Finding Acceleration of a Car Where Crumpling Occurs (Momentum)

[Rippling Hysteresis]

Homework Statement

Two automobiles of 540 and 1400 kg collide head-on while moving at 80km/h in opposite directions. After the collision the automobiles remain locked together. The front end of each automobile crumples by 0.60 m during the collision. Find the acceleration (relative to the ground) of the passenger compartment of each automobile; make the assumption that these accelerations are constant during the collision.

Relevant Equations

conservation of momentum, conservation of energy
W=F*d
F=ma
FT = impulse
W = change in energy

80km/h = 22.2 m/s
Through momentum: 1940(v_f) = 540 (22.2) + (1400)(-22.2) => v_f = -9.84 m/s
I figured the work that the energy lost in a collision is equal to the work done to crumple the cars. So W = K_i - Kf = [1/2 (540)(22.2)^2 + 1/2(1400)(-22.2)^2] - 1/2(1940)(-9.84)^ 2 = 384110 J

At this point I tried a couple things. Option 1: Total energy lost is W on car 1 + W on car 2= F (0.6) + F(0.6). So 1.2F =384110 J => F= 3.2 X 10 ^5 N.
Since the Force is equal and opposite, a_1=F/m_1 and a_2=F/m_2, so a1= 592 m/s^2 and a_2 = 229 m/s^2.

I also found a thread with what appears to be the same problem. It gave some suggestions with F = W/x and Ft = impulse, and calculating time and then velocity. I get the same results through that route.

But the answers are that a_1 = 850 and a_2= 130, so I guess I've gone wrong on both accounts.

Please help if you can.

 

[haruspex]

You calculated the force purely on the basis of work done in crumpling. So if the cars had bounced elastically off each other, what force would you have calculated?

 

[Rippling Hysteresis]

Hmm, good point! I would imagine I need to do something like F_net = F_crumpling + F_elastic, which is why the suggestion for the second approach probably involved time. I still get some wonky numbers, but perhaps it's closer. Or maybe my equation is wrong for force, but I see what you mean about needing the force that is always there, elastic or not, and then one that crumples.

Since force is constant = W/x, I can say W/x * t = m* delta_v, right? So t= m*delta_v *x/W. a = delta_v/t, so substituting for t just gives W/(m*x) to give the same result as before, so I'm assuming that's acceleration due to crumpling and I need to factor in the change in velocity to the ground.

Now here's where my intuition gets shaky. I think the F I originally calculated is due to the center of mass reference frame. So then I think I should factor in that the difference between the center of mass velocity (-9.85) and then ground is of course 9.85, and I should use THAT delta_v in the problem. At least it seems to get the right answer when I do that, but I don't know if I've convinced myself of the logic. delta_v=-9.85 and t was 0.02687, so a= -9.85/0.02687= -367. The other acceleration was also negative and -592, so the sum is -959, which is about right.

This makes sense sort of, especially given that the problem says to consider the center of mass reference frame, but can you help me iron out some of the kinks in logic?

 

[haruspex]

the sum is -959, which is about right

But not really close enough.
I think we have to neglect the mass of the crumple zone. But if we model a vehicle as an incompressible mass prefixed with a massless, compressible body then we get the awkward result that the front of the massless part instantaneously adapts to the final velocity... yet in principle it can't know yet what that final velocity is supposed to be!
No matter, we know what it will be.
Using that approach I got 127 and 855.

 

[Rippling Hysteresis]

Oh right, it is supposed to be around 850, not 950. Erg. Which step am I going wrong in if I go down that path then?

I had 1.2F =387000, so for car 1, a =387,000/(1.2*540) = 597
Then I'm getting t=m*delta_v * x /W. So for car 1 that's 540* (-32.1)((0.6)/387000 = 0.2687. From there, adjusting for the fact that we want this relative to the ground, not the reference frame (which is moving at 9.8 m/s), a= 9.8/0.2687 = 364

Both of these accelerations are to the left for car 1, which give -597 - 364. But that gives the undesirable answer.

 

[Frabjous]

I would suggest doing this in the zero momentum frame (accelerations will be identical in this frame with the ground frame)
In this frame the final velocities will be zero and the initial and final positions are trivial.
1) Calculate initial velocities for this frame.
2) Calculate the deceleration time for each passenger compartment with constant acceleration as an unknown variable.
3) Given the deceleration distance, calculate the constant acceleration for each passenger compartment.

Hint: start with a(t)=constant for each compartment and integrate.

 

[Rippling Hysteresis]

I would suggest doing this in the zero momentum frame (accelerations will be identical in this frame with the ground frame)
In this frame the final velocities will be zero and the initial and final positions are trivial.
1) Calculate initial velocities for this frame.
2) Calculate the deceleration time for each passenger compartment with constant acceleration as an unknown variable.
3) Given the deceleration distance, calculate the constant acceleration for each passenger compartment.

Hint: start with a(t)=constant for each compartment and integrate.

1) Car 1 is going 22.2 m/s and car 2 is -22.2 m/s, but the zero momentum frame is moving -9.9, so v1=-32.1 m/s, v2= 12.3 m/s w.r.t. the zero momentum frame

2) This might be where I'm getting confused. I would think here is where we say Integral (F dt ) = m*delta_v. Since F=ma is constant, then I believe we just have m*a*t=m*delta_v. For Car 1, t= delta_v/ a. So car 1's change in velocity is 32.1 m/s w.r.t. the zero momentum frame. 32.1/a = t?

3) OK, so here the distance would be 0.6m. I would think we'd use x=v_i *t + 1/2at^2 here with d=0.6. If I substitute 32.1/a into the equation I do get about 850. Just want to make sure it's not a coincidence. It seems to work for car 2 as well.

 

[Frabjous]

2) This might be where I'm getting confused. I would think here is where we say Integral (F dt ) = m*delta_v. Since F=ma is constant, then I believe we just have m*a*t=m*delta_v. For Car 1, t= delta_v/ a. So car 1's change in velocity is 32.1 m/s w.r.t. the zero momentum frame. 32.1/a = t?

If acceleration is constant, v(t)=v0+a0t
After deceleration v=0, so t=-v0/a0

 

[Frabjous]

3) OK, so here the distance would be 0.6m. I would think we'd use x=v_i *t + 1/2at^2 here with d=0.6. If I substitute 32.1/a into the equation I do get about 850. Just want to make sure it's not a coincidence. It seems to work for car 2 as well.

You are being a little bit sloppy with defining positions which means there are sign issues (one car has a positive acceleration, one car has a negative acceleration) You should probably draw a picture. There will be an x_i and x_f with a difference of 0.6m

Ignoring signs, if you do not fill in the numerical values, you would see that a=v_i^2/2d which is an equation that you might have seen before

 

[Rippling Hysteresis]

You are being a little bit sloppy with defining positions which means there are sign issues (one car has a positive acceleration, one car has a negative acceleration) You should probably draw a picture. There will be an x_i and x_f with a difference of 0.6m

Ignoring signs, if you do not fill in the numerical values, you would see that a=v_i^2/2d which is an equation that you might have seen before

OK thanks for the tips! I was rushing a little with typing, so I'll check the signs. Appreciate it.