Recent content by vxr

I saw this general formula: ##w_{0} = \sqrt{\frac{k}{m}}## In my case both masses after collision create connected system, so ##w_{0} = \sqrt{\frac{k}{m+M}}## Plugging it into ##\omega = \sqrt{\omega_{0}^2 - \beta^2}## gives : ##\omega = \sqrt{\frac{k}{m+M} - \beta^2} = \sqrt{80 - 21^2} <...

Without doing any proofs, it is better to jump off last moment, right?

So there are two cases: a). free fall (straight forward for me) b). ladder rotating and jumping off in last moment (I am interested in trying to understand this case) I believe I should take into account momentum at the time the man hits the ground in both cases? The smaller, the better. Or...

Still quite struggling with that one. Did I solve it correctly? Assume that ##s = 2.5m## ##v_{f}^2 = v_{0}^2 + 2as## ##a = \frac{v_{f}^2 - v_{0}^2}{2s}## ##a = \frac{16.6^2 - 0^2}{2*2.5}## ##a =~ 55 \frac{m}{s^2}## This is correct, right? Or.. Anyway, if that's correct then I suppose...

Hello. I am first year student, taking introductory physics course and next week I have a whole-semester exam. (kinematics, dynamics, rigid body, termodynamics, mechanical waves - that kind of basic stuff). I don't have a chance to directly ask my teacher following questions, so I figured who...

Well, if momentum is conserved, then I assume: Bullet's momentum: ##p_{1} = mv_{0}## Resting block's momentum: ##p_{2} = 0## Connected system's initial momentum: ##p = p_{1} + p_{2}## ##(m+M)v = mv_{0}## ##v = \frac{mv_{0}}{m+M}## But what I need to find is ##x## and ##t##. So if I...

Momentum of the system?

No idea. I have just assumed that it indeed does apply here. Can I get any advice how should I find the ##x##?

This is task from my textbook and it does not provide us with an answer. So I cannot verify if I did mistake. Can someone double check, please? My solution: ##E_{k_{0}} = \frac{(m+M)v_{0}^2}{2} \quad \land \quad U = \frac{kx^2}{2}## ##E_{k_{0}} = U## ##\Longrightarrow (m+M)v_{0}^2 = kx^2##...

Thanks.

Can someone double check my calculations? I will skip ##\theta## shift angle in all calculations for simplicity. ##x(t) = Acos(\omega t) \quad \land \quad \omega = \frac{2\pi}{T} \quad \land \quad A = \frac{1}{50}m = h## 1.1: ##x(\frac{1}{4}T)## = ? ##x(\frac{1}{4}T) = Acos(\frac{2\pi}{T}...

So I am almost sure I know how to solve this, just curious about the maximum velocity. Anyway, if you could double check my calculations, here it is. ##T = \frac{t}{n} = \frac{10s}{15} = \frac{2}{3}s## ##\omega = \frac{2\pi}{T} = 2\pi \frac{3}{2} = 3\pi## a). position at ##t = 0.8s##...

I have heard somewhere that someone who is studying basics of physics should pay a lot of attention and should spend a lot of time learning classical mechanics (kinematics, dynamics I believe), before proceeding further? By doing that it should be easier to grasp further chapters of physics. Is...

Good point. It is both. We had in total about 20 hours of lectures of pure theory. Problem is I am not from an english-speaking country, and our studies are taught in english. Even though I believe the teacher who held the lectures is a person of great mind, cited often around the world and...

After the foot is released from the gas pedal, the car will naturally start decelerating. From my understanding two things are causing this: air resistance and kinetic friction. From my understanding also, in a vacuum and in a place where gravity is relatively small, the car would not...