@Jasmine We normally take the energy at infinity to be zero. That means the energies of the orbitals are negative. So for example the energy of the hydrogen 1s is -13.6eV. We have to supply 13.6eV of energy to remove the electron from the 1s orbital to infinity.
So as we move from 1s to 2s to 3s etc the energy increases because it becomes less negative.
@Jasmine be a little cautious about kinetic energy of electrons in bound states. Remember that an electron in an atom is not a little ball whizzing round the nucleus. It is a fuzzy blob that is delocalised over the whole atom.
So the interpretation of kinetic energy for an electron bound to an atom is somewhat subtle. Normally it's best to just talk about total energy.
Yes. If we start with the electron in a 1s orbital then add 13.6eV to remove the electron the electron energy is now 13.6eV greater than it was before.
@Jasmine to separate the proton and electron we had to add 13.6eV of energy. But Einstein told us that mass and energy are related by his famous equation $E=mc^2$, so that 13.6eV of energy added a mass of $13.6/c^2$.
So a hydrogen atom has a mass that is smaller than the summed masses of a proton and electron by $13.6/c^2$. And it really does. If you measure the mass of a hydrogen atom to very high precision you find it really is smaller than $m_p + m_e$.
@Jasmine suppose we consider a classical system of a planet orbiting the star. Now there is no ambiguity about KE and PE because it's just classical mechanics, and we can ask what happens to the KE and PE as we increase the orbital radius.
If you increase the radius then the PE increases (becomes less negative) and the KE decreases. However the PE increases by twice as much as the KE decreases, so the total energy PE+KE increases (becomes less negative).
In Bohr's model as we move away from the nucleus the PE increases and the KE decreases just like our planet. The total energy increases because the PE changes by twice as much as the KE.
@JohnRennie adding energy to electron means taking it to higher energy level.. and removing means taking it back to energy levels closer to the nucleus..
Remember that if we start with an atom that has an electron in an excited state that atom will emit a photon and decay to the ground state. That photon carries away energy, so energy is removed from the atom as the electron moves to a lower energy state.
@user64829 while the rod is falling freely, before the end hits the stop, it behaves like a point mass, and the angular momentum of a point mass is $r \times p$
In this mass the linear momentum $p = mv$ and the distance $r$ is just $a$. So we get $L=amv$.
Because the support that the rod hits exerts an external force on the system so energy is not conserved.
Angular momentum is conserved about the point A because the distance of the force to point A is zero, so if we use A as our origin for taking angular momentum then the torque the support produces is zero.
Linear momentum is conserved if and only if no external force acts on the system
Angular momentum is conserved if and only if no external torque acts on the system
Linear momentum is simple because it's always just $mv$. But with angular momentum it is always defined relative to some point that we choose as our axis.
In this case we chose the point A as our reference because the force acts through that point so the torque due to the external force is zero and that makes the calculation easy.
I've completed this chapter but never got this kinda thing in my mind after completing the chapter and even doing complex questions I get this kinda stuff in my mind :)
In effect what you're doing is approximating $(1 + 0.2)^2$ by neglecting the quadratic term i.e. you're using $1 + 2*0.2$ as an approximation for $1 + 2*0.2 + 0.2^2$
This works only when the quadratic term is small compared to the linear term.
@sammygerbil in bs we are said to get the curl of a vector field if its zero. But in these scenarios how do you come to know if its conservative or not?
Or the line integral through closed curve is zero.
Can be assigned as $\vec F =- \frac{\partial v}{\partial x}$ etc. But how do you get them here . Whats the idea here?
What a complicated question! ... My knowledge of the use grad, div and curl in these situations is not very good. I am not sure how to tackle this question. I may have to leave it for John Rennie to deal with. Like you, I am a bit stumped with this question!
@Nobodyrecognizeable Yes. We must presume that the plane is frictionless (because we have not been told anything about the coefficient of friction) so the only force it exerts on the ball is the normal reaction force.
@sammygerbil just had to post it. Sometimes i might be very slow at replying because im replying from mobile. Anyway the perpendicular component is $ev_0sin(\theta + \alpha)$
I would draw the incline as horizontal. Then the initial velocity $v_0$ is inclined at $\alpha$ to one side of the vertical and the final velocity (after collision) is inclined at a greater angle on the other side of the vertical.
@sammygerbil got that. You guys all over the uk talk in english or there are other languages as well .
/*disclaimer /* ill just be doing trash talk for the rest . If you have something important or less important you may leave. But i request you to stay.
@Nobodyrecognizeable But we haven't got the factor $\sin\alpha$ right. I think when the question says below the original point of impact it does not mean vertically below (as I assumed) but down the incline. So we did not need to multiply $x$ by $\sin\alpha$.
@Nobodyrecognizeable English is by far the most prominent language, and everyone (except recent immigrants) is able to speak it. But there are 2 or 3 other minor British languages which are taught in schools. These are Welsh and Scots Gaelic and Manx (spoken on the Isle of Man) which is similar to Scots Gaelic.
@Nobodyrecognizeable Yes Easter Holiday is Good Friday and the following Monday. We also have state holidays such as New Year's Day and a few Bank Holidays including the 1st Monday in May.
@Nobodyrecognizeable Saturday and Sundays (the Weekend) are holidays for most people, and there is usually no school at the weekend, but some people continue working. Banks now open on Saturday mornings, and shops open on Sundays - about 20 years ago these didn't happen. Only small corner shops were allowed to open on Sundays.
@Nobodyrecognizeable thanks :) well the native bengalis celebrate that much better and in a beautiful way :) the whole celebration is beautiful as far I've seen them
@AbhasKumarSinha Guards are professional soldiers so they have good self-discipline. They would be shouted at by their commanders if they laughed but no I don't think they would be punished.
Q is an ellipse again with constant angular velocity. Calculating $\ddot r$ shows the force is central so angular momentum is conserved. So is total energy E. Options 2,5. That also agrees with A.
Therefore answer is A.
Q is a spring force orbit, ie $F \propto r$ not $F \propto 1/r^2$. The force is directed toward the centre of the ellipse, not toward one focus. $1/r^2$ orbits do not have constant angular velocity, they have constant areal velocity (Kepler's 2nd Law).
