Fast and Furious Scene Analysis

[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?