Fast and Furious Scene Analysis

[silento]

Homework Statement

In fast 5, there is a vault drag scene. A big rectanglar prisim vault is being dragged by a dodge charger. On the bridge, the vault was used as a ram against a police vehicle traveling in the opposite direction. When the vault made contact with the vehicle, the vehicle was launched into the air. The vault continues to the right. How do I find how high the vault was launched and by how much was the vault slowed by?
Here's some data:
Vault's weight= 14000 kg
charger's weight (the car dragging the vault)= 1900kg
police car's weight= 1900 kg
Vault's and chargers velocity= +80mph
police car's velocity= -60 mph

Relevant Equations

p=mv, m
car⋅v car_initial +m vault ⋅v vault_initial=m car ⋅v car_final+m vault ⋅v vault_final

The collision seems to be an inelastic collision meaning momentum is conserved however, energy is not due to likely thermal energy from the collision. Using conservation of momentum, we can maybe somehow find the initial momentum of each vehicle and set it equal to the final momentums? However, would the charger and the vault be treated as one body? Thank you!

 

[jbriggs444]

When the vault made contact with the vehicle, the vehicle was launched into the air. The vault continues to the right.

The collision seems to be an inelastic collision meaning momentum is conserved

That is not what inelastic means. In a closed system, momentum is always conserved. "Inelastic" means that kinetic energy is not conserved. It is entirely expected that in a car crash, some kinetic energy will be dissipated into heat, mechanical deformation, vibration and noise.

In a "perfectly inelastic" collision, vehicle and vault would end up moving together as a combined twisted mass.

Since neither vault nor vehicle had any vertical motion prior to the collision and since the vehicle has vertical motion subsequent to the collision, we can see that momentum was not conserved. If we pretend that the scenario is realistic, this could be due to an interaction with a concealed ramp or explosive device.

Given the before the crash data and an assumption of no other energy inputs, we could compute a maximum height to which the vehicle could have been launched. This would correspond to a "perfectly elastic" collision. Would that serve your purposes?

 

[silento]

Hello! Yes I could change the scenario a bit to match this. That would be great thank you

 

[PhDeezNutz]

thank you for making this thread so much. Screw that lame movie.

Also there was another part where Paul Walker (or his twin brother) broke into a vault in some Arabic destination sky scraper high rise looking for something (I forget what).

They saw a car in the vault and Vin Diesel said something like “who keeps a beast like this caged” then proceeded to lift the car from the back end while Paul Walker retrieved something.



Here it is.

I’m going to take a closer look at your question but I had to rant about fast and furious.

 

[PhDeezNutz]

To find how high the vault was launched we’d have to know more about its dynamic properties. It’s moments of inertia etc.

There was only initial travel in the horizontal direction and none in the vertical so a conservation of momentum equation could only be applied in one direction.

We’d have to consider conservation of angular momentum, torque, moments of inertia etc to find how high the vault was launched.

 

[silento]

To find how high the vault was launched we’d have to know more about its dynamic properties. It’s moments of inertia etc.

There was only initial travel in the horizontal direction and none in the vertical so a conservation of momentum equation could only be applied in one direction.

We’d have to consider conservation of angular momentum, torque, moments of inertia etc to find how high the vault was launched.

oh boy. that is a lot more complicated then I initially thought

 

[PhDeezNutz]

I believe there was another part where Dwayne Johnson jumps on a formula 1 car while it speeds away, then he falls off, and falls straight down.

Newton’s first law anybody?

 

[PhDeezNutz]

oh boy. that is a lot more complicated then I initially thought

It’s super easy when you are directing a movie and don’t have to obey laws of physics lmao.

 

[silento]

yeah I agree. What you think? should I just completely change the problem?

 

[PhDeezNutz]

yeah I agree. What you think? should I just completely change the problem?

Maybe constrain it to 1D motion and analyze how bad a bank vault would damage a car if they are both going the same speed?

 

[jbriggs444]

Hello! Yes I could change the scenario a bit to match this. That would be great thank you

Along these lines, let us clean up the question.

We start with a 14000 kg vault moving at 80 mph to the left and a 1900 kg police car moving at 60 mph to the right. We ignore the Charger that is towing the vault. Both vault and police car are moving on what we will take, for purposes of the collision, as an infinitely massive, frictionless road.

There is a collision between vault and police car. It may be perfectly inelastic, perfectly elastic or anything in between.

Question: What is the maximum vertical height which the police car can obtain as a result of the collision?

With the question in hand, we may proceed with the analysis.

It may simplify things if we adopt a frame of reference where the center of mass of the police car plus vault is initially at rest. Since the total horizontal momentum of the pair is conserved, this center of mass will remain in place (horizontally at least) throughout the scenario.

Questions for @silento:

1. What is the velocity of this new frame of reference relative to the road?
2. What is the velocity of the police car relative to the new frame of reference?
3. What is the velocity of the vault relative to the new frame of reference?

 

[silento]

I have not done anything with frames of references in awhile so please bear with me.
1. 80mph
2. 20mph?
3. same as 1?

 

[jbriggs444]

I have not done anything with frames of references in awhile so please bear with me.
1. 80mph
2. 20mph?
3. same as 1?

1. No. The new frame of reference is not moving at the same speed as the vault. It should be moving at the weighted average of the police car's velocity and the vault's velocity.

One way to approach this is to add the momentum of the vault to the momentum of the police car. The result is the total momentum of the pair together. Divide by the total mass of the pair and you have the velocity of the pair's center of mass. This will also be the velocity of a frame where the total momentum of the pair is zero.

1a. What is the momentum of the vault?
1b. What is the momentum of the police car?
1c. What is the total momentum of the pair?
1d. If you divide that by the mass of the pair, what do you get?

2. No. I suspect that you subtracted 60 mph from 80 mph to get 20 mph. That calculation would be wrong for at least two reasons. Let us circle back to this after you get question 1 under control.

 

[silento]

1a. Momentum of the vault is 14000kg*35.76m/s= 500640 NS
1b. momentum of the police car is 1900kg* 26.8m/s= 50920NS
1c. momentum total= 551560NS
1d. 34.689 m/s? (is that the units?)

 

[jbriggs444]

1a. Momentum of the vault is 14000kg*35.76m/s= 500640 NS

A Newton-second is correct choice of unit. It is not one that is often seen. Usually I see momentum left as kg m/s. But Newton-second is the same thing. [Newton-second would be a good unit for "impulse" and impulse is a sort of momentum]

I agree with the result.

1b. momentum of the police car is 1900kg* 26.8m/s= 50920NS

Yep.

1c. momentum total= 551560NS

We have to be careful here. The vault is moving one way. The police car is moving the other way. The momenta are in opposite directions. They will not add. They should subtract!

1d. 34.689 m/s? (is that the units?)

Let us re-do this one after addressing 1c as above.

 

[silento]

thank you for correcting me on my mistakes!
1c. Deeming the vault's momentum as the positive direction, net momentum would be 449720 NS
1d. 449720NS/(14000kg+1900kg)= 28.28 m/s

 

[jbriggs444]

thank you for correcting me on my mistakes!
1c. Deeming the vault's momentum as the positive direction, net momentum would be 449720 NS
1d. 449720NS/(14000kg+1900kg)= 28.28 m/s

Without carefully checking the math, that sounds right. So we have this frame of reference moving to the left (same way as the vault) at 28.28 m/s.

Let us proceed to question 2 now. How fast is the police car moving relative to this frame of reference?

 

[silento]

28.28m/s + 26.8m/s= 55.08 m/s. is this how you solve 2? since they're going in opposite direction so you would add them but it would be negative. -55.08m/s

 

[jbriggs444]

28.28m/s + 26.8m/s= 55.08 m/s. is this how you solve 2? since they're going in opposite direction so you would add them but it would be negative. -55.08m/s

Yes. Perfect.

Can you get the velocity of the vault in the center-of-mass frame now?

 

[silento]

is it 35.76 m/s-28.28m/s? so 7.48m/s?

 

[jbriggs444]

is it 35.76 m/s-28.28m/s? so 7.48m/s? (also please bear with me going to have to drive home from work 7 mins~) thank you

Yes.

Now we can compute the kinetic energy of the vault and of the police car in the center-of-mass frame.

4. What is the formula for kinetic energy?
5. What is the kinetic energy of the police car in the center-of-mass frame?
6. What is the kinetic energy of the vault in the center-of-mass frame?

 

[silento]

Formula for KE is 1/2mv^2 calculating now

 

[silento]

for the vault (1/2)(14000)(7.48)= 52360J
for the police car (1/2)(1900)(-55.08)=(-52326J)?
seems awefully close...did I do somethign wrong?

 

[jbriggs444]

for the vault (1/2)(14000)(7.48)= 52360J
for the police car (1/2)(1900)(-55.08)=(-52326J)?
seems awefully close...did I do somethign wrong?

Looks like you forgot to square the velocity.

With the unsquared velocity you are computing half of the momentum. It is no coincidence that the two momenta are equal and opposite in the center-of-mass frame.

Oh, good job on noticing that the two results were nearly equal and opposite. Sanity checks are important!

 

[silento]

for the vault 391652.8 J
for the car 2882116.08 J
there we go. silly mistake thank you

 

[jbriggs444]

for the vault 391652.8 J
for the car 2882116.08 J
there we go. silly mistake thank you

Without checking the math carefully, those two results look reasonable. One expects most of the kinetic energy to be in the faster moving body -- it is a ratio of M/m I believe. And those two numbers have that ratio.

Which is a good sanity check.

If we add those two energy figures together, we have the total kinetic energy available in the center of mass frame. That is the most energy we could possibly have after the collision. We achieve the greatest possible height if all of that kinetic energy goes into the vertical motion of the police car.

7. What is the total kinetic energy available? (391652.8 J + 282116.08 J)
8. What is the formula for gravitational potential energy?
9. How high can the police car if all of the kinetic energy is converted to gravitational potential energy?

There is a different way we can do 8 and 9 if you do not want to use the potential energy approach.

 

[silento]

If all of this KE is turned into gravitational then is this assuming the vault becomes stationary after the collision?
Out of curiosity what is the other approach?

 

[jbriggs444]

If all of this KE is turned into gravitational this is then assuming the vault becomes stationary after the collision?
Out of curiosity what is the other approach?

Right. The way you get the most kinetic energy into the car is by bringing the vault to a stop. This drains the vault of all of its kinetic energy.

Instead of considering gravitational potential energy, we could calculate the velocity of the police car if all of the kinetic energy of vault plus car ended up in the police car alone.

You'd get that using $KE=\frac{1}{2}mv^2$ and solving for $v$.

With the velocity in hand and assuming that velocity was vertically upward (say the car hit a 90 degree ramp) you can calculate how long it would take gravity to cancel that upward velocity.

You'd get that by taking $t=\frac{v}{g}$.

This is the time of flight until the apex -- maximum height. The average velocity during the upward flight is the average of the starting and ending velocities. The average of $v$ at the bottom and $0$ at the top.

That's just $\frac{v}{2}$.

Multiply time to apex by average upward velocity and you have the height. Which is what we wanted.

 

[silento]

if we were to assume the vault continues with a velocity of roughly half of its initial, would that mess up any of our calculations/scenario? could we just adjust the KE for the vault so that not all of it goes to the car? If we have to then we can keep it the way it is. Also, I enjoy that 2nd method you mentioned :)

 

[jbriggs444]

if we were to assume the vault continues with a velocity of roughly half of its initial, would that mess up any of our calculations/scenario? could we just adjust the KE for the vault so that not all of it goes to the car? If we have to then we can keep it the way it is. Also, I enjoy that 2nd method you mentioned :)

Yes. If the vault keeps some velocity, that needs to go into the energy accounting.

The reason we played games by switching to the center-of-mass frame was so that the vault could end up with zero horizontal velocity.

If we are going to specify that vault's final horizontal velocity then we have to change our approach slightly.

Horizontal momentum is conserved. Knowing the vault's final horizontal velocity tells us how much horizontal momentum is carried away by the vault. Conservation of momentum then tells us what the horizontal momentum of the car must be.

The vault then carries away some kinetic energy.
The horizontal motion of the police car carries away some kinetic energy.
What is left (if anything) can go into vertical motion of the police car.

 

[silento]

lets keep it the way we had it then. vault becomes 0. total KE would be 3273768.88 J

 

[silento]

3273768.88J= (1/2)(1900)(v)^2

 

[silento]

v=58.7 m/s

 

[jbriggs444]

v=58.7 m/s

Sounds right.

 

[silento]

t= 58.7/9.8, t=5.99s.
v/2= 29.35m/s
5 .99* 29.35m/s= 175.82 m :0