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jbriggs444
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Yep. Pretty high.silento said: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 said:t= 58.7/9.8, t=5.99s.
v/2= 29.35m/s
5 .99* 29.35m/s= 175.82 m :0
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.jbriggs444 said:In a "perfectly inelastic" collision, vehicle and vault would end up moving together as a combined twisted mass.
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.haruspex said: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.
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.jbriggs444 said:I still maintain that a perfectly inelastic collision is one in which the two masses end moving up at the same velocity.
haruspex said: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#
https://en.wikipedia.org/wiki/Coefficient_of_restitution
- e = 0: This is a perfectly inelastic collision.
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.
The Wiki article that I Googled up agrees with you:haruspex said: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.
Though this differs from the version that I internalized all those many years ago.https://en.wikipedia.org/wiki/Inelastic_collision said: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.
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.silento said:Angular Momentum= l x w
Put all the numeric values aside and work purely algebraically. It has many benefits.silento said: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
I can do steps 1 and 2 right now. However I am unfamiliar with step 3haruspex said: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:
- write the conservation equation for horizontal momentum
- 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
- write the expressions for the initial and final angular momenta of the vehicle about P
1. m1i⋅v1ix+m2i⋅v2ix=m1f⋅v1fx+m2f⋅v2fxharuspex said: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:
- write the conservation equation for horizontal momentum
- 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
- write the expressions for the initial and final angular momenta of the vehicle about P
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.silento said:1. m1i⋅v1ix+m2i⋅v2ix=m1f⋅v1fx+m2f⋅v2fx
2. got it
3. how do I do this?