# Tag Info

27

In a crude way, the answer is yes, and this effect was commonly exploited in old school sci-fi films by shooting scaled-down models and playing back the action in slo-mo. This general effect is also exploited in the world of physics and engineering, where accurate scaling factors are used to model the behavior of full-scale systems by using models in which ...

21

We can apply Dimensional analysis to answer this question. The obvious relevant constants are density $\rho \propto \frac{M}{L^3}$ free fall acceleration $g \propto \frac{L}{T^2}$ viscosity $\eta \propto \frac{M}{LT}$ With these two, we can simultaneously scale size $L$, mass $M \sim L^3$ and time $T \sim L^2$ while keeping density and acceleration ...

12

Since the electron and photon are quantum mechanical objects, the angle at which the electron and photon move after the collision is probabilistic. As you have pointed out, they do not actually "collide" but instead they interact. You can think of this process as the electron absorbing the photon at one point, and then re-emitting one at another. ...

10

No The water droplets have a certain size no matter the size of your stone, so if you record a small stone the water droplets will look larger relative to the stone, than if you recorded a big one. Also, the amount of droplets will be different, due to the energy differences as explained above. I still think it could be worth a try, even though it doesn't ...

8

At the moment when $m_1$ collides with $m_2$, net force along X-axis is $0$ so, momentum is conserved in that direction. But momentum is not conserved along Y-axis.

8

The basis for deriving the Poisson law is the assumption that the events occur with a constant rate, $\lambda$ (although one can generalize to a varying rate). We can then divide the time interval $t$ into $N$ intervals of size $\Delta t = t/N$, with the probability of an event occurring in each interval being $\lambda \Delta t$, and calculate the ...

8

The energy of the impact is proportional to the mass (for the same velocity). And $M \propto R^3$. But the area that receives the impact is proportional to $R^2$. So, I suppose that the splashing effects increases with $R$, (energy by area). One possibility is to increase the speed of the falling small sphere until get the same ratio. And adjust the speed of ...

7

No. The reason is the well-known Square–cube law. In fact, you have to deal with a linear-square-cube law: You'll have quantities that scale linearly (the 10 cm model will move with the apparent speed of the 1 km asteroid through your video), you'll have some that scale with the square (inner cohesion of bodies, repulsive forces in collisions), and you'll ...

5

I'm typing in mobile so I'm ignoring all the vector signs. The problem is to show that $$a^2 + b^2 = c^2 + d^2$$ and $$a + b = c + d$$ Gives you $$|a - b| = |c-d|$$ From the second eq you get by squaring both sides $$a^2 + b^2 + 2ab = c^2 + d^2 + 2cd$$ Using the first equation you then get $$ab = cd$$ Now subtract $2ab= 2cd$ on both sides of the first ...

5

Use the reduced mass identity to write the total energy as $$\frac 12 m_1 |{\bf v}_1|^2+ \frac 12 m_2 |{\bf v}_2|^2\\= \frac 12 \frac{m_1m_2}{m_1+m_2} |{\bf v}_1-{\bf v}_2|^2+ \frac 12 (m_1+m_2)|{\bf v}_{CofM}|^2.$$ Then, if energy is conserved (definition of elastic) we must have that $|{\bf v}_1-{\bf v}_2|$ is the same before and after.

3

The collision boundary differs from the classical mechanical billiard ball model in that there isn't a rigid boundary that defines whether the particles collide or not. In the billiard ball model, if "the center of another billiard ball is at a distance from that line of less than the radius of a billiard ball" then the balls clearly collide. If ...

3

Sure. For example, in a perfectly inelastic collision the velocities of A and B after the collision will both be equal to half the original velocity of A.

3

If it can be determined what the stopping distance was for the ship, such as by a measurement of the depth of penetration of the ship into the canal wall, and can ignoring the resistance of the water to the ship movement, one can estimate the average impact force using the work energy theorem, which states that the net work done on an object equals its ...

2

Momentum conservation is a vector equation, not a scalar equation; it should be $$m_1 \vec{v}_1 + m_2\vec{v}_2 = (m_1 + m_2)\vec{v}_3.$$ In some problems, where the motion is all along a given line, you can ignore this property. But if the motion is inherently 2D (i.e., it involves both $x$ and $y$ motion), then you can't. To solve this, you need to ...

2

You are right to be concerned. A better demonstration of momentum conservation would have been if the blocks had been sitting on perfectly frictionless ice, or even better: in space. The reason they chose this example because it shows how the famous Newton's cradle works. If you've ever seen a Newton's cradle in real life you know that after a while the ...

2

In your question you have a photograph of tracks in some kind of bubble chamber. This is an old technology, in which some kind of fluid is prepared in a supercritical chemical state, on the wrong side of a phase change. (In a "cloud chamber" the fluid is ready to condense liquid from vapor; in a "bubble chamber" the fluid is ready to ...

1

Electrons and photons do in fact have a spatial extension. It is not a spatial extension of the particles themselves though. They are point-like objects. On the contrary, the wavefunctions accompanying the electron and the photon are spatially extended. Because of this there is a continuous range of possible interaction (collision) outcomes. If the particles ...

1

here is found in the list a "picture" of an antiproton: The dark lines in this picture are produced by charged particles as they force their way through liquid deuterium. The highlighted track is an antiproton , produced in the decay of an antilambda into an antiproton and a $π^-$. In the top left corner of the picture, this antiproton ...

1

In the Poisson formula the average value $\lambda = s + b$ does not have to be small. E.g. if we have many particles passing through a detector and most of them do not contribute to the obtained signal (background or real), the conditions for a Poisson process are satisfied. This is exactly why we use this distribution to model the signal. Here an example: ...

1

The accelerated electrons scatter off the outer electrons of the gas molecules. If they hit one of those electrons hard enough to transfer ~several electron volts of energy to it, they can temporarily knock it out of its orbital, ionizing the gas. that loose electron is then available to be accelerated by the field and thereby participate in conducting ...

1

If Car 2 is immovable then the momentum that is lost by Car 1 in the collision is actually “transferred” to the Earth. Of course, the large mass of the Earth means this makes a negligible difference to the Earth’s motion through space.

1

Generally, a force is only going to keep the velocity magnitude of the target body constant, if the transferred momentum is infinitely small. This is actually a slightly convoluted way of saying: the force that is perpendicular to the body's velocity is generally going to keep the velocity magnitude constant only in a single moment. You can't generally ...

1

You are not deriving the conservation of kinetic energy from an elastic collision. That would be redundant since the definition of an elastic collision is one that conserves kinetic energy. Instead, you use the measurements in the table to show that this collision conserved kinetic energy--and was therefore elastic. Calculate the total kinetic energy before ...

1

Using $$\sin(2x)=2\sin (x) \cos (x)$$ If $\sin\left(\frac{\theta}{2}\right)>0$: $$\cos\left(\frac{\theta}{2}\right)\Bigg|\sin\left(\frac{\theta}{2}\right)\Bigg|=\cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)=\frac{1}{2}\sin\theta,$$ so the $\sin\theta$ cancels with the one on the denominator. If $\sin\left(\frac{\theta}{2}\right)<0:$...

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