Elastic Collisions in One Dimension

Transcript

0:00Introduction

Hello everyone! Small puzzle first. On a smooth table, roll a ball straight into an identical ball at rest. If the collision is elastic, the moving ball stops dead, and the other moves off at its full speed. Now roll a heavy ball into a very light one at rest. The light ball flies off at nearly twice the heavy ball's speed! How can it move faster than the ball that hit it? One simple result explains both, and it comes up often in JEE, so let's begin.

0:31Elastic collision: two laws at once

Look at this figure. Two balls, masses m one and m two; the small numbers below the letters label the bodies, okay? Top row, before: velocities u one and u two, along one straight line, which is what head-on means. The red arrows are the velocities. Middle row, at maximum deformation, the greatest squeeze: both move with one common velocity v. Bottom row, after: v one and v two. With me? Now, the word elastic. Tell me, in an elastic collision, what is conserved? Two things! First, momentum, as in every collision. m one u one plus m two u two equals m one v one plus m two v two. Okay? And second, kinetic energy: K E before equals K E after. That is exactly what elastic means. Right? Solve these two together, and a beautiful result comes out. v two minus v one equals u one minus u two. This line is the heart of today's lesson. Clear? See what it means. u one minus u two is how fast they come together; v two minus v one, how fast they move apart. So they separate exactly as fast as they approached. On the right of the figure, the restitution rule has a letter e: e equals one means elastic, exactly this; the other values of e come next lesson. Got it?

2:01The velocities after the collision

Solving fully gives the final velocities. v one equals, m one minus m two, times u one, plus two m two u two, all over m one plus m two. Okay? Now, a topper's question. Can you write v two, without solving anything again? Yes, exactly! Just swap the labels one and two: v two equals, m two minus m one, times u two, plus two m one u one, all over m one plus m two. One formula, not two. Right? One more picture: in the centre of mass frame, the frame moving with the centre of mass, each velocity simply reverses. With me?

2:42Special cases

Alright, now the fun part: special cases. Put m one equal to m two in the formulas we just found. What happens to the velocities? The first terms vanish, and the twos cancel! v one equals u two, and v two equals u one: equal masses exchange velocities. So the moving ball stops dead. First part of our puzzle, solved! Okay? Now the second part. Let m two be at rest, and m one much heavier. The heavy ball carries on at almost u one, but they must separate as fast as they approached, at u one. So how fast must the light ball go? Two u one, yes! u one to keep up with the heavy ball, plus u one more to move away from it. This double symbol means much greater than, and the wavy equals means approximately equal. So v one is about u one, and v two about two u one. Got it? And the opposite case: a very light ball hits a heavy one at rest. The double symbol now points the other way: much less than. The light ball bounces back, v one about minus u one. Right? Now real numbers. Say a ball of five kilograms, k g, moving at four metres per second, written m s to the power minus one, hits a fifty gram ball at rest, g for grams, head-on and elastically. v two equals two times five times four, over five point zero five, the total mass in kilograms: about seven point nine two metres per second, nearly twice four! Puzzle solved!

4:17How much kinetic energy is passed on?

Now, energy. When m one hits m two at rest, head-on and elastically, the fraction of its kinetic energy given to m two is four m one m two, over m one plus m two, the whole squared. Okay? For equal masses, four m squared, over two m, the whole squared, is one. All of it, like the exchange. Right? Now a classic trap. To slow down fast neutrons, which should a neutron hit: heavy lead, or a light proton? Lead, surely? No, no, careful! On a proton, of almost the same mass, the neutron gives away practically all its energy. Lead has mass number two hundred and seven, the A on screen: four times two hundred and seven, over two hundred and eight squared, only about one point nine percent. The neutron just bounces off lead. Clear? That is why light nuclei are used as neutron moderators, to slow neutrons down. With me?

5:17Example 1: 2 kg ball hits a 3 kg ball at rest elastically

Example one, an easy one. There is a two kilogram ball at five metres per second, hitting a three kilogram ball at rest, head-on and elastically. Find the velocities after. Pause the video and try it. Done? Okay, let's check. First, a quick guess. The lighter ball hits the heavier one. Does it keep going forward, or come back? It comes back! u two is zero, so the second term drops out: v one equals two minus three, over five, times five, minus one metre per second. So it moves back, at one metre per second. Okay? And v two equals two times two, over five, times five: four metres per second, forwards. Right? Always check. Kinetic energy before: half times two times five squared, twenty five joules, J for joules. After: half times two times one squared, plus half times three times four squared, one plus twenty four, twenty five joules again. Nothing lost. Perfect!

6:25Example 2: three balls in a line, how many collisions

Now a harder one, JEE level. Balls A and B, mass m each, and C, mass two m, lie in a line on a smooth floor. Ball A is sent at u towards B; all collisions are elastic. Find the final velocities and the number of collisions. Pause and try it. Done? Let's check, one collision at a time. Ball A hits B first, and they have equal masses. So what happens? They exchange velocities, exactly! The small letter below each v names the ball. So v A is zero, and v B equals u. Okay? Now B, mass m, hits C, mass two m, at rest. C is at rest, so one term in each formula drops: v B equals m minus two m, over three m, times u: minus u over three. And v C equals two m over three m, times u: two u over three. So B bounces back. Right? So, two collisions, and we are done? Think carefully. Where is ball A now? Oh no, careful! This is the trap. Ball A is still sitting at rest where it stopped, and B is coming back towards it at u over three. So B hits A again! Equal masses, so they exchange once more: v A becomes minus u over three, and B stops. Clear? So, the final velocities: A moves back at u over three, B is at rest, and C moves on at two u over three. They all move apart now: three collisions in all. Got it?

8:10Quick revision: separation speed, equal masses, energy share

So, quick revision. Elastic: momentum and kinetic energy both conserved, and v two minus v one equals u one minus u two. Equal masses exchange velocities; a heavy ball on a light one at rest sends it off at nearly two u one. Energy given to a body at rest: four m one m two, over m one plus m two, the whole squared; all of it for equal masses. Now try the practice problems on this page. All the best!