Taking the rifle and bullet as one system, the total momentum just before the shot was zero. By principle of conservation of momentum, the total momentum just after the shot was fired should also be zero. Since the bullet carries a forward momentum, the rifle must recoil in the opposite momentum in order for the total momentum to remain zero.

Do note that the momentum of the bullet and rifle are exactly the same magnitude. So the ratio of the bullet and recoil velocity is the reciprocal of the ratio of their masses.

The rifle in use in this video is supposedly a 0.700 Nitro Express, which apparently uses an extraordinarily massive bullet. This exerbates the recoil which caught many subjects in the video by surprise.

First, recall that impulse equals the change in momentum.

FΔt = Δp

With or without the cushion, the change in momentum of the pile driver was the same. It’s momentum just before impact was p, after the impact was 0, cushion or not.

With the cushion, however, the impact duration is lengthened (Time is bought by allowing the egg to back away from the pile driver) For the same change in momentum, the maximum force is reduced.

See the F–t graphs below. Keep in mind the area-under-the-graphs for both graphs should be exactly the same, since the impulses for both cases were the same. A longer impact duration allows the impact force to be smaller and yet bring about the same change in momentum.

When the air was still trapped in the balloon, the total momentum of the system (consisting of balloon and the air inside) was zero.

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When the air gushes out from the balloon, the air attained a downward momentum. In order of the total momentum of the system to remain zero, the balloon must attain an upward momentum.

That’s how jet propulsion works. The thrust of a rocket results from the high speed ejection of the rocket fuel.

The monkey hovers in the air because its weight is balanced by the tension force of the slinky. It remains hovering as long as the bottom end of the slinky continues to be stretched by the same amount. From simple mechanics we can understand why the monkey does not drop immeditely, because it takes time before the unstretching progresses down the slinky. However, why must the entire slinky collapse before the bottom end starts to unstretch?

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The above video shows that when a rubber band is used instead, it still takes sometime before the bottom end starts to unstretch. However, unlike the slinky, there is no need for the entire rubber band to collapse before the bottom end starts to unstretch. So what’s so special about the slinky?

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I was about to delete the above video when I noticed something interesting. By accident, a longitudinal compression wave was sent down the slinky just before it was dropped. So this video captured two “things” racing down the slinky. There is an ordinary longitudinal compression wave that travels at a speed as dictated by the mass and tension of the slinky. Hot on its heels is the collapsed slinky. The video shows quite clearly that the collapsed slinky actually travels faster than an ordinary compression wave would down the slinky. The “collapse”, being accelerated by both gravity and the tension force of the slinky below it, pushes into the slinky faster than a compression wave can propagate down the slinky.

That explains why the monkey does not drop before the slinky collapses totally. The medium itself (the slinky) travels faster than the wave. The entire slinky collapses before the disturbance can reach the monkey.

(The dynamics for this demo is almost identical to that of a roller coaster’s loop-the-loop. Just replace the string’s tension with normal contact force of the coaster’s ramp, and everything is identical.)

Many students have the misconception that as long as the pendulum does not come to complete rest in the midst of motion, it can successfully loop-the-loop. That is wrong. The pendulum moves in a circular path only if the string is taut. However, when the pendulum is moving through the upper half of the circle, gravitational pull is actually tending to pull the pendulum inside the circle. If the pendulum is not moving fast enough, the component of gravitational pull in the centripetal direction will be larger than the required centripetal acceleration for circular motion at that speed and radius. This means the pendulum is being pulled by gravity towards the centre of the circle too quickly. The string becomes slack. Once that happens, the pendulum is no longer constrained to a circular path but goes into a parabolic path instead.

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There is a minimum drop height in order for the pendulum to be moving fast enough to loop-to-loop successfully. The solution for the drop height is presented in this xmtutor video. So in theory, the limiting point for the experiment set up in this video is roughly when r = 25.0 cm.

For the lazy ball, its momentum changed from mv to 0. So the lazy ball’s momentum changed by mv towards the right. The impulse exerted by the blocks on the lazy ball is thus mv. By Newton’s 3rd Law, the impulse exerted by the lazy ball on the blocks is also mv.

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For the bouncy ball, its momentum changed from mv to –mv. So the lazy ball’s momentum changed by 2mv towards the right. The impulse exerted by the blocks on the lazy ball is thus 2mv. By Newton’s 3rd Law, the impulse exerted by the lazy ball on the blocks is also 2mv.

The fact that the bouncy ball underwent a larger change in momentum implies that a larger impulse between the ball and the blocks. That’s why a bouncy ball delivers more “punch”.

The terminal velocity of a falling body is attained when its weight is balanced by the drag force.

Let’s assume that the drag force is proportional to the velocity of the falling body. So we write the drag force as kv, where k is a constant that is dependent on how streamline the falling body is.

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We can then solve for the terminal velocity v_{t} for the single sheet of paper dropped horizontally:

kv_{t} = mg

v_{t} = mg/k

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When a stack of papers are dropped (instead of a single sheet), k is unchanged (since it is the same area pushing into air), but m is of course multiplied by the number of sheets of paper in the sheet. For a stack of 20 sheets, the terminal velocity would be increased by 20 times.

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For the single sheet dropped vertically, m is unchanged, but k is reduced dramatically (since now only a tiny area is pushing against air). Assuming the k is directly proportional to the cross-sectional paper, k would be reduced by more than 1000 times (because the cross-sectional area is reduced from 216 mm x 297 mm to about 0.05 mm x 297 mm). This implies that the terminal velocity would be increased by more than 1000 times.

This video highlights the fact that the friction between two surfaces increases if the two surfaces are pressed more strongly into each other. (f=μN)

For example, the grip that the stick exerts on the beans (f) is large only if stick is wedged tightly among the beans (leading to large N).

Same thing for the paper. Only if the paper is pressed hard horizontally against the stick will the vertical friction be large enough.

When we “pin” papers on the fridge using fridge magnets, it is not the magnetic force that holds the papers in position. In fact, the magnetic forces do not even act on the papers. What the magnets do is to ensure a large normal contact force , which then allows for a large friction.

Lastly, note that in all the examples in this video, friction was the force that resulted in motion. It is a common misconception that force opposes motion. This is wrong. Friction opposes relative motion between two surfaces, not motion per se.

The water exerts an upward pressure force on the card (F=P.A). If the pressure force is larger than the brass weight, then the card is pressed upward against the tube, and the pig is safe.

Since hydro-static pressure is proportional to the depth of the fluid (P = hρg), the pressure force decreases as the tube is raised slowly towards the water surface. There will come a point where the pressure force is less than the brass weight, the card drops away from the tube, and the pig is doomed.

Let’s do a rough calculation. The mass of the brass weights is 300 g. The inner radius of the tube is 8.6 cm.

P.A = mg

hρg.πr^{2 }= mg

h(1000).(3.14)(0.043^{2}) = 0.300

h = 5.17 cm

By equating the weight to the upward pressure force, we estimate the minimum depth to support the brass weights to be about 5 cm. Which looks about right in the video.

Below is a famous video of an Apollo 15 astronaut conducting the feather and hammer demo on the Moon.

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And then there is this BBC video of the same demo conducted in a GIGANTIC vacuum chamber.

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On one hand, gravitational pull on an object is directly proportional to its mass (F=mg). On the other hand, the acceleration of an object is inversely proportional to its mass (Newton’s 2nd Law resulting in F=mg). These two factors together means that every object accelerates at g=9.81 m s^{-2 }regardless of their masses.