New answers tagged

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"So trying to understand Momentum properly in terms of what kind of velocity we are talking about." There is only one kind of velocity. It is the instantaneous rate of change of some distance measurement. In each time frame, every particle in a system has defined properties of position and velocity. This described the kinematics of the system. A particle ...


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If the objects are accelerating they must have an external force acting on them and hence an impulse $I_{\rm external}$ acts for a time $\Delta t$. At the same time there are internal force acting on the bodies as a result of the collision. These result in an impulse $I_{\rm collision}$ acts for the same time $\Delta t$. Now it becomes a matter of ...


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One measure of the elasticity of a collision is the coefficient of restitution, which is given by $$e=\sqrt \frac{KE_{after}}{KE_{before}}$$ Where e ranges from 0 to 1. $e$=1 for a perfectly elastic collision and e=0 for a perfectly inelastic collision. If e is a constant then increasing the initial velocity should not change the elasticity of the ...


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Yes, the velocity can vary. This is the power of the principle. In Newtonian mechanics the idea is that if there is a cloud of particles all with forces between them and so all accelerating, nevertheless the total momentum of the cloud is unchanged (assuming no external forces). One way to think of a force is that it is the transfer of momentum between two ...


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No, the elastic collision equations can be solved analytically for any starting speed. The collision becomes inelastic when you are also imposing conditions at the end of the collision, such as the two moving with the same speed, etc.


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A simple example that shows why E is the correct option is the following: Consider a particle moving along x-axis with negative velocity. Now since the acceleration is positive, the speed initially decreases to zero and then increases again. A. Isn’t true as our initial velocity is negative. B. Isn’t true as we are initially slowing down. C. Isn’t ...


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No the $E$ option is correct one. Also $a=v/t$ is true only if initial velocity is zero otherwise the correct equation to consider is $$ a =\frac { \Delta v }{\Delta t} = \frac {v_f-v_i}{ t} $$ Here if $ \Delta v \gt 0 \Rightarrow a>0$.


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Two examples should illustrate why answer "E" is the correct answer. When you work a physics problem, you get to decide which direction is positive. Accordingly, when you are driving down the road, it is valid to state that the direction that is in front of your car is the positive direction. Example 1: You are starting from the "x=0" position, and the ...


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The formulas for the "midframe" are used in Loedel diagrams https://en.wikipedia.org/wiki/Minkowski_diagram#Loedel_diagram The corresponding formulas were derived from the Lorentz transformation by Mirimanoff in 1921: The Lorentz-Einstein transformation and the universal time of Ed. Guillaume and from the velocity addition formulas by Shadowitz: The ...


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This is how Rindler proves it: "For proof, consider a one-parameter family of inertial frames moving collinearly with S and S', the parameter being the velocity with respect to S. It is then obvious from continuity that there must be one member of this family with the required property (see Fig 2.4)." So we have the two frames S and S', and between them we ...


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When there isn't wind blowing, all the air molecules are moving in random directions, but with no average motion. For example, the $x$-components of velocity might be mostly within $-100$ to $100$ m/s. When air molecules of this kind hit you, they pick up energy, but they only slowly carry it away because each individual molecule doesn't go anywhere on ...


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These other answers are correct, but there is another way to think of it I want to share. Blowing wind causes forced convection. What this means is that the energy is carried away by the wind. The faster the wind blows, the faster it is able to carry that energy away from you. Since the heat loss depends on the temperature difference, the faster the wind ...


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While I am not sure of what "kinetic energy separate from velocity" means$^*$ (I will wait for clarification), there seems to be some confusion as to why wind makes one feel colder, even though the air is moving faster relative to your skin than still air would be. First, if we are stationary in a location with no wind and air that is at a temperature that ...


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Your science teacher was wrong. Wind feels colder to your skin on a hot day because it evaporates your sweat faster, carrying away more heat from your skin than can conduction to still air at the same temperature. Kinetic energy is equal to 1/2 * (mass) * (velocity^2), so the assertion that kinetic energy is not connected to velocity is also false. ...


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How do you measure mass at a single moment in time? You can put a mass on a spring scale and see how far the spring stretches, and compare that against known masses. You can apply a known force to it and see how fast it moves. However, mass is considered to be intrinsic to -- to things that have mass. Velocity is only velocity relative to something else....


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Average velocity is the total change in displacement divided by the time taken. It can only be defined over a time interval, so takes a finite interval to measure. E.g. if a ball travels $(3 \mathbf i + 2 \mathbf j)m$ in 2s, its average velocity is $(1.5 \mathbf i + \mathbf j)ms^{-1}$. Instantaneous velocity is the derivative of displacement with respect to ...


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$d\vec p$ is not actually a vector, but a differential volume element. It is a bit of sloppy notation because sometimes $d\vec p$ means the vector $(dp_x,dp_y,dp_z)^T$ and sometimes it is the volume element $dp_xdp_ydp_z$. You can also use $d^3p$ to denote a volume element but both are used often. A more proper way to derive $m^3$ is using the Jacobian. ...


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I am slowing down (decelerating) am I then accelerating backwards? The implication here is that the direction of the acceleration is in the opposite direction to the velocity. I think that you should avoid the word decelerate as it can be taken to mean a reduction in the speed of the car but this might not always be so. Suppose a car is moving at $\rm ...


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You experience a backward force when you hit the brakes, and so yes, you are accelerating backwards. Acceleration is a physical concept that is a vector. It also happens to be a word in English which loosely means ''to speed up''. Deceleration is an English word which is the opposite of acceleration in that sense. But that's clearly not the same as tacking ...


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Yes you are. A simple way to check this is to chose an inertial reference frame with the velocity the car has a at the instant. After doing this you would find that the direction of motion of the car is backwards. A thought experiment that you can perform is that you and your friend are moving in two different cars with the same velocity (wrt ground) and ...


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$$ v_f^2=v_o^2+2a(x_f-x_i) $$ Final velocity squared is equal to initial velocity squared plus 2 times acceleration times change in position (delta x).


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The equation of motion for such situation is: $$v_f^2 = v_i^2 + 2a(x_f-x_i)$$ Now for the concern you made in your comment: All equations I've found assumes a positive inital velocity :-( There are two things that I want to say: In the equation stated above you can see that direction of velocity (initial or final) has no use as they are squared. ...


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If you use the standard equations of motions then yes. You may substitute position, velocity or acceleration as negative and still get the right answer as long as you follow the sign convention.


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In general for relative motion problems at normal speeds you need two coordinate systems. (Assume all symbols represent vectors.) The position of a point in system one = r = R + r' where R is the position of the origin of system two as measured in system one, and r' is the position of the point in system two. (The primes indicate measurements in the ...


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I don't know what the discriminant rule is (I guess you mean the quadratic formula), but from \begin{align} 2E = m\left(k - r\omega\right)^2 + I \omega^2 \end{align} we can solve for $k$ directly: \begin{align} \left(k - r\omega\right)^2 = \frac{2E - I \omega^2}{m}\\ \rightarrow k = r\omega + \sqrt{\frac{2E - I \omega^2}{m}} \end{align} No quadratic formula ...


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All equations of motion can be solved using calculus. So the basics are as follows where $x$ is position, $v$ is velocity and $a$ is acceleration. $$v=\frac {dx}{dt}$$ $$a=\frac {dv}{dt}=\frac {d^2x}{dt^2}=v\frac {dv}{dx}$$ Observe- $$a(t) = C + Kv(t)$$ $$\frac{d\,v(t)}{dt} = C + Kv(t)$$ $$\frac{d\,v(t)}{C + Kv(t)} = dt$$ $$\int_0^t\frac{d\,v(t)}{C + Kv(t)} =...


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In classical mechanics the addition of velocities is based on Galilean transformations whereas in special relativity it’s based on the Lorentz transformations. Hope this helps.


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Yes. (by the way, don't watch to much discovery channel! It provides you with scientifically fake news, most of the time) Look at this picture: It contains all the info you need. Can you discover that info? Let's assume the body has the form of a ball. In the picture, one can read that the terminal velocity is the (in theory asymptotic) velocity when the ...


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When you draw the $x$ and $y$ axes you are really defining two directions and hence two unit vectors $\hat x$ and $\hat y$. A vector $\vec a$ of magnitude $3$ pointing in the $y$ direction you would write as $\vec a=3\,\hat y$. The magnitude is always a positive quantity. Now what about a vector $\vec b$ of magnitude $4$ pointing in the $(-x)$ direction ...


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Essentially, signs only belong to scalar values. A vector like $$\vec a=\begin{pmatrix}-1\\2\end{pmatrix}$$ has no sign in itself, only its coordinate values do. If you do see a sign on the vector, such as $$-\vec a=\begin{pmatrix}1\\-2\end{pmatrix}$$ then that corresponds roughly to "taking the sign out of the bracket", similar to what you would do in ...


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When we describe motion in one dimension we associate a sign with velocity, but in 2D, the perpendicular directions are independent of each other, so we need 2 numbers (corresponding with x and y direction). Here, with assumed x and y axes, you'll say velocity is positive in y direction and negative in x direction. It's like moving left and forward at the ...


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For the purposes of your question, acceleration is the process of going from one inertial frame to another. The differences you see between being in one frame and the other occur during acceleration. If you are 'looking at' events in a chosen inertial frame as you accelerate, you will see lengths changing - decreasing or increasing as the relative speed ...


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If they are massive objects travelling at $c$ then the Lorenz factor will diverge and they won't have a well defined energy. To find the energy at smaller but still relativistic speeds use $E=\gamma mc^2$ twice.


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Rapidity: $$ \omega = \cosh^{-1}{\gamma} = \cosh^{-1}{\frac E m}$$ is additive, so: $$ \omega = \omega_1 + \omega_2 $$ solves the problem. Note that thinking about what newtonian physics says is not helpful, nor is being imprecise with statements like "moving at about the speed of light". Generally people say "ultra-relativistic" for that concept. ...


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Well, let's look at the math: Clearly when $\vec{v} = 0$, then $\gamma =1$ and so we expect neither a shift in time or space. That's pretty self-explanatory, after all, it's just a stationary frame isn't it? What about when we start increasing $\vec{v}$? Well, $\frac{d\gamma}{d\vec{v}} = \frac{d}{d\vec{v}}\frac{1}{\sqrt{1-\frac{\vec{v}^2}{c^2}}} = - \frac{...


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What motion is? Well, it's a bit subjective matter. But to me it is more related with body trajectory in some reference frame


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