Hot answers tagged collision
4
As far as I know, nobody has ever done this, at least not at what we currently consider high energy. (Electron-electron collisions happen at low energy all the time, of course.) I doubt that anything interesting would happen, primarily because electrons are mutually repulsive, and they have a low mass. That means two colliding electrons would just bounce ...
2
Without friction, the forces during the collision (glancing or head-on) are applied exclusively through their centres of mass. (Illustration available on Wikipedia.)
The torque is given by $\tau=\mathbf r \times \mathbf F$ - but if the forces are applied through the centre of mass, then $\mathbf r$ and $\mathbf F$ are parallel, and hence $\tau=0$.
Without ...
2
This problem has a recursive flavor that we'll not try to avoid.
Conservation of momentum tells us that
$$m v_0 + (p+n-1)m v(n-1) = (p+n)m v(n).$$
Imposing the boundary condition $v(0)=0$ we find
$$v(n) = \frac{n}{n+p}v_0$$
as claimed.
Let $a_n$ be the time at which the $n$th bullet strike occurs.
We have $a_1=x_0/v_0$ and
$$v_0 (a_n - T) = v_0 ...
1
You're not doing anything wrong, the objects will have different momenta in different reference frames. What should be the same in every reference frame is the forces acting on the objects during the collision. The laws of physics are the same in every reference frame, but not necessarily the numbers that go into the equations.
By way of example, lets ...
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