Angles in elastic collisions

Рет қаралды 11,245

Garrison Turner

7 жыл бұрын

Пікірлер: 13

@SADDAMHUSAIN-ly4mm 2 жыл бұрын

Very good sir . I am seeing your lecture from india

@wrestlingbooster3138 5 жыл бұрын

Perfect video......Sir😁😁😁😁

@katsojohnobotsang2484 5 жыл бұрын

perfectly explained.....thank u sir

@mb-3faze 3 жыл бұрын

Beta is 90 degrees provided the masses are the same *and* the incident angle is not 0 degrees. If the incident angle is zero degrees and the masses are the same size and the initial condition states one mass is initially stationary, the equation still works but one of the velocities is zero (even though cos(beta) is 1). If the masses are not the same size the equation doesn't hold. If it did, when you dropped a ball from a height on to the surface of the large mass of the earth the ball would shoot off at 90 degrees (I.E. along the ground) which would be amusing but it would make basketball quite difficult to play. The general case with different masses is a bit more involved and revolves around the concept of 'reduced mass' - and although momentum is conserved, to work out the angles you have to assume that kinetic energy is conserved (which it isn't in the real world) and it gets a whole lot messier.

@juangokuz0578 5 жыл бұрын

make more videos like that

@flixerstudios1862 9 ай бұрын

Where did we use the fact that they do not collide head on? If the would collide head on then there would be no angle between them.

@wajeehulhassanmalik6260 4 жыл бұрын

Saved my ass phewww Tysm 🙌🙌

@rkkhatri9565 6 жыл бұрын

good explanation sir .thank u

@garrisonturner5670 6 жыл бұрын

Thank you!

@khormunhuai5901 3 жыл бұрын

When you carry out the practical experiment, using your math model, it only valid for result 90 degrees after collisions. When it come to different collision angles, your model is not valid.

@garrisonturner5670 3 жыл бұрын

The crux here is the argument is for masses that are equal. When you use unequal masses, you will not necessarily obtain 90 degrees between the two objects after the collision. However, unequal masses is a more arduous task to perform calculation-wise. But the initial prescription of the momentum being conserved and the kinetic energy being conserved still holds, so you can do the calculation with the same set-up, just don't cancel the masses in the beginning.

@khormunhuai5901 3 жыл бұрын

@@garrisonturner5670 the practical experiement I mentioned is carried out using the same ball masses. The arguement here is that the equation you are pointing tells us the resulting collision angle is always 90 degree with unknown initial collision angle. which is not match to the practical experiment. Lets say, if the balls are colliding at the angle difference of 5 degrees, they dont move 90 degrees apart after collision. Imagine you collide a ball at 0 degree difference, you guess it, they saperate at 180 degree (scalar) or 0 degree (direction vector)

@garrisonturner5670 3 жыл бұрын

@@khormunhuai5901 There are several nuances here: one thing I also did not mention explicitly along with equal masses is the fact that one of the objects is stationary before the collision. Under those two conditions, equal mass and one object stationary, then the options are 0 degrees after collision (if the collision is head on; its 0 because the incoming object stops and transfers all of the momentum to the second object), or 90 degrees after collision. If you ever watch billiard balls in simple collisions (not taking into account spin on the cue ball or friction with the surface of the table), then you will always obtain 90 degrees for the angle between the balls after the collision if they are not head on. This is a well-documented result in elementary physics texts.