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Q: When we write that, do we suppose a collisionless or collisional nature of the fluids? A: It's the energy-momentum tensor for a perfect fluid Chapter 2.26 Q: If this description corresponds to collisional fluids, why cosmological simulations are N-body simulations (collisionless) and are not simply based on hydrodynamics? A: Cosmological simulations are ...


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The second solution will mathematically satisfy the conservation equations, but corresponds the objects not actually colliding. Or they ``ghost'' and fly right through each other. :)


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If you have a planet of mass $M$, then its self-gravitational binding energy is roughly $-GM^2/2R$ give or take a small numerical factor. So, for the Earth, this would be $-2\times 10^{32}$ J. Something colliding with the Earth, which has a similar mass and size, would do so at velocities of tens of km/s at least. I think the minimum closing velocity would ...


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I have found mistake in my calculatuion. First j = −(1 + e) vab1 · n j/= 1⁄ma + 1⁄mb + (rap × n)2 ⁄ Ia + (rbp × n)2 ⁄ Ib is how we calculate the impulse. taken from here http://myphysicslab.com/collision.html. the expression (rap × n) is a cross product of vectros, i calcualted it as dot product. and the expression jn is NOT a scalar value. it is ...


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Given the mass m of the ball, the incident normal speed v, and the coefficient of restitution $\rho$, Then the integral of F over the duration of the collision $\Delta t$ is $$\int_0^{\Delta t} F dt = \frac{m(1 + \rho)v }{\Delta t}$$ assuning no rotational effects are incurred. This follows from the fact that at any instant the acceleration of the ball away ...


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Suppose someone suggests that following a perfectly elastic collision, two billiard balls are each traveling twice as fast as they were before (and opposite to their original directions). You can't prove him wrong using conservation of momentum, but you can prove him wrong using conservation of energy. Therefore conservation of energy has implications that ...


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Andromeda and the Milky Way belong to a group of galaxies called the Local Group. The two galaxies are the largest galaxies in the group, so to a pretty good approximation their interaction can be treated as a two body problem, with the other galaxies in the group producing only minor perturbations to their motion. So as you suspected, it isn't the case ...


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Anything over 500 miles in diameter, give or take is almost always sphere-shaped, the primary variation being rotation speed, which can give a flatness to the object, for example, Jupiter is visibly flattened by it's high rotational speed. The problem with building a strange shape by very large collision is that the heat generated in a collision of that ...


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Depends on the nature of the collision. If there is a mechanism that takes energy from the system, i.e. a deformation, than energy is lost. You could think of your example as the center of mass of your rod as a point mass that starts rotating on a massless string once is passes the pivot. As usual, energy and momentum are conserved. You could do the ...


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I will yet add another formalism: Lets start with the hamiltonian form of Hamilton's Principle. Let $c: \mathbb R \longrightarrow T^*Q; t\mapsto (q(t),p(t))$ be the trajectory of a particle in the phase space of the configuration space $Q$, we define a subset of $Q$, $C$, where no contac occur between the particles, and $\partial C$ is te set of ponts wer ...



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