Recent content by maxim07

okay I see, then to get the change in internal energy ΔE = 1/2m(v+2u)2 - v2) = 2muv if you consider u << v, where (v+2u) is the velocity of the particle after the collision

Okay, but then for 2 b wouldn’t I get Δp = -8 - 2 = -10 Kgm/s, am i supposed to be getting Δp = 6 Kgm/s as this is -2mu = -2(1)(-3) as the question requires?

1 a) u = 2 ms-1v = -2ms-1 b) Δp = -2mu = -4 kgms-1 c) 0 d) F = dp/dt from Newton’s 2nd law, is it conserved because there’s and equal and opposite force acting on the hall by the wall ? 2 a) u = 2ms-1, v = -5 ms-1 b) -7 kgms-1 c) 22.5 J 3. -3 kgms-1 4. 22.5 J i think v in 2 a is definitely...

Probably going to need someone to show me the first lines of workings for this, still can’t seem to make it work

So applying energy conservation in an elastic collision I’d have 1/2mv2 + 1/2Mu2 = 1/2mvf2 + 1/2Mu2 if I consider the energy of the piston too. Where vf is the final velocity of the particle. But then I’d get vf= v so that must be wrong

Greta thanks solved it

Ah okay, so then I get 1 + 2cos Φ - Acos α= 0 and -2sin Φ + Asin α = 0

The advice made sense I just thought it only referred to the part of the question about momentum not internal energy, does the limit u <<v mean I need to use the binomial expansion since for kinetic energy you have v2, which in this case would be (v+u)2 ?

So if your just considering the additional momentum could you write but then how can I use that to find the internal energy

tanθ = -a/b ?

But if you didn’t know the end result where would the sqrt(5 + 4cos(φ)) come from ?

Here is the solution, I understand how they got E, but I don’t see how they could get E’ from cosine addition formulas? I don’t need to know how to do it with complex numbers.

I understand up to Asin(1+2cosφ). The max value of (1+2cosφ)^2 = 9 so you have to divide I0 by 9. I understand that now, but if the (1+2cosΦ) is the amplitude then the sinx must get averaged, but I thought that would introduce a factor of 1/2?

Here’s an image of the equations better layed out here’s the solution, I don’t understand how they are squaring it to get the time average