55

You are indeed correct that the point where the coin hits the train must momentarily stop because of the change of direction. However, this does not mean that the train will stop! It will slightly decrease its speed due to the kinetic energy of the coin, but this will be unmeasurably small. What will happen is that the point of contact will stop (kinda, see ...


41

What does change the velocity of an object? Force of course. (Acceleration is the change of velocity). The question is how much time does it take to make that change? A long time? Infinitely small time? Momentarily? Well, it depends on the acceleration itself! If you accelerate in your car, it'd take a significant amount of time to go from $0$ to $100\ \...


15

I think this is a great question. I'm not 100% sure this is a complete correct answer, but I'm at least well over 50% sure. Here are the paths of the penny (in green) and the train (in red): First observation: The penny might never have velocity zero; at time $T$ it can jump discontinuously from a positive to a negative velocity. Second and more ...


9

It looks like there's a hole in your reasoning. If I understand correctly, you're saying this: At some point, the penny's velocity must be zero. At some point, the penny's velocity must be equal to the train's velocity. Therefore, at some point, the train's velocity must be zero. Both 1 and 2 are true (assuming that the collision is exactly head-on). But 3 ...


7

Two objects in contact have the same velocity No, not always. Take two balls and make one roll faster than the other and make them collide. They do collide with different velocities! But, in your case, if the coin hits the train and exerts a force capable of accelerating it in the opposite direction of its movement until its velocity becomes zero then yeah,...


5

For an object rotating about an axis, every point on the object has the same angular velocity. The tangential velocity of any point is proportional to its distance from the axis of rotation, i.e. $\mathbf{v_{\perp}} = \boldsymbol{\omega} \times \mathbf{r}.$ So it depends what you mean by "the velocity of the string."


4

We are interested whether a given force $$ {\bf F}~=~{\bf F}({\bf r},{\bf v},{\bf a},t) \tag{1}$$ has a velocity-dependent potential $$U~=~U({\bf r},{\bf v},t),\tag{2}$$ which by definition means that $$ {\bf F}~\stackrel{?}{=}~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}. \tag{3} $$ If we define the potential ...


4

For different points of the string to have same velocities, their direction of movement and speeds would to be the same as well. If the string is taut then all points are moving in the same direction at any given time, but it's easy to show that the speeds are not the same: If you have your 2 meter long string, the endpoint will move the total distance of 2·...


4

Constant velocity and speed have no meaning unless you specify the frame of reference with respect to which it is measured or observed. (The only exception is the speed of light). If you see an object “moving”, then it is moving with respect to your frame of reference. If you see it as “still” it is still with respect to your frame of reference. But in ...


4

This is actually in interesting question, and definitely not worth all of the down votes in my opinion. You're actually asking a different question than you realize. It's not like people were sitting around saying "We need a suitable definition for velocity. What should it be?" This doesn't make sense. Without a definition how would anyone know to even ...


4

It depends. In principle it could, in any realistic situation it is not even close. As you said, it comes down to the fundamentals. When you throw a coin at a train, the train will exert some force on the coin via electrostatic repulsion. On the atomic level this is due to the electrons in the coin getting too close to the electrons in the train and ...


3

The Maxwell-Boltzmann graph represents just this. $\hskip2in$ Source: (wewwchemistry.com) It can represent different temperatures too. You see a probability density (simplified here to # of molecules) against speed or kinetic energy. The area underneath gives the proportion of molecules at a speed range. With altitude changes, the pressure would change. ...


3

Your example of pushing a car at a constant velocity is doing work, because you are applying force on the car. The difference is that in that case, there is a friction force that also does work, in direct opposition to the work provided by pushing. The work done on the car by you is counteracted by the work done on the car due to friction, and thus net ...


3

The four-velocity is the tangent four-vector of a timelike world line, that is $U^\mu = dx^\mu / d\tau$, and this definition applies to any coordinate system. The difference of the coordinates $dx^\mu$ is a vector, i.e. an invariant, and together with the proper time $\tau$, an invariant by definition, produces the four-velocity $U^\mu$, which is an ...


2

You cannot tell the difference. In fact many would say there is no difference. If you're really in a universe where there's nothing except the object like you say, there would be no difference between just sitting still and moving at a constant speed. There are no stars to see whizzing by, no atmosphere to make you feel the wind, nothing to make the light so ...


2

Yes, integrating with respect to one $v_k$ will give a right-hand side that depends on every $v_l$ with $l\neq k$. But notice we have one equation for every possible value of $k$, and in the end, the left hand side of the equation is always the same, and so should be the right hand side. Thus, we conclude that the function at the right hand side cannot ...


2

The principle that two objects in contact have the same velocity is only true if they are in contact for a period of time. In the simplest modeling of the situation, the coin and the train are only in contact for an instant, where an impulse is transferred to the coin to send it clattering out of the way. The rule that two objects in contact have the same ...


1

Model the train as a tiny mass (the part that the coin is going to hit) connected to a great mass (the rest of the train) by a spring (representing the elasticity of the metal that’s hit). In this model, the tiny mass will indeed equalize its velocity with that of the coin, at some point being at rest relative to the rails, before elasticity causes the coin ...


1

This may be easier to understand assuming that the train and the coin interact during some short and measurable time via forces, caused by elastic (spring-like) deformation of the metal surfaces involved. In other words, imagine there is a tiny short spring between the coin and the train. Both train and coin are subject to the same interaction force so ...


1

Depending on the 'type' of velocity. • Angular velocity will be constant for every particle / unit length of the string i.e. 5/2 revolutions s-1 • Linear velocity will change according to the distance from the point of rotation V= r. w, where V is the linear velocity, r is the length of the string from the point of rotation and w is the angular velocity


1

To simply put, Kinetic energy can be calculated by the basic process of computing the work (W) that is done by a force (F). If the body has a mass of m that was pushed for a distance of d on a surface with a force that’s parallel to it. $W=F.d=m.a.d$ The acceleration in this equation can be substituted by the initial $(v_i)$ and final $(v_f)$ velocity and ...


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