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

 

[jbriggs444]

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

Yep. Pretty high.

 

[silento]

Ok so just making sure I got my head wrapped around this. the reason for our manipulation of the center of mass frame was to get a way for the vault to have 0 final momentum? Keep the COM in place?

 

[silento]

I hate to be bothersome but I just can't wrap my head around this. How are we able to assume that the vault becomes stationary after the collision? It has a much larger momentum so wouldn't that just overcome the opposing momentum of the car and keep going? Is the height I found assuming that the car is able to withstand the impact and all of that energy is put into its vertical velocity? also, we would have to be assume perfectly elastic collision?

 

[haruspex]

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

Not necessarily. That would be true if the two mass centres were travelling in the same line. Since one was thrown into the air, that is not the case.
More generally, perfectly inelastic means as much KE is lost as possible, given the rest of the circumstances.
In the present case, it is feasible that the two would lock rigidly, but because of the torque they would rotate together in a vertical plane. (There would be a vertical impulse from the ground on the rear wheels of the vehicle with the lower mass centre, just sufficient to prevent it from penetrating the road.)
But from the description, the vehicle with the lower mass centre stayed earthbound. So instead, treat the flying vehicle as a rectangular block struck off-centre. Now the impulse from the ground is at the front of each vehicle.

Edit:
You have three unknowns: the two post collision horizontal velocities and the angular velocity of the skied object. To proceed, consider conservation of both horizontal momentum and angular momentum about the point on the road where the objects meet. Since that point is below the mass centres, will need to include the contributions of horizontal momentum to that angular momentum.
The third equation expresses that the front wheels do not penetrate the road.

You will have to make up variables for the length of the skied object and the two mass heights, then vary these to maximise the height achieved.

Within the description given, I do not see any other mechanism by which vertical motion can be produced.

Edit 2: In the above, I mentioned relative heights of mass centres, overlooking that the exterior shapes matter. I.e. what is the height of those in relation to the main impact point? In principle, both objects could rear up. Since they don’t, maybe this misrepresentation is ok.

 

[jbriggs444]

Not necessarily. That would be true if the two mass centres were travelling in the same line. Since one was thrown into the air, that is not the case.

I still maintain that a perfectly inelastic collision is one in which the two masses end moving up at the same velocity. One can envision cases where they do so after becoming separated by significant distances -- e.g. paying out a cable to harvest any residual rotational kinetic energy.

I agree that if the police car departs vertically while the vault remains behind that the collision was not perfectly elastic.

I see a number of possibilities for the police car departing vertically.

1. This is a three body collision. Even a frictionless roadway can impart vertical momentum or angular momentum about a horizontal axis.

If, for instance, the "splat" when the two moving bodies meet results in the downward extrusion of some attached pieces then the roadway could resist this, resulting in an upward impulse on one or both bodies.

2. If the impact is off-center in the vertical direction then the result can be an upward impulse on one participant and a downward impulse on the other.

3. Even if the impact is on-center for both objects, one could engineer the colliding faces to have any chosen angle. For instance, an upward-sloping lower ramp on vault meeting a downward sloping overhanging ramp on car.

In order to achieve a maximally elastic interaction, I would envision the collision as an elastic bounce of a small "car" from a precisely angled surface on the "vault". The vault would be rigid enough so that the downward impulse on the vault would have time to reflect from the surface of the roadway and arrive back in time to complete the collision with the "car". The car would be ejected vertically while the vault would come to a stop in the center of mass frame. Vertical momentum would not be conserved due to the interaction with the roadway. Angular momentum about a horizontal axis would be minimized by keeping the collision at or near the individual centers of mass. We have freedom to choose creative mass distributions for both objects to help arrange this.

I am considering an almost completely abstract interaction between arbitrary shapes with idealized properties. This may put us at cross purposes since you seem to be contemplating something more realistic.

 

[haruspex]

I still maintain that a perfectly inelastic collision is one in which the two masses end moving up at the same velocity.

A standard counterexample is an oblique impact. Perfectly inelastic in that case, as I was taught, means that after the collision they have the same velocity components along the line of their mass centres at the instant of collision.

"A perfectly elastic collision is one in which the coefficient of restitution is zero."
https://www.vedantu.com/jee-main/physics-oblique-collisions#

• e = 0: This is a perfectly inelastic collision.

https://en.wikipedia.org/wiki/Coefficient_of_restitution

I do see some opinions online that it implies the bodies stick together, so spin, but that is a separate matter. Having collided perfectly inelastically, their relative motion is normal to the line of their centres. Sticking together means that some of the remaining KE is seen as rotational. For point masses (all the mass concentrated at the centre of a rigid ball) it does not reduce the KE further. Otherwise we must consider moments of inertia.

Oh, and cars are designed to be highly inelastic, but not sticky.

 

[silento]

A standard counterexample is an oblique impact. Perfectly inelastic in that case, as I was taught, means that after the collision they have the same velocity components along the line of their mass centres at the instant of collision.

"A perfectly elastic collision is one in which the coefficient of restitution is zero."
https://www.vedantu.com/jee-main/physics-oblique-collisions#

• e = 0: This is a perfectly inelastic collision.

https://en.wikipedia.org/wiki/Coefficient_of_restitution

I do see some opinions online that it implies the bodies stick together, so spin, but that is a separate matter. Having collided perfectly inelastically, their relative motion is normal to the line of their centres. Sticking together merely means that some of the remaining KE is seen as rotational, but it does not reduce the KE further.

Oh, and cars are designed to be highly inelastic, but not sticky.

Angular Momentum= l x w
Linear Momentum= mv

Both cars start out with linear momentum.
Linear Momentum of Vault= 500640 NS
Linear Momentum of Car= -50958 NS

 

[jbriggs444]

A standard counterexample is an oblique impact. Perfectly inelastic in that case, as I was taught, means that after the collision they have the same velocity components along the line of their mass centres at the instant of collision.

The Wiki article that I Googled up agrees with you:

For two- and three-dimensional collisions the velocities in these formulas are the components perpendicular to the tangent line/plane at the point of contact.

Though this differs from the version that I internalized all those many years ago.

 

[jbriggs444]

Angular Momentum= l x w

A more general formula for angular momentum in three dimensions is $\vec{r} \times \vec{p}$. That is the cross product of linear momentum $\vec{p}$ times the offset $\vec{r}$ of the particle's position from the chosen reference point.

For instance, if we measure angular momentum of a vehicle with 50,000 NS of forward momentum about a point 100 meters left of the road, that would give us 5,000,000 kg m²/sec of vertically pointing angular momentum even though nothing is rotating.

In the case of an object rotating about a reference axis, it comes to the same thing. The angular momentum [pseudo-]vector then ends up pointing up along the axis.

 

[haruspex]

Angular Momentum= l x w
Linear Momentum= mv

Both cars start out with linear momentum.
Linear Momentum of Vault= 500640 NS
Linear Momentum of Car= -50958 NS

Put all the numeric values aside and work purely algebraically. It has many benefits.
I would use M, m for the large and small masses, U, u for their initial horizontal mass centre velocities, V, v for the final ones (all positive to the right), and ω for the angular velocity of the lofted vehicle, anticlockwise positive. You can choose whatever you want instead, but please state your choices.

For simplicity, I would take the height of the mass centre of the vault to be the height of the collision, H, and use h for the mass centre height of the vehicle, and I for its moment of inertia about its centre.

Next steps:

1. write the conservation equation for horizontal momentum
2. define P as that point in space which is at the height of the collision point and directly above the vehicle's front wheels; draw yourself a diagram
3. write the expressions for the initial and final angular momenta of the vehicle about P

 

[silento]

Put all the numeric values aside and work purely algebraically. It has many benefits.
I would use M, m for the large and small masses, U, u for their initial horizontal mass centre velocities, V, v for the final ones (all positive to the right), and ω for the angular velocity of the lofted vehicle, anticlockwise positive. You can choose whatever you want instead, but please state your choices.

For simplicity, I would take the height of the mass centre of the vault to be the height of the collision, H, and use h for the mass centre height of the vehicle, and I for its moment of inertia about its centre.

Next steps:

1. write the conservation equation for horizontal momentum
2. define P as that point in space which is at the height of the collision point and directly above the vehicle's front wheels; draw yourself a diagram
3. write the expressions for the initial and final angular momenta of the vehicle about P

I can do steps 1 and 2 right now. However I am unfamiliar with step 3

 

[silento]

Put all the numeric values aside and work purely algebraically. It has many benefits.
I would use M, m for the large and small masses, U, u for their initial horizontal mass centre velocities, V, v for the final ones (all positive to the right), and ω for the angular velocity of the lofted vehicle, anticlockwise positive. You can choose whatever you want instead, but please state your choices.

For simplicity, I would take the height of the mass centre of the vault to be the height of the collision, H, and use h for the mass centre height of the vehicle, and I for its moment of inertia about its centre.

Next steps:

1. write the conservation equation for horizontal momentum
2. define P as that point in space which is at the height of the collision point and directly above the vehicle's front wheels; draw yourself a diagram
3. write the expressions for the initial and final angular momenta of the vehicle about P

1. m1i⋅v1ix+m2i⋅v2ix=m1f⋅v1fx+m2f⋅v2fx
2. got it
3. how do I do this?

 

[haruspex]

1. m1i⋅v1ix+m2i⋅v2ix=m1f⋅v1fx+m2f⋅v2fx
2. got it
3. how do I do this?

For the angular momenta before collision, they're the linear momenta multiplied by the height above P. Except, because of the (standard) sign convention I specified, we need to insert a minus sign.
Since I said to take the mass centre of the vault at the height of the collision point, its horizontal momentum contributes nothing to its angular momentum about P.
The vehicle's mass centre is (h-H) above P, so its initial angular momentum is $-(h-H)mv_i$. (I don't know which you are calling mass 1 and mass 2.)
After the collision, its a.m. is $-(h-H)mv_f+I\omega$, where I is the vehicle's moment of inertia about its centre of mass.

What criterion needs to be satisfied to ensure conservation of the vehicle's a.m. about P? Is the criterion met?