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Hadronization still doesn't break conservation of energy and momentum. So getting the total energy, or the total momentum, of the quark-antiquark pair is easy: just add up the total energy and momentum of all the reaction products. To get the individual energy or momentum of one particle, i.e. just the quark (or antiquark), we rely on the fact that they ...


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To complete David's answer, here is a two jet event in LEP, an e+e- collider, the ALEPH detector. The experiment is running on the Z mass. The central part of ALEPH consists of several different tracking detectors. The points where charged particles interact with the tracking detectors (hits) are shown as squares and the tracks fitted to the hits are ...


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Finding the missing equations Coordinate transforms just complicate the issue. The heart of the matter is that in n dimensions you have n degrees of freedom for the velocities of the COM of each sphere, and you only have n momentum conservation equations plus one energy conservation equation. That means you need an additional n-1 equations to solve the ...


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Often when two object collide it is often represented as an instantaneous impulse exchange. However in reality this happens continuously. Namely both objects are not completely rigid and will deform during the collision, storing energy in the elastic deformation (like a spring) and dissipating energy with any inelastic deformation. During such a collision ...


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Find the center of mass (CM) of the two connected particles. Then, determine the distance (r) from the center of mass of the part of the spring where the third particle hit it. Then, if you could allow a collision time and force while in contact, and assume the collision to be frictionless (this is might be hard for point particles and thin spring, cause ...


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The simple answer is that not only momentum but also energy needs to be conserved. This puts constraints on the number of balls that can be activated in the cradle. Note that this does not always give a unique solution either. But it enforces that $ n $ balls to $ n $ balls is a "stable" solution.


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The equation of motion for the ball from the time it bounces till the time it hits the ground again is $$ y = v_0t - \frac{1}{2}at^2 $$ where ground level is $y=0$, and $v_0$ is the velocity going up after adjusting for the coefficient of restitution, and $t$ is the time since the bounce. This equation will take the ball through its peak and back to the ...


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Your formula's wrong. You've got $v=\frac12 at^2$, whereas that's the formula for $y$=height. Velocity's actually $v=at$ (with $a=9.8\mbox{m/sec}^2$).


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Mandelstam $t = (p_1 -p_1')^2$ corresponds to the square of the momentum transfered between the two scattering particles, in an elastic scattering process. If you look at the scattering in the centre of mass frame, like you suggest, then clearly they cannot transfer any energy between them, since that would vilolate conservation of momentum.


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I come to think that when a mass collides with another, both of them should always have equal velocities post-collision. To paraphrase Feynmann, no matter how beautiful you may believe your reasoning to be about this, if it doesn't agree with experiment, it is wrong. If the two objects 'stick together' after the collision (the collision is totally ...


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What is conserved during a collision is not velocity, but momentum (mass times velocity). If the collision is elastic, kinetic energy is also conserved: https://en.wikipedia.org/wiki/Elastic_collision. If the collision is inleastic, only momentum will be conserved: https://en.wikipedia.org/wiki/Inelastic_collision The resulting equations (which you can ...



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