45

I've confirmed the experiment, using a McD_n_lds paper drinks cup and a beer can hollow plastic ball of about $5\mathrm{g}$, of about the same diameter as a ping pong ball (PPB): The observed effect depends largely on the cup being soft and permanently deformable (like an object made of blutack or playdough), so its collision with Earth is inelastic. A ...


18

As mentioned in the comments above, the ball in the cup is similar to Galilean Cannon. The maximum height to which the ball can bounce $h_{max}$ can be estimated using the law of energy conservation: $$(m+M)gH=mgh+E_{cup}+E_{water}+E_{heat},$$ where $m$ is the mass of the ball, $M$ is the mass of cup+water, $H$ is the initial height from which the ball was ...


8

Recall that if a ball normally hits a wall elastically, its velocity will be exactly reversed. Suppose the whole system hits the ground with speed $v$. Now, as the cup and the water hits the soft mat, their speed quickly reduces, and may start moving upward (depending on how soft the mat is) before the ping-pong ball is affected by a reaction force. Suppose ...


6

My hypothesis why the ping pong ball receives a large upward impulse: The floating ping pong ball is displacing some water. The amount of displacement does not change much during the fall. As the cup hits the floor the deceleration of the quantity of water gives a short pressure peak. Because of that pressure peak the water that is in contact with the ping ...


5

It should have said that ''kinetic energy'' is not always conserved. When you throw some sticky dough at a wall the kinetic energy is converted completely to heat and sound energy and the dough sticks to the wall. Momentum is conserved though. When you throw the dough the dough went one way and the entire Earth went the other. When the dough stuck to the ...


4

Momentum should not be confused with energy. In some cases momentum is conserved but energy is definitely not. This is indeed a confusing way of putting things. The first sentence is clear and unequivocal. But the latter sentence should really read: Momentum is always conserved but kinetic energy is not always conserved($^+$). The latter part of this ...


4

The $23.75J$ energy can't have gone anywhere. What this means is that your construction is impossible. If there is no friction, no deformation, etc, then in an elastic collision the projectile cannot bounce back with $5 m/s$. It must go faster. The exact speed you can solve for using exactly conservation of energy & momentum.


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, ...


2

For a rigid object, the equation of motion for the center of mass $\mathbf x_\text{cm}$ of the object is described by Newton's second law $$\mathbf F_\text{net}=m\ddot{\mathbf x}_\text{cm}$$ Since your new composite system (ring + particle) has no forces acting on it after the collision, we have $0=m\ddot{\mathbf x}_\text{cm}$, i.e. the center of mass cannot ...


2

Assume that the water in the cup is compressible and inviscid, experiencing one-dimensional flow and thereby satisfying the one-dimensional Euler equations. Initial conditions, velocity =$\sqrt{gh}$ downward and pressure =1 atm, are both uniform. The bottom of the cup is struck from below in such a way that the velocity of the water is reduced and the ...


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 ...


2

AFAIK collision is defined by two objects coming into contact, hence the name. This may be of assistance in terms of determining an inelastic versus elastic collision: https://en.wikipedia.org/wiki/Coefficient_of_restitution Other than that, here are your two situations that I can think of and can work with - Case 1 - Elastic $$m_{\text{cart}}v_{\text{cart}...


1

What exactly you mean by launching a ball "into" a cart is important. If this means that the rubber ball lands on a horizontal surface of the cart and bounces back out, the cart doesn't have to move at all (the momentum exchange can be purely in the vertical direction for a perfectly elastic collision) while it does with the sticky ball. If what ...


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 ...


1

This is the energy balance approach: Since friction with the tubular structure is negligible and the inertia (mass) of the tubular structure is negligible, the sum of the kinetic- and potential energies of the ropes must be constant. As the tubular structure descends, the total length of the vertical sections of rope is constant, and their total potential ...


1

This is a really good question and can be solved using simple conservation laws of physics. Firstly , assuming R > R/2 > Rm , we can totally neglect the radius of the planet(Rm). I'm trying to split the components of velocity along radial and tangential axis. before collision, Radial Velocity(towards the planet) of the system is only from asteroid, i.e ...


1

Perhaps the simplest example is two identical objects moving toward each other with the same speed and sticking together at rest after colliding. Initial momentum = 0. Final momentum = 0. So momentum is conserved. Initial kinetic energy is nonzero but final kinetic energy is 0.


1

Yes, the momentum of a rigid body can be changed without doing any work on it theoretically. An example of such a case is uniform circular motion where the body moves along a circle with constant speed. The velocity of the body is always tangent to the circle so its direction is changing, and because the body's velocity is changing its momentum is also ...


1

There is no answer that conserves energy. First, from elasticity, we know that the normal component of the velocity stays constant. We then have two unknowns: the new tangent velocity and the new rotation speed. Conservation of angular momentum around the point of contact (all forces act through it, so the net torque around it will be zero) provides one ...


1

The other questions you asked were inspired by an animated movie. A lot of what you see there is not explainable by physics - Many of the things you see just couldn't happen that way. So specific calculations might not be as useful as understanding some ideas. And you have extracted a reasonable physics question from it. A punch does come from an object, a ...


1

Before the bullet hits, the torque on the block is $0$. The reaction force of the horizontal surface is uniformly spread over the bottom of the block. As the block rotates, the reaction force shifts to the edge of the block. This creates a torque opposing that of the bullet. The assumption is that the bullet comes to rest very quickly. The change in angular ...


1

I have a spreadsheet which calculates the result of a 2 body collisions in 2D. I found that with equal masses, one starting at rest, and the two separating at 90 degrees, energy was conserved. With masses that are different it is almost always possible to choose an exit velocity (magnitude and direction) for one that will put it at 90 degrees to the other (...


1

Laconic answer - YES, they will collide, given enough time. In general, if event has non-zero probability, then it will happen sooner or later, just time interval will vary until that event happens. Explanation : We can try to imagine colliding particle system as an ideal gas of molecules, so this will give a bonus that we can apply kinetic theory of gasses,...


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