
$$a_l=+\frac{m_h}{(m_h+m_l)}\frac{(v_{h0}-v_{l0})^2}{2d}e^{-\frac{(v_{h0}-v_{l0})t}{2d}}\tag{12}$$
Note that both accelerations start out large in magnitude, and then rapidly decay exponentially with time.