I'm working on a small java program, and I was wondering if there was an equation to calculate the approximate velocity of a falling body, in function to its mass $M$ and the time $t$ since the beginning of the fall (example; Body $A$ of a mass of 57kg has a velocity of $y$ after 11 seconds of fall) I googled it and the only equations I found were either in function of time or mass, but never both. Sorry if this comes out as a rudimentary question, I'm a computer science student, my last physics course was in high school.
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I'm far from being a physics expert and figured this would be a good place to ask a beginner question that has been confusing me for some time.  According to Galileo, two bodies of different masses, dropped from the same height, will touch the floor at the same time in the absence of air resistance.  BUT Newton's second law states that $a = F/m$, with $a$ the acceleration of a particle, $m$ its mass and $F$ the sum of forces applied to it.  I understand that acceleration represents a variation of velocity and velocity represents a variation of position. I don't comprehend why the mass, which is seemingly affecting the acceleration, does not affect the "time of impact".  Can someone explain this to me? I feel pretty dumb right now :)
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I remember the first time I saw this formula, $F = ma$, and it seemed perfectly correct and natural to me (in the sense that it is very straight-forward and logical, not that it is easy by any means). But then, I wondered, what if an object is moving with a constant speed? Maybe a car. If it travels with constant speed and hits me, it still hurts. So it's force must not be $0$.  One more thing related to this problem. If the same car is pulling another car behind it, if the car goes with constant speed, it implies that the first car doesn't exert any force to the second car at that moment, since it is not technically pulling it anymore (espacially if we don't take into consideration the friction).  So my question would be the following: How do these two phenomenons describe the same law of motion?   PS: If gravity wasn't behaving as some kind of acceleration pulling me constantly to the Earth's core, then if I jumped upwards, would I start slowly getting away from Earth?
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I hear people say that , $\mu$, is used for two body problems, but what sort of problems? Is it so that the two body system can be treated as one body or is it to simplify calculations regarding how the two bodies act upon each other?  I can follow the derivation (which I shall go through shortly) in a mathematical sense but I can seem to be able to grasp the ideas behind why the steps were taken.  Derivation:  Step [1] - From Newton's laws of motion: $$F_{12}=-F_{21}$$ It follows that: $$m_1a_1=-m_2a_2$$ Step [2] - Simple rearrangement of the result from step 1 to make $a_1$ the subject (it seems a reasonable thing to do but I don't know why it's done). $$a_1=-\frac{m_2}{m_1}a_2$$ Step [3] - The relative acceleration, $a_{rel}$, can be found by subtracting the acceleration of one body from the other (I get how this is done and get that it would give the relative acceleration BUT I don't know why you would need the relative acceleration). $$a_{rel}=a_1-a_2=(1+\frac{m_1}{m_2})a_1=\frac{m_2+m_1}{m_1m_2}m_1a_1=\frac{m_2+m_1}{m_1m_2}F_{12}$$ Step [4] - According to Wikipedia the result from the last step can be used thus (I have no idea what's going on in this step): $$\frac{m_2+m_1}{m_1m_2}F_{12}=\frac{F_{12}}{\mu}$$ Thus: $$\mu=\frac{m_1m_2}{m_2+m_1}$$  Again, I can do the maths but my problem here is conceptual I believe. I have included the derivation in my question because I believe that it's possible that if you were able to shed some light on why particular steps were taken that this may help.
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Time dilation  using Schwarzschild metric for a non rotating spherical body is: $$t_0=t_f\sqrt{1-\frac{2GM}{rc^2}}$$  For such a non rotating spherical body, what would the time dilation of a clock in vacuum free-falling from infinity be?  (If the answer is non-trivial; a high level outline of the calculation would suffice / be appreciated)  Edit: I am currently working on an iOS app that is trying to model the mechanism underpinning relativity.  So, far the mechanism that I have created is shockingly simple and shockingly good at conforming to Relativity.  However, I am trying to break it.  I am trying to find any possible areas where the two may diverge.  I have noted that using my model a clock in freefall will experience no time dilation, i.e. $t_0=t_f$ and I want to make sure Relativity agrees.  I have noted the gravitational component of time dilation above.  Since my clock is moving one might also expect a kinematic time dilation.  I can calculate the velocity of my clock: $$E_k=\frac{1}{2}mv^2$$ $$E_p=\frac{-GMm}{r}$$ $$v=\sqrt{\frac{2GM}{r}}$$ Plugging this velocity into the kinematic time dilation equation: $$\Delta t'=\frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$$ $$\Delta t'=\frac{\Delta t}{\sqrt{1-\frac{2GM}{rc^2}}}$$  At this point one might make the observation that the kinematic dilation is the inverse of the gravitational dilation and therefore conclude that: $$t_0=t_f$$
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On this site, the Phys.SE question  has been asked. See also  Phys.SE question. The 'answer' provided on this forum has been that the curvature of spacetime explains both. The answer is still cryptic for me as I am more a concrete thinker.  Newton said $F=ma$. I can use this formula to measure inertial mass. Experimentally I can measure the motion of an object while applying a constant force to it. Newton also said $F=\frac{GMm}{r^2}$. In this case, what simple experiment will allow the measurement of gravitational mass?
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Average velocity:  $$v_{\rm avg,1}=\frac{v_{\rm final}+v_{\rm initial}}{2}$$   and average velocity:   $$v_{\rm avg,2} =\frac{\rm total\;displacement}{\rm time \;taken}=\frac{\Delta x}{\Delta t} $$  What is the difference between them and when do we use them?
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Starting from D'Alembert's general wave equation we can find the propagation speed of an impulse along a tightrope as a function of linear mass density, $$\boxed{v= \sqrt{\frac \tau \mu}} \tag 1$$ where $\tau$ it is tension of the rope (in Newton) and $\mu=m/\ell \,$ it is the mass of the rope per unit of length.  Are there simple methods for students of an high school question who don't know calculus (derivatives and partial derivatives, of course) to be able to get the relation $(1)$?
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We know that the equation of motion of particle can be derived from the respective action. But in the book I am reading, the author is saying:     ... timelike worldline of a massive particle is parametrised by proper time, $x^a=x^a(\tau)$. The velocity (tangent) vector is $u^a \equiv \dot{x}^a(\tau)$. This is normalised as $u^au_a =-c^2$.   How can one derive e.o.m?
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The concept of velocity is by definition the movement divided by the interval of time between initial position and final position.  If $f(t)$ is the position of a particle at time $t$; the velocity in the interval $[t_0;t_1]$ is $\dfrac{f(t_1)-f(t_0)}{t_1-t_0}$  The problem is that in a single instant there is no movement and the time is not changed; so no velocity.  I can consider $\lim_{t_1 \to t_0} \dfrac{f(t_1)-f(t_0)}{t_1-t_0}$, but mathematically it is only the limit of average velocity function and doesn't represent velocity at instant $t_0$  What are your views about this problem ?
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Suppose we had a spring-mass system where the spring isn't assumed to be massless (has mass $M$) and is of length $L$. One end of the spring is held fixed and the other end I guess is left to freely oscillate. Here, I am told that the spring is assumed to uniform and stretches uniformly.  If I want to find the kinetic energy of the spring, we have to set up an expression for it  $$dT_{\text{spring}} = \frac{1}{2}u^{2} dm$$  where $dT_{\text{spring}}$ is the kinetic energy of an infinitesimal part $dm$ somewhere along the spring and $u$ is its corresponding velocity. Since the spring is uniform, I can find its mass density  $$ \lambda = \frac{dm}{dx} \longrightarrow dm = \lambda dx = \frac{M}{L}dx$$  so that $$dT_{\text{spring}} = \frac{1}{2}u^{2}\frac{M}{L}dx $$  The one step I am not understanding is how $u= \frac{x}{L}v$, where $v$ is, I think, the velocity of some point that has been displaced by the stretching of the spring (please correct me here if I'm wrong). Why is the velocity of a piece $dm$ linearly proportional to $v$ and how can I derive that expression mathematically, i.e. if $u = \alpha v$, how do I find $\alpha$ and why? Something is not registering in my head and I feel like it has to do with the fact that the spring is assumed to be uniform. That then begs the question: what if it wasn't? What would I do in that case?
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We have two inertial coordinate systems, $K'$ and $K$. $K$ is moving with infinitesimal velocity ${\epsilon}$ relative to $K'$. Using Galilean relativity we can transform this into $v'=v+{\epsilon}$. Here Lagrangian is only a function of speed $L=L(|v|^2)$. $L'$ may differ from $L$ only by a total derivative of function of coordinates and time, i.e. $L'=L+\frac{d}{dt}f(q,t)$  $"$We have $L'=L(|v|^2+2{v}·{\epsilon}+{\epsilon}^2)$. [1. ] Expanding this expression in powers of ${\epsilon}$ and neglecting terms above the first order, we obtain $L(|v'|^2)=L(|v|^2)+\frac{\partial{L}}{\partial{|v|^2}}2v·{\epsilon}$ [2.] The second term on the right of this equation is a total time derivative only if it is a linear function of the velocity $v$ $"$  I understand nothing in the quoted part. How do we get [2.]? Why does $\frac{\partial{L}}{\partial{|v|^2}}2v·{\epsilon}$ being a linear function of velocity imply that it's a total time derivative?
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Good evening, I'm trying to calculate what kind of impact force a falling object would have once it hit something. This is my attempt so far:  Because $x= \frac{1}{2} at^2$, $t=\sqrt{2x/a}$ $v=at$, therefore $v=a \sqrt{2x/a}$ $E_k=\frac{1}{2} mv^2$, so $E_k=\frac{1}{2} m(2ax)=m \cdot a \cdot x$ Since $W=E_k=F_i s$, $F_i=E_k/s=(m \cdot a \cdot x)/s$  For an object weighing about as much as an apple, $0.182$ kg, falling $2.00$ m straight down and creating a dent of $0.00500$ m, this would result in:  $$F_i=(m \cdot a \cdot x)/s$$  $$F_i=(0.182 \cdot 9.81 \cdot 2.00)/0.00500=706 \, \text{N}$$  Does this make any sense? I wouldn't be surprised at all to find out I'm missing something here.   Any input would be appreciated,   thanks in advance!
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I am trying to calculate the minimum stopping distance of a car once the brakes are applied. I know that $F = ma$, and the braking force is $F = \mu N = \mu m g$, so $$a = \mu g.$$ Next, by applying the kinematics equation $$v_f^2 - v_i^2 = 2 a x$$ I found $$v^2 = 2 a d$$ where $a$ is the deceleration, $v$ is the initial velocity, and $d$ is the stopping distance. Then $$d = \frac{v^2}{2a} = \frac{v^2}{2 \mu g}.$$ Is this reasoning right?
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Let us consider a homogeneous rope hanging from the ceiling. I will call the vertical direction $x$ and the horizontal displacement $y$. When we apply the second Newton's Law to a portion of mass $\Delta m$ and proceed in the same way we do for a  we get $$\mu\frac{\partial^2 y(x,t)}{\partial t^2}=\frac{\partial}{\partial x}\left[T(x)\frac{\partial y(x,t)}{\partial x}\right].$$ The difference now is that the tension is not constant. Defining $x=0$ at the free end of the rope and orienting it upwards we have the tension $T=\mu gx$. Therefore, $$\frac{\partial^2 y}{\partial t^2}=gx\frac{\partial^2y}{dx^2}-g\frac{\partial y}{\partial x}.$$  So, my issue is that I have seen in a couple of physics books () that the wave speed of this rope is simply  $$v=\sqrt{\frac{T}{\mu}}=\sqrt{gx}.$$ In my opinion these books are either:  i) Cheating the students: they know it is wrong but assume it is right just to show some nice features (the speed would increase as the wave goes upwards).  ii) Had misconception assuming that the usual wave equation are still valid.  iii) Is doing some obscure approximation, which is another cheat if you do not reveal it.  My questions are:  1) Is there an expression for the wave speed for the "wave equation" above?  2) Is there an approximation leading to the wave speed given above?
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I have a document I am trying to copy to learn TeX. Here is what I have encountered: How can I have the "or" in the equation. This is what I have right now:  \[ f = ma; \]  But $a$ is the change in velocity,  \[ f = m \frac{dv}{dt};\] \[ f = m  \frac{d^2y}{dt^2};\]   Edit: Trying out Stefan's answer
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Can anyone please give me a proof for $k/m=w^2$ in simple harmonic motion?  I have tried energy conservation and Newton's laws as follows :  In the case of a mass-spring system,  $$F=ma =-kx\\ F=ma = mr\omega^2 $$  hence   $$\frac km  = \omega^2$$  Or   $$\frac12 mv^2 = \frac12kx^2$$  $$\frac km  =  \frac{V^2}{x^2}   =   \omega^2$$  therefore $k/m  =  \omega^2$  Are these valid and correct? I have been stuck in this for the  entire day.
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In my Computer science class I was given a problem where I have to simulate a bouncing ball using "real physics". I have been trying to find a equation that will simulate the height of the bounce given a gravity and an arbitrary mass. And I will need to calculate the next bounce and it's height. A lot of the equations I've found require a time. But I don't really care about a time, all I want is to get the next height after the previous bounce until it finally hits height of 0 or close to it. I haven't taken a physics class since high school( 5 years ago) and that was basic physics.