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# 999 Conservation of Angular Momentum

# 122 Balloon Helicopter

# Rifle Recoil (YOUTUBE)

# 121 Pile Driver

# 120 Rocket Balloon

# 117 Lazy Ball vs Bouncy Ball

# 061: Magnetic Cannon

# 058 William Tell Overture and Newton’s Pendulum

# 041 Perfectly Inelastic Collision

# 028 Can you blow your own sail?

Category: 03 Dynamics

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As each blade push air backward and out through the hole in the blade, the ejected air pushes the blade forward (Newton’ 3rd Law). This is how a forward thrust is produced to propel each blade forward, resulting in a torque that causes the blades to spin.

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Because the blades are tilted at an angle, the ejected air is also pushed downward and out through the holes in the blades. This implies the ejected air must be pushing the blades upward. (Newton’s 3rd Law). This is how an upward thrust is produced to lift the balloon.

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The two main forces acting on the helicopter are the upward lift and the downward weight of the helicopter. In this video, we can see that initially, the lift force was barely able to support the weight of the helicopter, resulting in the helicopter hovering at about the same height. However, as the helicopter ejects air from the balloon, its weight decreases. So the **net** upward force becomes larger and larger, and the helicopter accelerates upwards. Right at the end, the balloon ejects the air in a spurt, resulting in a kick in the lift force. This coupled with rapid decrease in mass of the helicopter, results in the obvious jerk up in the last part of the motion.

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.

*F*Δ*t* = Δ*p*

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.

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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.

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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 2*mv* towards the right. The impulse exerted by the blocks on the lazy ball is thus 2*mv*. By Newton’s 3rd Law, the impulse exerted by the lazy ball on the blocks is also 2*mv*.

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”.

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The interesting part is when a magnetic ball collides into two stationary metallic NON-magnetic balls (@ 0:48).

Clearly, the outgoing ball had a much greater momentum and energy than the incoming ball. So is there a double violation of the conservation principles (of momentum and energy)? Gosh.

Nope. Nope.

First for momentum, notice two balls recoiled to the right after the collision. (The recoil was at quite a high speed, but friction brought them to rest quickly) So even though the outgoing ball had a large leftward momentum, after subtracting the rightward momentum of the recoil, the total momentum is still equal to the initial leftward momentum.

As for energy, note that the magnetic field must have an associated magnetic potential. Since the field is attractive, the balls must be losing magnetic potential energy as they come closer. So even though the outgoing ball had a large kinetic energy, after accounting for the loss in magnetic potential energy, the total energy is still equal to the initial total energy.

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If ONE BALL rams into the stack, why must ONE BALL come out the other end at the same speed? Why can’t TWO BALLS come out at HALF the speed, for example? (This outcome also conserves the total momentum!) And how about all FIVE BALLS swinging away at 1/5 the speed?

True enough. There are infinite number of outcomes that fulfil the Principle of Conservation of Momentum! However, since these collisions are elastic, the total kinetic energy should be unchanged after the collision as well. With this additional consideration, we realize there is only one possible outcome. The same number of balls always come out the other end at the same end, conserving both total momentum and keeping total kinetic energy unchanged.

Now are you ready for Psy?

The Principle of Conservation of Energy and the Principle of Conservation of Momentum are universal laws applicable to ALL situations, including collisions. So the total energy and total momentum are always conserved in ALL collisions.

There is however nothing which dictates that Kinetic Energy must be conserved.

– In an (perfectly) elastic collision, all the KE is conserved.

– In an inelastic collision, some KE is “lost”^{1}.

– A perfectly inelastic collision is the one that loses the most KE^{2}.

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A “classic” collision is that of a body colliding into another stationary body of equal mass. If the collision is perfectly inelastic, the outcome is that of both bodies traveling at half the speed after the collision. Note that this outcome results in the maximum loss of total kinetic energy AND conserves momentum.

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Another “classic” collision is that of two equal masses travelling at the same speed colliding head on into each other. If the collision is perfectly inelastic, the outcome is that of both bodies coming to rest after the collision. Again, note that this outcome results in the maximum loss of total kinetic energy AND conserves momentum.

In fact, it can be shown mathematically that for a two-body head-on collision, having both bodies travelling at the same speed after the collision is the outcome that maximises the loss of kinetic energy. So having two bodies “stuck together” after the collision is the tell-tale sign for a perfectly inelastic collision.

1. The kinetic energy of the colliding bodies could have been passed on to (1) vibrational energy, manifesting as heat and sound, or (2) potential energy in the molecular bonds, manifesting as permanent deformations in the colliding bodies.

2. It is not possible to always possible to lose ALL kinetic energy in a collision because of the Principle of Conservation of Momentum. If the total momentum before the collision was not zero to start out with, the total momentum after the collision cannot be zero either.

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Basically, the air gave the cart a backward force when they are pushed forward by the fan, and a forward force when they collide into the sail. However, since these two forces need not have the same magnitude (they are not action-reaction pair), it is unclear which direction the net force will be.

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The principle of conservation of momentum can make the analysis easier. Just use the fact that the total momentum of the cart plus the air as a system must remain as zero!

With the original sail, the air got pushed out around the sides of the sail. Since the net change in momentum of the air is zero, the cart cannot have any change in momentum.

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With the improvised sail, the air got turned around and back. Since the air have acquired a backward momentum, the cart must acquire a forward momentum.

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In practice of course it will be silly to blow your own sail forward. It is much easier (and more efficient) to simply blow backward.