Recent content by PeroK

What specifically is not clear? Once you get the idea that light has momentum and that the centre of mass should not move, the kinematics are quite straightforward, are they not?

I think the binomial method is quite general and easy to remember: $$x^2 + y^2 = \frac 1 2(a^2 + b^2)$$$$x^4 + y^4 = \frac 1 8(a^4 + 6a^2b^2 + b^4)$$$$xy = \frac 1 4(a^2 - b^2)$$$$x^2y^2 = \frac 1 {16}(a^4 -2a^2b^2 + b^4)$$Then, we have: $$4(x^4 - 6x^2y^2 +y^4) = 4(x^4 +y^4) -24x^2y^2$$$$ =...

I saw my method more as jujitsi than brute force!

Note that: $$(m+n)^3 + (m - n)^3 = m^3 + 3m^2n + 3mn^2 + n^3 + m^3 - 3m^2n + 3mn^2 - n^3 = 2(m^3 + 3mn^2)$$Where the RHS is twice the sum of the binomial terms with even powers of ##n##. In fact, using this method, we can write down the solution for a higher power with minimal effort: $$x^5 +...

Whatever you call it, "should have" is not a simple tense, but involves some sort of condition. And should is a "modal" verb. Along with would, could, must and ought.

How do you know when you burn coal the energy comes from the coal? Maybe the coal energy is destroyed and new heat energy is created mysteriously from somewhere else and just happens to balance the energy lost by the coal? How is that an enlightening or helpful way to look at it? Maybe when...

Just to add something simple but useful. The proposition is identical to saying that if, for all ##z## in a neighbourhood of 0, we have: $$c_0 + c_1z + c_2z^2 +c_3z^3 \dots = 0$$Then all the coefficients are zero. And, as any neighbourhood of zero contains a real, open interval of zero, then...

One thing to make a mental note of (and nevr forget) is that an equation of the form:$$x + \frac a x + c = 0$$is a quadratic equation in disguise!

The fundamental confusion here is between position as a mathematical concept, which is well-defined in whatever mathematical model is being used; and, position as a physical measurement. If we consider the measurements of position of planets, stars and other galaxies, we can see that their...

If you did, you would be analysing the free rotation of the rod about the pivot point after it's been given an initial impulse by the bullet.

The impact of the bullet will give the rod an initial angular velocity. After that gravity will take over. The centre of mass remains almost exactly below the pivot point for the duration of the collision. Here's another example. If you hit a golf ball, it goes in a parabolic trajectory...

After the collision, gravity will act through the centre of mass and the rod will swing back and forward as you would expect. But, that's beyond the time of the initial analysis.

Imagine that the rod was only very loosely connected to the ceiling when it gets hit by a bullet! What would happen? The rod would get ripped out of position and probably go flying off to the right somewhere. If that doesn't happen, then you know that there must have been a significant force...

The rod gains kinetic energy, so that is not the reason. You cannot assume energy is conserved. Perhaps it is and perhaps it isn't. You have to look for quantities that you know are conserved. The rod is connected to the ceiling, which may apply a force with a horizontal and vertical...

The probability of getting two blue balls is the same, regardless of whether we consider B1-B2 different from B2-B1 or not.