Recent content by Steve4Physics

The centre of mass is some point on the string. At this point, 2 forces act: - tension: 120 N towards the 3 kg mass; - tension: 120 N towards the 2 kg mass. Since these forces have equal magnitudes and opposite directions, the net force = 0 Tension in the string is an internal force of the...

Yes. The system described consists of 2 masses connected by a string with no net external forces. So the CM has a constant velocity of ##-1~(\text{m/s})~\mathbf{\hat y}##. BTW I'm assuming the original question is badly worded so that: "At any instant the velocities [are 5 m/s in opposite...

It's not an assumption. Using the Post #1 diagram we see that the momentum of the 3 kg mass is 15 kgm/s in the -y direction and the momentum of the 2 kg mass is 10 kgm/s in the +y direction. It’s natural (and IMO convenient) to move into the CoM frame. Then only simple calculations are needed...

The system's centre of mass (CoM) has some velocity. If you find this and transform the velocities to the CoM frame, the figures are then consistent with uniform circular motion about the centre of mass. This gives a tension of 120 N, as required. Note: the unit 'newton' is written with a...

I'd like to add this to what @jbriggs444 has said. No. You are wrong! Try working through the following carefully. Step-1 Question: what is the probability of a gas particle having a speed of exactly 50.1 m/s? You might say I haven’t given you enough information – but I have! The answer...

@Mike_bb, can I throw this in... Suppose you have ten 6-sided dice, each marked with the values: -3, -2, -1, +1, +2, +3. You throw them. A = the number of -3s B = the number of +3s. You will probably find ##\frac AB## is not 1. But if you have ##10^{23}## dice you will find that ##\frac...

@Mike_bb, I’m coming into this thread rather late but would like to add some random thoughts which might help. It looks like you are dealing with an ideal gas made of identical particles (i.e. with equal masses). (Note that if you have a mixture of different gases, e.g. hydrogen and oxygen...

@eneacasucci, I'd like to add this to what @Charles Link has said. That's not what the graph shows. The graph shows the potential energy of an electron due to the electric field produced by a row of positive ions. You might find it useful to sketch a graph of potential energy vs. separation...

I agree. It would appear that ##- \frac {\partial F}{\partial r}## should be ##- \frac{dU} {dr} ##, which is the force, ##F(r)##. (Since ##U(r)## is a function of ##r## only, I don’t think a partial derivative is appropriate.) When ##F(r)>0## (for ##r<r_0##) we have repulsion. When ##F(r)<0...

Yes, but the equilibrium position is not relevant. It’s essential to realise that to find the minimum value of ##F##, we must consider the case where ##m_1## gets as far right as possible (giving maximum spring extension). We need the spring tension to just reach the minimum value needed to...

It explains why your approch in Post #7 is wrong. In Post #7 you wrote : There is no force balance*. Forces aren't balanced because ##m_1## is accelerating. *Edit. Let me qualify that - the forces on ##m_2## are balanced at the moment it is about to start moving.

When ##m_1## first comes to rest, its acceleration will be maximum (not zero). So the net force on it will be maximum. Have you covered simple harmonic motion yet? Think of a comparable system - a mass hanging from a vertical spring and oscillating vertically. When is the spring tension...

Totally unfamiliar territory for me but, out of curiosity (and the desire to delay starting some repair/painting work), I did a bit of a search. If you haven’t already found it, this link https://www.vasp.at seems a good place to start. It provides, for example: - a forum you could use to ask...

Your correction repeats the mistake! You mean ##\arccos (\frac 23)## to the vertical - as you correctly show in your diagram. A general point: it's helpful (to whoever is marking your work) to include few words of explanation. Then, even if you make some arithmetic/algebraic mistake(s), the...

I think you’ve got it but I would have said it slightly differently, such as: "The bob+string can not reach the top of the circle because the string goes slack before reaching the top. Whereas a bob+rigid massless rod would reach the top because the rod cannot go ‘slack’ – the rod will be in a...