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If the spring locks when the ball is at rest in the lab frame, then by the arguments you give it follows that the spring must not be compressed at all. This is indeed the case. As the ball slows down, the block begins to speed up. Eventually they are traveling at the same speed, at which point the spring has reached its maximum compression. As the spring ...


4

However, how does a person outside the car rationalize this phenomenon occurring? When the car initially accelerates, the person outside the non inertial frame of the car sees the pendulum bob as "staying in place" while the car accelerates forward, i.e., the bob does not accelerate with the car. The person explains it as due to the inertia of ...


4

The potential energy in a uniform gravitational field is: $$ U = mgh $$ "g" is Earth's average gravitational acceleration : $$ g = 9.80665\,{\rm m/s^2}\approx 10\,{\rm m/s^2}$$


3

If you have an external impulse, you don't have an isolated system, and you shouldn't be chuffed if you discover that energy has been exchanged with the surroundings via the momentum-changing interaction. Your particular example — a rod which receives the same impulse at the middle or at the end — is a famous example of an inelastic-collision experiment ...


3

I'm trying to figure out why they considered maximum static friction to act on the left block and then used that to find forces on the 4 kg block For $2kg$ : What is the magnitude of the external force acting on the $2kg$ block ? It is $10N$ and the maximum static friction that could act on it is just $8N$ , so we must consider this exact value of static ...


3

The problem is exactly how billiard balls work (in 1 dimension, it's a much easier game like that). The spring is replaced by the elasticity of the ball, which doesn't need to lock, because that's how billiards works: all the cue ball's E and p go to the target ball, leaving the cue ball at rest.


3

If the spring locks when the deformation is maximum the collision is not ellastic. At that point the relative velocity of the two objects is zero so they continue to move with the same velocity. This is a model of non-elastic collision. The collisions are ellastic when the spring does not lock and restore its potential energy into kinetic. For actual balls ...


3

The Work-KE theorem says the net work on a system is equal to its change in KE. But the net work includes work done by conservative and nonconservative forces, and the work done by conservative forces is, by definition, the negative of a change in PE: $W_{net} = \Delta KE = W_C + W_{NC}$ $=-\Delta PE + W_{NC}$, so $W_{NC} = \Delta (KE + PE) = \Delta E_{mech}...


2

The first law of motion states that anything that creates a change in the rectilinear motion of a particle is a force. This is, however, only true in an inertial reference frame; in fact, this can be thought of as defining an inertial reference frame. Temperature isn't a force, it's just a macroscopic phenomenon that has to do with microscopic forces. ...


2

The straw has a (surprising) amount of compressive axial strength, but very little strength away from the axis. In other words, it will bend easily if there are sideways forces. If you place it under moderate compression it is okay, but adding a small sideways force to it at the same time will cause it to buckle. You may not have much information about the ...


2

Remember the rule: only impulsive forces can balance impulsive forces. You know for sure that the collision force between the bodies is impulsive. Draw the free body diagrams of the two bodies during collision, and see which other normal forces are balancing the collision force. Those normal forces will be impulsive. In your example, the collision force on ...


2

The problem was twofold: I should have used norm2 instead of abs. So, this line: to_add=(rj-ri)*big_g*masses(i)*masses(j)/((abs(rj-ri))**3.0d0) should be changed to to_add=(rj-ri)*big_g*masses(i)*masses(j)/((norm2(rj-ri))**3.0d0) My data was messed up, so I tried switching to some one else's, but I forgot to change the units when I did. As such, it is ...


2

The idea that the force between two spherical bodies goes as $1/r^2$ is only valid outside of the bodies. Once you're inside the bodies, things are different. If the bodies are of uniform density, the "shell theorem" applies, and the force goes to zero as $r$ goes to zero. (It's not obvious that this would be true, but if you work out the math, ...


2

Impulse is defined as $F \Delta t$. So if you have a force $F$ which is applied to any object for a time $\Delta t$ you have an impulse regardless of the nature of the force. Whether it be gravitational or electrical or mechanical. Any force which causes a change in momentum and acts for a certain time can be termed an impulsive force. I think this term was ...


2

The different parts of the universe we inhabit are causally-connected in the sense that one part (a cause) can interact with another part (the effect), by a variety of means at the atomic level. The details of those physical interactions don't matter for the moment. What does matter here is that none of those interactions are instantaneous; they are all ...


2

There are infinitely many such shapes that work, even ones without anything like spherical symmetry at all. To give a simple example, recall the method of images for spheres in electromagnetism. When you move a point charge near a conducting sphere, it induces a non-symmetric charge distribution on the sphere. But the electrostatic field of that charge ...


2

This answer is trivial, but Keplerian orbits are assured as long as the gravitational fields are spherically symmetric and thus decay as $1/r^2$. There are an infinite number of mass distributions that can generate a given field. For example a cubic shell of side $l$ with a mass density $(l/2)/((l/2)^2+x^2+y^2)^{(3/2)}$ on each square surface will produce a ...


2

After the collision, the two blocks have the same velocity $v'$. Conservation of momentum: $mv = 2mv'$, so $v' = v/2$. Conservation of energy: $\frac{1}{2} m v^2 = \frac{1}{2}(2m)v'^2 + PE_{spring} = \frac{1}{4}mv^2 + PE_{spring}$ So, $PE_{spring} = \frac{1}{4}mv^2$. (Half the incident energy gets stored in the spring.)


1

You are conflating work with energy, but there is a slight difference. In fact: Work=change in Energy, i.e. U=ΔK. Work is the accumulated force magnitude over distance along a path, so that an element of work dU=<F, ds>=F.ds.cos(θ) , where ds=elemental displacement along trajectory curve S , F=applied Force along S. We integrate this and then call the ...


1

Such a collision is called superelastic. This can only occur if one or both of the colliding particles has some way to store internal energy. For example, imagine that you coated your billiard balls with a contact explosive that gave them an extra "boost" of energy when they collided. More realistically, some atomic physicists conduct experiments ...


1

I believe your confusion comes from thinking the ball is accelerating down the ramp. The horizontal acceleration is measured when the ball is in equilibrium with the parabolic block. The ball is actually stationary with respect to the block. Looking at the $m\ a=\text{$\Sigma $F}$ in the y direction: $m\ a_y=m\ g - n \cos (\theta )$ but in equilibrium $...


1

Imagine only the pulley moving up by a distance of $x$, while the string remains as it is. Now from both sides of the pulley, you'll see that $x$ length of the string has been freed. So, in order to make the string taut around the pulley, the block will have to move left by a distance of $2x$. It follows that if the block moves a distance $x$, the pulley ...


1

Multiply both sides of your equation by the mass and you have equality of the magnitude of the angular momentum at the perigee and the apogee. At these points the velocity is perpendicular to the position, so $|\mathbf{L}|=|\mathbf{r}\times m\mathbf{v}|=mvr$. In modern language, Kepler’s Second Law expresses the conservation of angular momentum, not the ...


1

I) Equations of motion Kinetic energy : $$T=\frac{m}{2}\left(\dot{x}^2_1+\dot{x}_2^2\right)$$ Potential energy $$U=\frac{k}{2}\left(x_1^2+(x_2-x_1)^2+x_2^2\right)$$ with Euler Langrage you get: $${\ddot x}_{{1}}+{\frac {2\,kx_{{1}}-kx_{{2}}}{m}}=0\tag 1$$ $${\ddot x}_{{2}}+{\frac {2\,kx_{{2}}-kx_{{1}}}{m}}=0\tag 2$$ II) Equations of motion: Normal mode In ...


1

You should notice that the other normal coordinate is implied to be fixed at zero while you consider the motion along the normal coordinate $n_1$. The normal coordinates of two particles (or blocks in this case) can generally be written as \begin{align} n_1 =& a_{11} x_1 + a_{12} x_2, \\ n_2 =& a_{21} x_1 + a_{22} x_2.\label{eq: n1n2}\tag{1} \end{...


1

The difference is in the coordinate system. You are doing your calculation with $x=0$ as the point where the string is in its natural length. The equilibrium position, in that case, is $(m+m_c/2) g/k$. However, your professor is taking that equilibrium position $x_o=(m+m_c/2) g/k$ as his origin $x' = 0$ in the shifted coordinate system. You can arrive at ...


1

As you showed, the maximum frictional forces are enough to counter the external forces, so the center of mass of the system is definitely not moving. Since the string is inextensible, this means that the two blocks can't move in the opposite direction either. So the only natural thing to assume is that the blocks are at rest. For the first block to be at ...


1

I start with a description in terms of the inertial coordinate system: As you state, the magnitude of the required centripetal force to sustain circular motion is given by: $$ m\omega^2r$$ If the actually exerted centripetal force is less than that the object will recede from the axis of rotation; if there is a surplus of centripetal force the object will be ...


1

But as t goes to infinity, v approaches 0, suggesting the distance is finite, and x goes to infinity which is contradictory. The position goes to infinity before the velocity starts decaying to zero fast.


1

The answer depends on the material of the body you are exerting force on. If it is a rigid body, then its shape is, by definition, constant. Assuming here for simplicity that your force vector passes through the COM, so that there isnt any torque on the body, then every particle will experience the same force in the direction you are pushing, as soon as you ...


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