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Try with the following steps: Write down the equation of motion $x(t)$ and $y(t)$ for a generic particle shooted with angle $\beta+\alpha$; and with an initial position such that it start from the two lines as you wrote above. Given this trajectory $x(t)$ and $y(t)$, compute the tangent vector of the position as the function of the time $t$ (which is the ...


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It's not as simple as that. You'll have to obtain velocity and displacement by integrating your given acceleration and using correct boundary conditions. For example: Suppose the acceleration is given by A(t) = 2t [m/s²] and the problem states that the particle starts its movement from rest and from the origin of your coordinate system, so that X(t=0)=0 ...


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The velocity after a time t1 of accelerating is the starting velocity of the deceleration phase. Thus $$a t_1 = b t_2$$ (not worrying about the sign here. I suppose you could) Further you have $$\frac12at_1^2+\frac12bt_2^2=s$$ Now you have two equations with two unknowns. Solving: rearrange first equation $$ \frac{a}{b} = \frac{t_2}{t_1}\\ t_2=\frac ab ...


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The problem is in assuming the force is forwards at a contact point right underneath the axis of the wheel, together with the no-slip condition. This leads to zero friction force, since their is no friction when their is no slipping. A solution can only be found when deformation of the wheel is taken into account, allowing the force to act on a different ...


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$\int^t_0 A x^2 dt = x_0 + A x^2 t$ is incorrect. You are assuming $x$ as a constant. $x$ is a function of time x(t). Try $\dfrac{dx}{dt}=Ax^2 \implies \dfrac{dx}{x^2}=Adt$. Now integrate both the sides in appropriate limits. $$\int_{x_0}^{x(t)}\dfrac{dx}{x^2}=\int_0^t Adt$$ $$\int_{x_0}^{x(t)}x^{-2}dx=\int_0^t Adt$$ ...


3

First problem: you say $v(t) = A x^2$, but that is a function of position, not time. Putting the definition right: $$ v = \frac{dx}{dt} = A x^2 $$ You can regroup terms on the same variable: $$ \frac{dx}{x^2} = A dt$$ And then do the integration: $$ \int \frac{dx}{x^2} = \int A dt$$ This is homework, so I will leave the integral limits and the ...


-1

The inverse tangent of the coefficient of static friction is $17^{\circ}$.


5

Since the gravitational force only pulls the ball down, but not back or forth, it will not experience any acceleration changing its forward velocity but only downward acceleration. Thus, the ball will return to the thrower. You can also imagine the train to have no windows and be moving extremely smoothly. The thrower won't know if the train is moving or ...


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Unfortunately I cannot comment due to insufficient reputation, so here a comment on the question. There are three cases: $\frac{1}{2}mv_A^2>2mgR$ In this case the pearl has a velocity $v>0$ in the top point and will continue its movement. $\frac{1}{2}mv_A^2<2mgR$ In this case the pearl won't reach the top and will oscillate around point $A$. ...


2

The particle continues moving until it reaches the maximum height. At maximum height its velocity vanishes. In return, it again takes $T1$ to travel from $B$ to the ground. So it takes $\tfrac{T2 -T1}{2}$ for it to travel from B to the maximum height. $V_0$ denotes for initial velocity of particle on the ground and $H(B)$ the height of B, ...


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Lots of good answers here, but most of them are pretty mathematical and not very intuitive. Lets consider a realistic example. You're on the moon with a six shooter and some extra bullets. You are in a uniform field, and you make two point masses travel through the same distance by dropping a bullet with one hand and firing at the lunar surface with the ...


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It had a tangential velocity and its weight was enough to provide the centripetal force for motion.


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Work done is also defined as change in kinetic energy of the body. Since F is constant force so F/m=a is a constant acceleration of m. So, $$v^2-u^2=2ad$$or$$mv^2/2-mu^2/2=2mad/2$$which is the work done by the force. The body has travelled d distance with accleration a in the force field assuming u was a constant velocity when it entered the field and v is ...


1

Without any math and considering only Newtonian model here, I would say that if you move the inertial system at the same speed and direction as your mass point is moving, than you have no initial movement of the mass point and the total force used for acceleration will be the same as if you calculated or measured it in the original inertial system.


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Well, you simply need to accept that work is given by Force time Distance, and it doesn't matter how long it takes. For example, the work done on a mass $m$ lifted a distance $h$ against gravity with an acceleration $g$ is given by:$$W=F\times h=mgh$$ If you are told that someone is going to drop a $1$ kilogram mass on your head from a height of $10$ ...


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Well, the reason it doesn't matter is that work is defined as $$W = \int\vec{F}\cdot\mathrm{d}\vec{s}$$ so if you keep the force the same and the distance the same, this remains the same, regardless of what you do with the initial velocity. Of course, that definition probably isn't particularly satisfying. So consider this: when an object is subject to a ...


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As you describe, the definition of work is just: $W=F d$. What you are confusing maybe is the rate of work $P$ and the force $F$. When you move fast, $P=Fv$ is larger, however the travelling time is shorter. let's consider we are moving in a constant velocity. Then: $$W=Pt=Fvt=Fd$$ Independent of velocity.


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As you note, for a constant force acting on an object which moves in one direction, the work done is equal to $Fd$. One can see from the equation that work is not dependent on time, but only on force and displacement. In order to conceptualize this, you could think about the energy involved in the situation you describe. When work is applied by an external ...


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Consider two masses M and m in circular motion with same velocity,v. Both has acceleration v^2/R. The forces acting on the two masses are different. Force will become more on the greater mass. But acceleration of both are same. Because, if you put M and m in the following relation, you get same v^2/R. $$(mv^2/R)/m=v^2/R$$ since we know $$F/m=a$$ where ...


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You can think of acceleration from a purely mathematical context - it's the rate of change in velocity. (If you're familiar with calculus, you can say that the acceleration is the derivative of velocity.) Because of this, you don't need any mechanics to determine acceleration, so the mass is irrelevant. More concretely, a mathematician could calculate the ...


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The definition of acceleration is rate of change of velocity. If you know velocity as a function of time, then you know acceleration. No information concerning mass is required.


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Mass and acceleration are two independent variables. You need to consider them together to arrive at a force with the direct relationship F=ma. In other words, if it's mass doubles, so does the force and the acceleration is the same. Gravity works exactly this way. If you are considering a fixed force from something other than gravity (a force of constant ...


-1

The criteria is $$ \mbox{center to center distance} \le \mbox{radius 1} + \mbox{radius 2} $$ BAM!


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I think you have too many parameters, and not all of the necessary ones To simplify your thinking: Change to a frame of reference in which one billiard ball is initially at rest at the origin, and the second is moving at velocity $V$ from right to left along the straight line $$y=k, \,k>=0$$ A collision will take place if and only if $k<2\sigma$. ...


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It depend on which "many relationships" you are referring to. A better model would be a Kepler orbit with an eccentricity ($\epsilon$, a measure of how elliptical an orbit is) smaller than one. This model describes a two body problem. But the downside of this model is that it can not explicitly be expressed as a function of time. Namely you would have to ...


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The two relations that you give, are strictly valid only for circulation motion. You give an example of circular motion with variable radius, but, that is not circular motion anymore. Also, how would you define the centre of the motion, and thus angular velocity or cetripetal accelaration? If you have a well-defined case, you can always derive similar ...


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Take for example the equation for the velocity: $$ v = r\omega $$ In circular motion the velocity is always normal to the radius. In non-circular motion, e.g. an elliptical or hyperbolic orbit, the velocity is not normal to the line joining the object to the focus of the orbit. However the velocity can be expressed as a vector sum of a normal, $r_\perp$, ...


3

There is one formula relating the speeds of any two "platforms" (say $P$ and $Q$) between each other: $$V_{P}[ Q ] = V_{Q}[ P ].$$ And there's of course the well known symbol for "speed of light (in vacuum)", as determined of light signals exchanged by members of any one platform between each other: $c$. The speed of any one platform ($Q$) as determined ...


1

There are two formulas for adding velocities. The first is typically called Galilean relativity and the second special relativity. The first is simple, if you stack your tank on top of a train then the speed of the shell is the sum of the velocities, $v_1+v_2$. These things can be added as much as you like. You have an aircraft carrier moving at $v_1$ ...


2

Energy-momentum conservation is a stronger statement than the statement* that the inner product $p_\mu p'^\mu$ is conserved. It states that the sums are conserved individually/coordinatewise - $P_1+P_2=P_1'+P_2'$. As I see it, the conservation of the inner product is a statement about change in reference frames, whereas the conservation of energy and ...


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See page 3 where $\Omega(t)$ is defined by: $$ \frac{dpos(t)}{dt} = \frac{\Omega(t) r}{K_r} $$ so the linear velocity of the vehicle in km/h is given by: $$ V(t) = \frac{3.6 r}{K_r} \Omega(t) \tag{1} $$ At the bottom of page 16 the constraint is introduced: $$ \Omega(t) \le \frac{K_r}{3.6 r} 50 $$ Comparing this with equation (1) shows that the number ...


0

I'm not sure if it would necessarilly lead to such an instability, as Joce says for a first order method you'll need a really small time step to maintain any accuracy, but at the moment your acceleration is very wrong. What you effectively have here is ${\bf a} = - \frac{GM}{r^2} {\bf r}$ when you want ${\bf a} = - \frac{GM}{r^2} {\bf \hat{r}}$ (or ${\bf ...


0

The time discretisation you have chosen is an explicit Euler scheme. In order for it to be stable, you need the time step to be low enough, see e.g. wikipedia. You could use an implicit method, or increase the order of the method, but in any case there will always be a numerical drift in the orbits, proportional to the numerical accuracy. This can be ...


1

If you were to draw out the velocity field, you would see that it is going outward. In fact if you were really astute, you would notice that your velocity has the same form as the electric field from a wire. This should help you visualize the field. Because of the symmetry, it makes sense to work in cylindrical coordinates. The position along the axis will ...


0

There is a pseudo force acting on $m$ because of the accelerating $M$ which will be equal to $$F'=ma$$ where $a$ is the acceleration of $M$. To find $a$, $$F=(M+m)a$$ Thus,$$F'=\frac{mF}{M+m}$$ But, there is a frictional force of $\mu mg$ on the opposite side and the net force is $0.25g$ on $m$ (which you already calculated). So,$$0.25g=F'-\mu mg$$ I think ...


1

Your equations of motion are: $$ \Delta y = v_{0}\sin{\theta} \Delta t - \frac{1}{2}g\Delta t ^2 \\ \Delta x = v_{0}\cos{\theta}\Delta t $$ You can rearrange to eliminate $\Delta t$: $$ \Delta t = \frac{\Delta x}{v_0 \cos{\theta}} $$ Substituting this into the $\Delta y$ equation yields: $$ \Delta y = \Delta x \tan{\theta} - \frac{g(\Delta ...


1

You have a differential equation that says \begin{equation} a(x) = -0.01*w = \frac{d w}{d t} \end{equation} What you did with the change of variables is correct, so $w$ cancels on either side. Otherwise you have a first order differential equation to solve.


1

Looks like the book forgot a term in the integral $$\int_0^t \frac{F_0}{M} e^{-b t'}dt' = -\frac{F_0}{Mb} \left. \left( e^{-bt'}\right)\right|_0^t = \frac{F_0}{Mb}\left( 1 - e^{-bt} \right) $$ For $t\rightarrow \infty$ this goes to $F_0/bM$ and for $t=0$ it goes to $0$.


0

Oke in order to understand this more intuitively lets calculate the following example. We have 3 rockets of 100 kg side by side floating in space. Rocket 1 and rocket 2 have the same speed and rocket 3 has a speed $v_{int}$ different from rocket 1 and 2. whe call rocket 1 our observer and where gone try to get rocket 2 to they same speed as rocket 3 In ...


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Can I suggest a different way of doing this? The first thing you should always do with experimental data is graph it to get an idea what it looks like. The graph of your data looks like: The blue dots show the data points, and they obviously lie on a straight line (give or take experimental error). So I used linear regression to fit a straight line to ...


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$$K.E.= \frac{1}{2} p v$$ since: $$K.E.=\frac{1}{2} m v^2$$ and $$p=mv$$



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