New answers tagged

1

Time dilation is best thought of an apparent effect, an illusion caused by mixing a historical or classical view of time with a fully relativistic one. Historically people have thought of time as a global concept, formalised by Newton with his notions of Absolute Space and Absolute Time. But, strictly, this is an assumption useful in Newtonian mechanics ...


0

Consider a projectile horizontally launched at some vertical height. Now, consider another identical projectile dropped from the same height. Simple Kinematics tells us that the time of the projectiles in the air will be the same. The horizontal motion slows down as the vertical speed increases. If the projectile is a pulse of light, then always: $$ \sqrt{...


1

One of the postulates of special relativity is that the speed of light is the same in every intertial reference frame. For this to be true, either lengths have to contract or time has to slow down, or both. If you assume that the speed of light is affected by the speed of the inertial system, then this will conflict with the postulate. From Symon, Mechanics,...


-1

As the marble gains the velocity of 8m/s only after 1sec, it can't cover distance of 8m in first second.


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My question is: What factors will determine or contribute to the amount of kinetic energy lost during collision of these two bodies? I believe the primary factor is the degree to which material deformation results in internal dissipative (heat producing) forces. Some kinetic energy is lost in the form of sound (unless the objects collide in a vacuum) due to ...


1

Suppose the two objects are ball bearing assemblies, with the inner race wider than the rest, so that only that inner races touch the frictionless surface. If they have only translational movement before the collision, they start to spin after it (unless their path is perfectly radial). So part of the energy is rotational after the collision. The principle ...


2

Kinetic energy losses will be determined by a material property known as the coefficient of restitution. This number is equal to 1 for a perfectly elastic material having no internal friction losses which dissipate the kinetic energy, and is less than 1 for materials which are dissipative. The classic example of a dissipative material is one exhibiting ...


2

I think what you mean is "How to quantify the energy lost during an inelastic collision?" So to begin with, if you consider a 2 body system, during collision all the forces are internal. So you can conserve the momentum. And depending on the nature of the surfaces on which collision occurs, each material has its unique "Coefficient of ...


1

It cannot be done with a mathematical model, because swing bowling (quite distinct from spin, btw) depends on the layer of turbulence adhering to the edge of the ball, and we cannot accurately give a mathematical model of turbulence. However, the basic physical principles can be explained. First, consider golf balls. Golf balls are dimpled because it was ...


0

Providing “swing” refers to spinning of the ball, the explanation you are looking for is the Magnus effect. A spinning object moving through a fluid (in this case a ball moving through air) will deviate from a parabolic path: a ball spinning clockwise as viewed by the bowler will curve to the right, and the converse for an anticlockwise spin. There is also ...


2

First, note that this is not a question which is fundamentally electromagnetic in nature, so you could gain some intuition by imagining a block sliding down a rough inclined plane. Imagine what the motion of the charge would have looked like in the absence of the electrostatic attraction from the positive charge. After you provided some kinetic energy in ...


5

Dimensionally-inconsistent equations like this are intended to be understood in a dimensionless way in some system of units, such as $$\frac{x}{1\text{ m}}= \left(\frac{ t}{1\text{ s}}\right)^2 -2\left( \frac{t}{1\text{ s}}\right) -3. $$ In this case, the the unit of length is the meter and the unit of time is the second, as stated in the problem.


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The phrase "[...] where $x$ is in meters and $t$ is in seconds" means that $x$ is a pure number which is equal to the position coordinate of your object divided by one meter. The same goes for the time. That is, you can think of your equation as being $$\left(\frac{x}{1\text{ m}}\right)=\left(\frac{t}{1\text{ s}}\right)^2 - 2\left(\frac{t}{1\text{ ...


0

I disagree with part of the answer given by trula. I agree that the time taken in the laboratory frame will be t = 100m / .9c But, the time in the particle frame would be t' = t / γ


-1

I suppose the 100m is in the laboratory System not in the system of the particle. then in the laboratory id takes the time t=100m/0.9c and its own time would be $$t'=\gamma*t$$


1

$\frac{\vec v_{CM}}{c}=\frac{ \sum \vec p_{i}c }{ \sum E_i }=\frac{\vec P_{SYS}c}{E_{SYS}}$. On an energy-momentum diagram, this says add up all of the 4-momenta to get the 4-momentum of the system. The spatial velocity of the system 4-momentum is essentially the slope of that 4-momentum vector: the ratio of the spatial-components (the vector sum of ...


-1

Since they arrive home $10$ minutes earlier than usual, then Mr B must have met Mr A $5$ minutes earlier than usual i.e. at $5$ minutes to $3$. From this you can work out how long Mr A had been walking.


0

Is your goal simply to write down an equation that relates $v_x$ and $v_y$, without reference to $x$, $y$, or $t$? If so: Such an equation can’t exist, for the simple reason that $v_x$ is a constant while $v_y$ varies over the trajectory. If you found some relationship $f(v_x, v_y)=0$, you could differentiate it with respect to $t$: $$ 0 = \frac{d}{dt} f(v_x,...


2

You are right to question this. Technically the system consisting of the coconut and the firecracker form a closed system$^*$, so that the entire momentum of this system is then conserved (assuming nothing flies into the air of course). If you just considered the coconut as the system, then the force from the explosion is external (since the explosion is not ...


0

You must use the relationship between v, u , a and s (since you are given the distance travelled, final speed and the initial speed) i.e. v²=u²+2as


1

Since the cloud is given to be homogeneous, its density has to be uniform throughout, at any point in time. Let $\rho_1$ be the density of the outermost shell and $\rho_2$ be the density of an arbitrary shell of radius $R$ $\rho_1 =\rho_2$ $\therefore \frac {V_1}{m_1}=\frac {V_2}{m_2}$ $\therefore \frac 1{m_1} \frac {dV_1}{dt} =\frac 1{m_2} \frac {dV_2}{dt}$ ...


0

Your CM is revolving around a given axis in a circular motion, but each particle isn't just doing a circular motion, so you can't apply that formula for its velocity. The particles are moving around the CM in some arbitrary fashion. The velocity of each particle, as seen by your observer $O$, will be $\vec{v}_i=\vec{V}^{}_{CM}+\vec{v}_i'$, where $\vec{V}^{}_{...


0

How exactly is this time dependent? The problem has been formulated so that the comet is a point particle moving through the gravitational field produced by the Sun and by Jupiter. We can work in the center of mass frame of the Sun, so the sun isn't moving, and therefore the gravitational potential of the Sun is time independent. However, in this frame, ...


0

This is a great question! The limit on the right-hand side is from $0$ to $t$ because that is the interval of time we are considering (It is defined by the problem). In that interval of time, the speed changed from 0 to $u_{x}$, hence the limits on the left-hand side. Good luck with these things!!


2

The slowing-down of your car is a function of all the resistive forces acting upon it: that is, all forces that are trying to dissipate the car's kinetic energy relative to the surface of the Earth. Hence, depending on the exact nature of those forces, the slow-down time, and the deceleration profile, can vary. That said, we can nonetheless come up with some ...


0

Think of the limits as $u_f$ and $u_i$ on the left and $t_f$ and $t_i$ for the right... since you are integrating over $u$ and $t$, respectively. Thinking about units would also be helpful. What should the units of the right-hand-side be? How would you handle ordinary kinematics with constant acceleration $dv=a\ dt$?


0

I see this equation $$x_{\text{final}} = x_{\text{initial}} + v_{\text{initial}} \, t + \frac{1}{2}\, a \,t^2.$$ Excellent! I want to find the free fall distance ... therefore setting $a \equiv g \approx 9.8~{\rm{m/s^2}}$ ... for 2 seconds [if] $v_{\text{initial}}$ is zero and initial position is zero. Alright; but note that $t$ in the above formula is ...


0

I think you meant to ask what it means that your object's position versus time function is second order. So to answer in general terms, when your position versus time has a first order relationship $ f(x) = vx + c $ or $ s(t) = vt + c $ where c is the starting position at $t=0$ it means the object is moving at constant speed $v$ away from starting position $...


1

It's pretty much straight-forward conclusion of vector and calculus algebra. Position vector is defined as : $$\mathbf r = x\, \mathbf{\hat i} + y\,\mathbf{\hat j} + z\,\mathbf{\hat k}$$ If you value picture, then this is position vector in 3D space : Speed is position vector derivative against time by definition, so : $$ \begin{align} \mathbf v &= \...


1

You have graphed position s as a function of time: $s(t) $in Physics notation. But you call time $t$ your independent variable $x$, and your $f$ refers to the mathematical shape of the function rather than the resulting Physical quantity position $s$, which is the dependent variable, a.k.a. $ y$ in maths. The derivative with respect to t would be the ...


0

The logic goes something like this: A position vector consists of 2 or 3 components expressed in a Cartesian coordinate system. The time derivative of each component results in the speed along each of the components Combined all the speeds define the velocity vector. $$ \boldsymbol{v} = \pmatrix{ \dot{x} \\ \dot{y} \\ \dot{z} } = \frac{\rm d}{{\rm d}t} \...


0

$\def\th{\theta} \def\ra{\rightarrow}$Suppose the ball is thrown from $(0,h)$ to $(d,H)$ under the influence of gravity in time $t$ and that the initial velocity is ${\bf v}_0 = (v_0\cos\th,v_0\sin\th)$. (In what follows we assume $d>0$ so $-\pi/2<\th<\pi/2$.) We have \begin{align*} d &= v_0 t\cos\th \\ H &= h + v_0 t\sin\th-\frac1 2 g t^2. \...


0

The velocity at the contact point I is: $$v_I=\omega_1\,a_1-\omega_2\,a_2$$ thus the components of the velocity in $R_0$ coordinate system are: $$\vec{v}_0=v_I\,\begin{bmatrix} \cos(\varphi)\\ \sin(\varphi) \end{bmatrix}$$ with $\varphi=\int\,\omega_3(t)\,dt=\omega_3\,t\quad $ if $\quad\omega_3=$constant


3

The article specifies the equation dealing with kinetic energy is looking at the relative kinetic energy. For a perfectly inelastic collision, the bodies are not moving relative to each other, so the relative kinetic energy is $0$. Thus there is no contradiction. To add more detail to this, the best thing to do is to work in the center of momentum frame, ...


0

When ${\omega}_1=0$, ${\omega}_2=(\frac{a_1}{a_2}+1){\omega}_3$ (the factor $1$ appears because ${\Sigma}_2$ makes one extra turn after having rolled one turn around ${\Sigma}_1$ in the $R_0$ coordinate base). So ${\omega}_3=\frac{{\omega}_2}{(\frac{a_1}{a_2}+1)}$. When ${\omega}_1\neq0$, we have to add this to ${\omega}_3$, so: ${\omega}_3=\frac{{\omega}_2}{...


1

One could solve Newton's equations with drag: $$m\ddot{\mathbf{r}} = \mathbf{F}_{drag}(\mathbf{v}) + \mathbf{F}_{gravity}.$$ The drag force is typically directed opposite to the velocity and proportional to its magnitude: $$\mathbf{F}_{drag}(\mathbf{v}) = -\frac{\mathbf{v}}{v}F_{drag}(v). $$ The simplest model with $F_{drag}(v) = \nu v$ is definitely ...


1

The answer is yes, traveling at constant speed is the unique way that minimizes travel time. I do hope there is an easier way to show this, but this is the best I can do at the moment. The problem is the following: find the function $f(x)$ that minimizes $$T = \int\limits_0^X{\frac{dt}{dx}dx} = \int\limits_0^X{\frac{1}{f(x)}dx},$$ subject to the constraint $$...


0

Using the Pythagorean theorem on distances is the wrong idea for finding the angle. With the data you have listed , you have a unique solution for $v_{y0}=\text{your }v_y$ and $v_{x0}= \text{your }V_x$. Then find the magnitude and angle of velocity from the components.


0

It seems you didn't notice that the enemy camps are right at the foot of cliff . Thus,in order to hit the camp,the catapult should at be at a such a distance that by the time the projectile reaches the edge of cliff it's horizontal component of velocity becomes zero. Because if there is even a small horizontal component , the huge hight will ensure that ...


0

From your reference, $\text{Fig.3}$, it can be seen that in fact $v(t)$ starts off negatively ($<0$). Which is as it should be because the vector points downward. If so, then $\frac{\text{d}v(t)}{\text{d}t}$ must be positive because the derivative of a $\cos$ is a $-\sin$. The acceleration vector points upward during the collision, so its scalar is ...


1

Velocity tells you how fast the position changes. Intuitively the velocity vector should have the following properties The direction of the vector should tell you the direction of movement The magnitude should tell you how fast the position is changing. Or to be more precise, the magnitude should tell you the speed. When you have two position vectors $\vec ...


1

It would be possible to answer this question in different ways. Let me try to use an argument which unifies the 1D and multidimensional cases. Among the possible ways to define the velocity at time $t$ in 1D is that number that, when multiplied by a small time $\Delta t$, will provide the best (linear) approximation of the position at time $t + \Delta t$. ...


1

To start understanding vectors intuitively, just think of them as mathematical objects that combine dimensions together. If you do understand why the velocity is the change of position with respect to time in one dimension, think of two separate velocities, one in the x-axis, and one in the y-axis. In time t, the given object moves a distance $v_x \times t$ ...


0

@Semoi's graph is correct for plotting velocity versus time. Acceleration is the change in velocity. Many a times like when in uniform circular motion the speed remains constant but due to the change in the direction of velocity the body is said to be accelerating. Same can be said for your case. Lets say that the body's initial velocity(at t= 0) was 5 m/s $\...


1

(1) "The v-t graph will show that the body is not accelerating" The body is accelerating because its velocity is changing even though its speed is constant. (2) "Now suppose a horizontal v-t graph is given. Is it possible to tell whether the object was accelerating or not?" I assume that $v$ is the body's speed (that is the magnitude of ...


3

You probably denote the absolute value of the velocity by $v_{net} = |\vec v|$. Why don't you plot each velocity component of the vector $(v_x, v_y)$ separately. This would describe the situation much better. Please note, unless you tell use the time interval during which the acceleration takes it is impossible to calc the acceleration. As the time interval ...


0

From the exercise, we can know that the equation of motion is $$\vec{r}=x\vec{i}+y\vec{j}.$$ Derivative with respect to $t$, we have $$\vec{v}=\frac{d\vec{r}}{dt}=v_x\vec{i}+2xv_x\vec{j}\quad(*)$$ where $$v_x=\frac{dx}{dt};\ v_y=\frac{dy}{dt}=\frac{dx^2}{dx}\frac{dx}{dt}=2xv_x.$$ Since the speed of the particle is a constant, we have $$ \sqrt{v_x^2+v_y^2}=...


0

In your first calculation, you found the average velocity. To find the acceleration you need the final velocity.


0

The derivation of $F=ma$ comes from $F=\frac{\mathrm{d}p}{\mathrm{d}t}$. So if mass of system changes we can say $$F\neq m\frac{\mathrm{d}v}{\mathrm{d}t}$$ because it's equal to $m\frac{\mathrm{d}v}{\mathrm{d}t}+v\frac{\mathrm{d}m}{\mathrm{d}t}$ according to $F=\frac{\mathrm{d}p}{\mathrm{d}t}$.


0

This is due to the fact that for a small interval of time the distance is approximately the same as displacement. Just like in a curve, when we zoom into it small change in position is actually the distance covered. When divided by the same time change "$dt$" the magnitudes of the rates of change of displacement and distance which are velocity and ...


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