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15

By the Newtonian definition, the photon wouldn't count due to its zero mass, but this is a relativistic collision, so you need a relativistic definition of the center of mass. Relativistically, the c.m. frame is the one in which the total momentum four-vector of the system is purely timelike. No, this does not coincide with the electron's frame. What you're ...


11

The reference frame of the center of mass is, by definition, the one where the total $3$-momentum vanishes. It exists almost always also for massless particles as I go to discuss. The total $4$-momentum $P$ of a system of $N$ free particles is the sum of their $4$-momenta of the particles, i.e., $$P = \sum_{a=1}^N P_{(a)}\:,$$ where each $P_{(a)}$ is a ...


2

Let's make a concrete example with numbers: Suppose that $v_a = 6m/s$ and $v_b = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6, v_{cm} = p/M = 2$ . According to the conservation of energy and momentum: Kinetic energy and momentum are conserved only in a perfect elastic collision, if the bodies stick together the collision is inelastic an ...


2

In this type of collision where you have what amounts to a very quick change in velocity, the force is called an impulse force and it is best to think of the equation a little differently. For example, instead of: $$ \sum F = \frac{\Delta mv}{\Delta t} $$ Think of $\int F \mathrm{d}t$ being equal to the change in momentum, that is: $$ \Delta mv = \int ...


1

The force can be surprisingly large, but $\Delta t$ is not zero, and the force is not infinite. Make some estimates: the duration of the collision is so short that our eyes and brain cannot perceive it. Make an estimate for an upper limit for the duration. (There's no right answer, but a lot of wrong answers. For example, I would think that a duration ...


1

what is the force that the first marble applied one the second marble? The collision is almost instantaneous. Wouldn't that make the force in ΣF=Δmv/Δt insanely large because Δt is so small? Suppose two steel balls A, B of equal mass (m = 0.1 r = 0.03 m) collide and B is at rest: Ball A will exert on b the Impulse of a Force $J$ and its velocity, ...


1

The problem with your solution is that the inelastic collision and assumption that kinetic energy is conserved are mutually exclusive. You can see that in your math when you try to solve for $v_2$. Rewriting equation $(1)$ gives $v_1=\left(m+M\right)v_2/m$ which inserted into $(2)$ yields $$ m \left(\frac{m+M}mv_2\right)^2 = \left(m+M\right)v_2^2.$$ This ...



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