Momentum Conservation in Two Dimensions

Transcript

0:00Introduction

Hello everyone! Let's start with a puzzle from the snooker table. One ball sits still, and another hits it a little off-centre, not head-on. Players notice something strange. Very often, the angle between their two paths comes out close to ninety degrees! Why ninety? Why not sixty? Today we will prove it, with a few lines of vectors. And momentum in two dimensions comes up often in JEE, in collisions and explosions. So, let's begin.

0:32Momentum is a vector: conserve each component

So, now the bodies move in a plane, so momentum is a vector. This big Greek letter, capital sigma, means the sum over all the bodies. The small arrows on top mean vectors: u is the velocity before, and v the velocity after. With m the mass, and no external impulse, the sum of m times u vector equals the sum of m times v vector. Okay? Now tell me. In a plane, how many ordinary equations does one vector equation give? Two! Yes, exactly! Along x: the sum of m times u x equals the sum of m times v x. The small x below the letter means the x component. Right? And along y, completely separately: the sum of m times u y equals the sum of m times v y. Two directions, two equations. Clear?

1:21Bodies that stick: perfectly inelastic

Now, the simplest case. Two bodies collide and stick: perfectly inelastic. They move on as one body, with one common velocity. So what is that velocity? Total momentum, over total mass! Correct! Capital V vector equals m one u one vector, plus m two u two vector, over m one plus m two. The small one and two tell you which body. And the top is a vector sum, so add it by components. Got it? And kinetic energy is lost. This triangle is the Greek letter delta, and delta K is the loss of kinetic energy. It equals half times mu, this Greek letter, times the modulus of u one vector minus u two vector, squared. The straight bars are the modulus, the size of a vector, and the small two up top means squared. With me? And mu is the reduced mass from the last chapter: m one times m two, over m one plus m two. Same as in one dimension, but now the relative velocity is a vector. Okay?

2:26The 90 degree rule: equal masses, one at rest

Now, our snooker puzzle. Look at the conditions. The collision is elastic, so no kinetic energy is lost. The masses are equal, both m. One ball is at rest. And the hit is not head-on. Got it? First, momentum. m u vector equals m v one vector, plus m v two vector. This double arrow means implies. Cancel m, and u vector equals v one vector plus v two vector. Right? Next, energy, since it is elastic. Half m u squared equals half m v one squared, plus half m v two squared; no arrows, just speeds. Cancel half m: u squared equals v one squared plus v two squared. Clear? Now the clever step. Square the momentum equation: take the dot product of each side with itself. u squared equals v one squared plus v two squared, plus two times v one vector dot v two vector; the dot is the dot product. Compare this with the energy line. What must be zero? Yes, exactly! v one dot v two equals zero. So the two velocities are perpendicular: the balls move off at right angles. Our snooker puzzle, solved! In a head-on hit, one of the velocities is zero instead. That is why we said not head-on. Got it? And the speeds. v one and v two are at right angles and add up to u, so they make a right angled triangle with u as the longest side. So v one equals u cos theta, and v two equals u sin theta, where theta, this Greek letter, is the deflection of the first ball. Okay? Now, a trap. Suppose a heavy ball hits a light ball at rest, elastically. Do they also fly off at ninety degrees? No, no, careful! We cancelled m only because the masses were equal. Unequal masses, or energy lost, and the ninety degree rule breaks. Check all four conditions first. With me?

4:19Example 1: east and north bodies stick, common velocity

Now, Example one, an easy one. There is a two kilogram body, k g on screen, moving east at three metres per second, written m s to the power minus one. It sticks to a three kilogram body moving north at two metres per second. Find their common velocity. Pause the video and try it. Done? Okay, let's check. First body: two times three, six, towards the east; the cross means times. Second: three times two, six, towards the north, both in kilogram metres per second. Okay? So the total momentum is six plus six, twelve, right? No, no, careful! This is the trap. One is east and one is north, at right angles, so you cannot add them like plain numbers. Keep the components: six comma six, in brackets. Clear? Now divide by the total mass, two plus three, five. V vector is one point two east and one point two north, in metres per second. Right? So the speed is one point two times root two, six root two over five; this tick is the root sign. The wavy equals sign means approximately: about one point seven zero metres per second. Equal parts east and north, so the direction is north-east. That's it!

5:39Example 2: glancing hit of equal balls, was it elastic

Now a harder one, JEE level. Here, a one kilogram ball moving at four metres per second along x hits another one kilogram ball at rest. Afterwards, the first moves at two metres per second, at sixty degrees to x; the small circle means degrees. Find the second ball's velocity. Was the collision elastic? Pause and try it. Done? Let's check. Both masses are one kilogram, so they cancel. Along x: four equals two cos sixty, plus v x. Two cos sixty is one, so v x, the second ball's x velocity, is three. Okay? Now along y. Before the hit, how much momentum is there along y? Zero! Nothing moves along y before the collision. So zero equals two sin sixty, plus v y, and two sin sixty is root three. So v y is minus root three: the second ball goes below the x axis, cancelling the first ball's upward part. Right? So v two vector is three i hat, minus root three j hat; the little caps mean unit vectors along x and y. Its size is root of nine plus three, two root three, about three point four six metres per second. Its angle below x is tan inverse of root three over three, the minus one up top meaning inverse: thirty degrees. With me? Now, was it elastic? Compare the kinetic energies, before and after. K is kinetic energy. Before: half times one times four squared, eight joules, J for joules. After: half times one times two squared, two, plus half times one times twelve, six. Eight joules again. No energy lost, so yes, elastic! Perfect! Now look at the angles. Sixty degrees above x, and thirty below. Angle between them: sixty plus thirty, ninety degrees! Equal masses, one at rest, elastic, so right angles. Exactly our snooker rule! And a topper's check: the deflection is sixty degrees, so v one is four cos sixty, two, and v two is four sin sixty, two root three. Both match! That's it!

7:54Quick revision: components, sticking bodies, 90 degree rule

So, quick revision. Momentum is a vector: conserve x and y separately. Bodies that stick: the common velocity is total momentum over total mass, and energy is lost. Elastic, equal masses, one at rest: they move off at ninety degrees. Now go and try the practice problems on this page. All the best!