13
After the 1st hour, the first train would have moved 45 km, the second train will just start moving. Therefore, the distance between the two trains is 45 km.
In the next hour, the second train moves 60km, the first train moves only 45 km, so the second train would have caught up by 15 km. The distance between the two trains is now 30km.
Repeat this for ...
10
What you are dealing with here is typically called relative velocity. Honestly, I've never heard the term net velocity before and it's doesn't seem right to talk about several velocities "acting" on an object.
Rather, an object has just one single velocity. We can talk about its velocity vector components if we will, but that's all.
Relative ...
8
You can plug in $45$ km/h when using the relative speed. Relative speed is the rate the relative distance changes. Let's say the train that remains stationary in the relative frame is train $0$, and the other we call train $1$. Now go back to the original frame, and set the origin of time $t=0$ to be the time that train $0$ starts and train $1$ is $45$ km ...
7
Common usage often does not differentiate between terms which have precise scientific definitions.
Examples are mass & weight, gas & vapour, and the example that you have given, speed & velocity.
This can also spill over into the words used by scientist who will use the term weighing rather than massing although of course many scales compare ...
6
Speed is all the police officer cares about when stopping you for speeding. "He's going 100 km/hr! Stop him!"
But ask NASA whether they care about not just speed but also direction of near-going comets. "It's heading towards Canada with 3000 km/hr!"
Both versions are in use in technical work in different fields. It just depends on the ...
6
It does not. The diagram is incorrect, as it assumes that the path becomes a straight line after it reaches $y = 100 \text{m}$ from the ground. The path still remains a parabola, so the final angle($\theta_f$) is different from the initial angle at which the bullet was shot ($\theta_i$)
To solve the question, find the range $R$ for any arbitrary $\theta_i$ ...
6
Let's call the two points $A$ and $B$, and we'll take them to be a distance $dx$ apart as measured in your frame.
In your frame you measure the other spaceship to take a time $dt$ to get from $A$ to $B$, and since the other spaceship is travelling at a velocity $v$ relative to you that time is given by:
$$ dx = v~dt \tag{1} $$
The proper time, $d\tau$, for ...
5
The car, the person throwing the ball, and the ball --- these three constitute a whole system as long as they are moving together. And when is that? When the person sits on the car and keeps the ball in his hand, the three move together with the same velocity.
Now as soon as the person throws the ball, he no longer has contact with it. Hence, that person and ...
5
Alright I have a possible solution, and I would like to hear others weigh on it. Bottom line: this process via this vertex is not allowed.
With a vector coupling and a process $M\to m \gamma$, this violates the Ward Identity. i.e. $q_{\gamma}^{\mu} \cdot {\cal{M}}_{\mu}\not=0$ where $q_{\gamma}$ is the momentum of the outgoing photon and ${\cal{M}}_{\mu}$ is ...
4
The OP's answer is sound, but I will summarize the answer to his
Is there something inherently wrong with this diagram?
to refocus the question.
The answer to this is yes, the vertex supporting it is not gauge invariant, and fits into the broad rubric of SM-compatible interactions:
There are no flavor-changing neutral (& photon) currents!
Call the M ...
4
There are three useful frames of reference we could use to study this problem: one that holds the train station static, one that holds the first train (Train A) static, and one that holds the second train (Train B) static.
From the frame of reference where the station is static/fixed:
For the first hour, Train A moves at 45 km/h, while Train B moves at 0 ...
4
There is no paradox, because your 5 equations are not independent.
Fill in (1) in (3) to obtain (2)
(2)+(3) = 2*(4)
(5) is left as an exercise to the reader.
You also see this in your example. Combining three equations inevitably leads to one equation becoming $0=0$. Proving that the three equations are not independent.
So, you have only 2 equations. One ...
3
From the point of view of the bystander next to the track, the ball is stationary until it is hit, after which it goes at 400km/hr, so its change of speed arising from the collision is 400km/hr.
From the point of view of the passenger, the ball is coming towards them at 200km/hr before the collision, then rebounds at 200km/hr in the other direction after ...
3
I need a length in order to calculate this. I can't simply plug in the
45 km from above because that would be the time in which the second
train got to the 45km mark, but the first train would have moved away
from that point.
You just found here a very good way to solve the problem. But you're right, since the trains are constantly moving, you cannot just ...
3
The high school kinematics problems normally assume one or two dimensional travel on a Euclidean plane. The earth is NOT a Euclidean plane, leading to the zero displacement issue. I suggest you reformulate this problem for your high school student such that you restrict the automobile to a flat two dimensional surface and have the automobile travel in a ...
3
Consider a particle undergoing circular motion around an origin $O$ in a plane (2D). By the definition of circular motion, the position vector relative to $O$) is
$$
\vec{r} = r \cos (\omega t) \hat{x} + r \sin (\omega t) \hat{y} ,
$$
where $\omega$ is the angular velocity (the angle $\theta = \omega t$, analogous to $x = vt$ for rectilinear motion).
To get ...
3
A simple mathematical basis for determining relative quantities
Because all positions, velocities, and speeds are relative (i.e., there is no absolute frame that we know about), we need a structured approach for describing these quantities. We can use vectors to describe these quantities. Without getting into formal linear algebra, let's first consider ...
3
Note that, deliberately, this is not a complete answer to the question asked.
You are treating $\cos \theta$ as a constant. It isn't. What is constant (or almost constant) is the horizontal separation (call it $2X$) of the tops of the strings. Then if $y$ is the vertical distance of the point where the strings meet below the line joining the tops of the ...
2
When one says, The car is moving with $200\ \text{km/sec}$. That means, they are only concerned with the magnitude of the velocity that is speed.
Whether it's important to specify the direction is a matter of problem. In some cases, the direction plays a role, and in some, it doesn't.
2
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\newcommand{\e}{\bl=}
\newcommand{\p}{\bl+}
\newcommand{\m}{\bl-}
\newcommand{\gr}{\bl>}
\newcommand{\les}{\bl<}
\newcommand{\greq}{\bl\ge}
\newcommand{\leseq}{\bl\le}
\newcommand{\plr}[1]{\left(#1\right)}
\newcommand{\blr}[1]{\left[#1\right]}
\newcommand{\lara}[1]{\langle#1\rangle}
\newcommand{\lav}[1]{\langle#1|}...
2
If the car is constantly accelerating after throwing the ball up, then the ball will retain the horizontal velocity which it gained at the moment it was thrown up(due to N2L) and if the car accelerates,
the ball will move backwards with respect to the guy in the car,
and for the guy on the ground, the velocity will be the same as it did when the ball was ...
2
I'm not sure I really understand your train example, but the principle of relativity specifically states that the laws of physics in your local neighborhood do not depend on your state of motion. As a consequence of this, certain things are invariant with respect to who is observing them (i.e. all observers will agree on): the speed of light, the spacetime ...
2
There is a mechanism between each set of 'leaf pairs'
that allows the pair above to rotate slightly further than the one underneath.
The image was from this youtube clip 'How to make a Helicone'.
2
At small velocities $v\ll pT$, your model becomes the model of viscous drag (or we may say that it generally has a viscous term):
$$
\dot v = -T^{-1}v.
$$
The solution of this model is an exponential decay:
$$
v = v_0 e^{-t/T}
$$
You can easily see that the question when $v$ completely vanishes makes little sense in this model. Usually, we either speak of ...
2
The relationships $s = s_{initial} + ut + {1 \over 2} a t^2$ and $v = u + at$ assume constant acceleration. (You left out the initial position $s_{initial}$ in your relationship.) Your results $a = {2s \over t^2}$ and $v = {2s \over t}$ are correct, assuming the initial position $s_{initial}$ and initial velocity $u$ are both zero.
$a$ is constant. But $v$...
2
You cannot use vector addition for velocity like that. Imagine two horses running in the same direction at the speed $v$ pulling the same wagon. Does the wagon move at speed $v$ or $2v$? In your example, vector addition would work for forces but not velocity. Moreover, you have to take into account that $\theta$ also changes with time.
Start from the right ...
2
the "work" that done due to the tension force $~T$ is
$$T\,\cos(\theta)\,\delta x=T\,\delta u\quad \Rightarrow\\
\delta x=\frac{\delta u}{\cos(\theta)}$$
1
How relativity works in your example:
The situation ("nature") is absolute: when describing what happens either from the perspective of a passenger or that of a bystander, we do not get the feeling that they live in two different worlds - intuitively, it is clear that we just look to the same event (a train hitting a ball) and that no observer ...
1
Assuming the blocks are elastic, when the lighter block hits the heavier one, the two momentarily compress then uncompress and come apart. It is that interaction that transfers momentum and energy from the lighter block to the heavier one.
The situation you describe in which the velocity of the lighter block exceeds the velocity of the heavier one is true up ...
1
Yes your predictions are absolutely right. The ground observer seeing the ball thrown upwards in a car would see the velocity of the ball to be moving with $(10 \hat{i} + 3 \hat{j}) \frac{m}{s}$ and a person in the car will be seeing it as $3 \hat{j} \frac{m}{s}$. Yes if the car was moving at uniform velocity, the observer would be still seeing it as $(10 \...
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