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Acceleration in One Dimensional Inelastic Collision

  • Thread starter Calvin Pitts
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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

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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

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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

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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.
 
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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.
 

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