Draw the billiard balls in contact, with one centre higher up than the other. Draw a line connecting the centres of the two balls. The initial velocity of the incoming ball is not along this line, it is perhaps horizontal. You have to resolve its velocity into components along ang perpendicular to this line.
@sammygerbil how would what angle the balls will go so how can write the conservation of momentum along the line of action. Should I assume that angle $\theta $ or so ?
It is best to do the main part of the calculation as a head-on collision along the line of centres. Momentum conservation : $mU_1=mU_1'+mU_2'$ where $U_1=V_1\cos45$ is the initial velocity of ball 1 along the line of centres and $U_1'$ its final velocity. $U_2'$ is the final velocity of ball 2 in this direction - its initial velocity was zero.
$V_1'$ here is the final velocity of ball 1 along the line of collision. This ball still has component of velocity $V_1\sin45$ perpendicular to the line of collision.
No. We have conserved momentum along the line of collision. Momentum is also conserved perpendicular to this direction. That means the velocity of ball 1 in this direction remains the same. This component is not affected by the collision.
Maybe. What I did was find the angle between the final velocity and the component perpendicular to the line of collision. This angle is $\theta$ where $\tan\theta=\frac14$.
The final velocity component parallel to the line of collision (opposite) is $\frac{V_1}{4\sqrt2}$ as we calculated above. The perpendicular component (adjacent) is $\frac{V_1}{\sqrt2}$ which was not changed by the collision. The angle between these is $\theta$ where $\tan\theta=\text{opposite/adjacent}=\frac14$.
No they are not referring to a point charge but to the charge distributions on the surfaces of the plates. So in all cases the electric field is horizontal and equal to $q/2e0$ where $q$ is the charge density on the surface of the plate.
The charge distribution on each surface of each plate causes a uniform electric field in the horizontal direction to the left and right of the surface.
The solution assumes that the charge on each surface is +ve, then the direction of $E$ is away from the surface. The charge might actually be -ve but you will find this out by getting a -ve sign from the calculation.
When you have an isolated conducting plate there is an equal surface charge on the other plate. So the electric field from each of the 2 surfaces cancel out inside the plate.
@harambe When you have an isolated conducting plate there is an equal surface charge on the other plate. So the electric field from each of the 2 surfaces cancel out inside the plate. If the surface density on each plate is $q$ then the total electric field outside the conductor (from both surfaces) is $q/2e0+q/2e0=q/e0$.
@harambe Yes, that is a common source of confusion. ... My last comment applies to an isolated plate. In this case we have 3 plates which are not isolated. So the charge on opposite faces of the same plate need not be the same.
Even on a thin sheet, the charge is distributed equally on both sides.
If the conductor surface is a plane, eg the xy plane, with the conductor extending in the -z direction, then there is an electric field $E=\sigma/2\epsilon_0$ extending in the +z direction outside of the conductor. In order for there to be no electric field inside the conductor there must be some more charge elsewhere, eg another surface parallel to the xy plane on which the surface density is also $\sigma$.
@sammygerbil During deexcitation from certain orbit of hydrogen atoms only 3 photons are obtained. The energies of these photons are $\pu{10.2 eV}$ ,$\pu{12.088 eV}$ and $\pu{1.88 eV}$. Minimum number of hydrogen atoms required for these photons are?
I think your reasoning is correct. 12.088eV and 10.2eV cannot come from the same atom, whereas 1.88eV could come from the same atom as either of the other two. So the answer is two atoms.
@Abcd Whoops! Sorry, the 1.88eV photon cannot come from the same atom as 12.088eV because 12.088+1.88=13.968eV which is more than 13.6eV. So the 1.88eV photon can only come from the same atom as the 10.2eV photon. But the answer is still 2 atoms required.
:46965706 Good. I was in the process of drawing the diagram when you said you got it, and I didn't want to delete the diagram, so I completed it and posted it anyway just in case.
@user477343 according to johnrennie anybody can ask any physics problem here. It doesn't mean anybody will flag you if you ask any other subjects. But physics questions are most likely to be asked here.
So a question for example: how does a black hole look like?
This is physics related, but also a bit cosmology/astronomy related. Is this question okay?
(...how does a black hole look like? I imagine, given that it has an intense gravitational pull, asteroids will constantly block its view from us... but yet again, it is known they exist)
@user477343 but yeah technical questions will be appreciated first. Or mostly problem solving questions. But explanation type questions wouldn't be flagged but yeah they will not be the priority.
@Nobodyrecognizeable okay. Some technical questions might require parametric equations. (I assume you are familiar with these.) Are questions like the following alright, then? "A car travels 50km p/h off a cliff that is 100m high. Given the car was driving along a flat horizontal surface perpendicular to the wall of the cliff, how far away from the cliff will the car land?"
Of course, in questions like these, I will include my attempts, so as to not exploit anyone here.
@user477343 black hole is a region whose gravity is so strong that even light can not escape. So we can not see anything but complete darkness through a certain region. But (as I have a little knowledge as a freshman) some of them radiate so we may see some radiation in infrared or so.
@Nobodyrecognizeable I undertook teacher training for secondary school in 2000/2001 but I had difficulty finishing the training (you have to be very well organised, which I am not) and you have to be good at handling disruptive students (I'm not good with that either).
@sammygerbil, suppose that a semiconductor contains equal number of donor and acceptor impurities do they cancel each other in the electrical effect if so what is the mechanism if not why not?
@user477343 No I was a mature student who is now sort of retired. Apart from answering questions here I do some voluntary work fixing bicycles. I have some money saved up from working years ago, so I do not need to work.
@TheSimpliFire ok but if T is not the minimum temperature then how does it relate to the problem?
Well to be technical, I would say in my example question, "...off a cliff that is 100m high from the ground .... a flat horizontal surface perpendicular to the wall of the cliff and of which is parallel to the ground joined to the vertex that the cliff extends upwards from...." :D
@blue_eyed_... Sorry I am not sure about that. I guess that they do cancel out if the donors and acceptor atoms are in the same region, or where two different regions overlap. Better to check with John Rennie about this.
@user477343 by priority it means if somebody asks some problem solving questions he will attended first then an explanatory question will be attended. Although we have a ready explanation type chat room h bar.
@TheSimpliFire Yes that makes sense. So we should ignore the relation m2= T * m1 ? If you wish to keep this condition you need to make it dimensionally correct - eg m2 / m1 = T / T1.
Can anyone help me with a simple physics question?
I am not sure what am I supposed to find in the question
A plane mirror of semicircular shape (r=1m) is placed in y-z plane with diameter along y-axis and centre at origin. An object is placed on the axis of mirror at a point (5,0,0). At a distance of 10m from plane murror there is a wall also parallel to y-z plane. Then what is the ratio of spot of light on wall to that of mirror?
@sammygerbil, Two particles P and Q carrying equal charges after being accelerated through same pd enters a region of uniform magnetic field and describe circular paths of radii R1 and R2. What is the ratio of mass of P to Q? Is the answer R1/R2?
@Jasmine Yes that is a confusing question. Have you missed out some words? I am expecting there to be something between "what is the ratio of ..." and "spot of light on wall". For example, "the ratio of the distance of the spot of light from the axis of the mirror to that on the mirror"?
@blue_eyed_... The particles have the same charge and are accelerated through the same PD so they acquire the same KE : $\frac12 m_1v_1^2=\frac12 m_2v_2^2$. The magnetic forces on them are $F_1=Bqv_1, F_2=Bqv_2$. These are the centripetal forces. The radii of the circles are given by $F_1=m_1v_1^2/R_1$ and likewise for $R_2$.
Therefore $\frac{m_1v_1}{R_1}=\frac{m_2v_2}{R_2}$. Squaring and dividing by the KE equation we get $\frac{m_1}{R_1}=\frac{m_2}{R_2}$. So yes your answer is correct : $\frac{m_1}{m_2}=\frac{R_1}{R_2}$.
Correction : That last line should be $\frac{m_1}{R_1^2}=\frac{m_2}{R_2^2}$. So your answer is not correct. You should have $\frac{m_1}{m_2}=(\frac{R_1}{R_2})^2$.
@sammygerbil you have two occupations. One in the morning you do the volunteering and in evening physics question solving really what a variety of lifestyle(perhaps 2nd one is bit more obsession I think). :p
@sammygerbil sorry if i hurt you. It was never posted to offend you.
@user477343 There's a chemistry chatroom as well. The questions I had posted were from my JEE tests. At first, there was only this room, then other rooms were created to avoid confusion and separate the questions. That makes learning easier. So, students like me usually hang out in these rooms.
@Jasmine I think I have made sense of your mirror question. The plane mirror stands on the floor, eg against a wall. There is a parallel wall opposite 10m away. Halfway between the two walls on the floor is a point object (a source of light). The reflection of this light in the mirror casts a semicircular area of light on the wall. I think you have to find the ratio of area of the semicircle on the wall to the area of the mirror.
@sammygerbil, A charged oil drop in Millikans apparatus is observed to fall in air through a distance of 1mm in 27.4 sec in absence of electric field. The same drop is held stationary when a field of 2.73*10^{4} N/C is applied between the plates. How many excess electrons does the drop acquires? (given coefficient of viscosity of air = 1.8*10^(-5) Ns/m^2). Please give some hints
@sammygerbil, in an evacuated tube electrons are accelerated from rest through a potential difference of 3600V and then travel in a narrow beam through a field free space before entering a uniform magnetic field of 2*10^(-3) T, the Flux lines of which are perpendicular to the beam. In the magnetic field the electrons describe a circular arc of radius 0.1m, calculate e/m of the electron
