Principle of conservation of momentum is the reason behind recoils. During the firing of the bullet, we have the rifle and the bullet pushing each other. As far as the bullet-rifle system is concerned, this is a pair of internal forces. Without any external force acting on the bullet-rifle system, its total momentum remain unchanged. Since we started with the rifle and bullet both at rest, the total momentum must remain at zero.

The bullet, though small in mass, does carry a large forward momentum thanks to its speed. This forward momentum must be matched by a backward momentum of the same magnitude.

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.

Same impulse. But the collision with the wall is a stiff and snappy one: large F small Δt. Whereas the collision with the sheet is a soft and draggy one: small F large Δt. Nevertheless, the resulting Δp is exactly the same.

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Bending the knees upon landing lengthens the impact duration significantly, resulting in significantly smaller F. Do realize that the impulse is not reduced. It is just spread over a longer duration.

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Check out the timers at the bottom left corner. Notice how quickly the car body comes to a rest compared to the passenger. Air-bag or not, the impulses experienced by the the passenger are the same. But hitting the air bag is a lot less forceful than hitting the rigid steering wheel because of the longer impact duration.

When you’re hit by a water cannon, you’re continuously bombarded by godzillion number of H2O molecules at any one instant. The force exerted by each molecules is tiny and fleeting (lasts only for an instant). But collectively, they exert a large and continuous force.

To calculate the force exerted by a water cannon, we turn to calculating the force experienced by the water cannon instead. When the water hits you, it loses momentum. The rate at which it is losing (or changing) momentum is equal to the force that you are exerting on the water. By N3L, the water is exerting an equal but opposite force on you. Ouch.

During the collision, the bouncy ball’s momentum changed from mv to –mv, corresponding to a momentum change of 2mv (rightward).

The lazy bag’s momentum changed from mv to 0, corresponding to a momentum change of only mv (rightward).

Assuming that the duration of collision was about the same, this implies that the tower pushed the bouncy ball (rightward) harder than it pushed the lazy ball. Because F=Δp/Δt.

By N3L, the tower was pushed harder (leftward) by the bouncy ball than by the lazy ball.

Sometimes I wonder if we can send some of our students to the International Space Station (ISS) for a homestay. With gravity and friction out of the way, it is pretty obvious that all a mass wants to do, is to be stick to its current motion. No need for any teaching, every student understands immediately maintaining motion requires no force.

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Better still, throw the student out of the ISS. With nothing to push/pull against, the student immediately learns that his current motion cannot be changed without an unbalanced force.

This reminds me of the scene in the movie “Mission to Mars”, where the crew were forced to abandon ship and take refuge in the REMO (“Resupply Module”) orbiting Mars. The plan was for Woody to launch himself at the module and attach a tether to it.

It was Woody’s inertia that killed him. Unable to conjure any force to act on himself, he was doomed to carry on moving at his current velocity. (In the movie it was suggested that eventually Woody was accelerated by Mars’ gravity and burnt up as he sped through the Martian atmosphere.)

What is the force that throws you back into your seat when the car races off the line? And what is the force that throws you forward when the car brakes suddenly? What is the force that pins you against the door when the car makes a sharp turn?

The answer to all the above? None. It is the lack of force that “throws” you around in the car.

Watch the video below. See whether you can explain the motion of the tissue box placed on top of the dashboard.

Sitting on top of the dashboard, the only horizontal force acting on the tissue box is friction.

At around 0:12, the car made an emergency brake and slowed down abruptly. The tissue box hardly slowed down since the frictional between the tissue box and the dashboard falls far short of the required retardation force. As the tissue box remained in motion while the car comes to a screeching stop, it appears to jump forward.

At around 0:15, the car was rammed from behind by another car and lunged forward. The tissue box (which had already come to rest a split second earlier, thanks to the retardation force provided by the implacable windshield) could not keep up with the car. As the tissue box remained at rest while the car lunged forward, it appears to jump backward.

So there is no “phantom” force. It is just the inertia of the tissue box.

Elastic or not, total momentum is always conserved in any collision. So

What is special about an elastic collision is that the total KE before and after the collision is unchanged. So

Dividing (2) by (1):

Notice that u_{1} – u_{2} represents the relative speed at which m_{1} approaches m_{2} before the collision, while v_{2} – v_{1} represents the relative speed at which m_{2} separates from m_{1} after the collision.

Physically, it means that in any elastic collision, the two bodies approach and separate from each other at the same relative speed.

Mathematically, it means that an elastic collision can be calculated more easily using equations (1) and (3), rather than equation (2).