The Third Law and Conservation of Momentum

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Fire a gun, and it kicks back hard into the shoulder. But the gun only pushed the bullet forward; nobody pushed the gun back. So why does it fly back? Hold that thought. The third law and the conservation of momentum answer it, and both come up often in JEE problems. So, let's begin.

0:23Newton's third law: forces come in pairs

So, the third law: if body A exerts a force F on body B, then B exerts minus F on A, always. The arrow on F means a vector, the minus sign means the same size in the opposite direction, and the double arrow means implies. Look at the picture: a book on a table. The book pushes down on the table, the red arrow, and the table pushes up on it with force N, the blue arrow. Equal and opposite: an action reaction pair. Okay? Now think. These two forces are equal and opposite. So, do they cancel each other out? No, never! The two forces of a pair act on different bodies: one on the table, and one on the book. Forces can cancel only when they act on the same body. Clear? Two more facts. The forces of a pair are of the same kind, both gravitational, or both contact forces, and so on, and they act at the same instant. Right? Now, a trap. On the book, the blue arrow N is the normal force from the table, and the green arrow W is its weight, the earth's pull. Equal and opposite. So, are N and W an action reaction pair? No, no, careful! On the book, N equals W, but they are not a pair! Both act on the same body, the book; they are equal only because the book is in equilibrium. Read the figure: the reaction to N is the book's push on the table, and the reaction to W is the book's pull on the earth. So always check which body each force acts on. Got it?

2:01Conservation of linear momentum

Now, the second big idea. For a system of bodies, capital P with an arrow is the total momentum. The Greek sigma means sum: add m sub i times v sub i over all the bodies; m sub i is a mass, and v sub i, with its arrow, a velocity. With me? Now think. Inside the system, the bodies push and pull on each other: these are internal forces. By the third law, what happens to them when we add them all up? Yes, exactly! They cancel in pairs. Each internal force has its partner inside the system, so only external forces remain. F with an arrow and the small letters e x t below is the net external force; it equals d P over d t, the rate of change of the total momentum. Okay? So, if the net external force is zero, the zero with an arrow on screen, the total momentum stays constant. That is the conservation of linear momentum. Note the condition: zero net external force. Right? Now a very useful point. It also holds for one component alone. For example, if only vertical external forces act on the system, then the horizontal component of the total momentum stays constant. Clear? And one more. During a short impact or an explosion, the impulsive internal forces are far larger than an ordinary external force like gravity, so momentum just before and just after is conserved approximately, even though that external force acts.

3:40Recoil and explosions

Got it? Now two standard results. First, recoil. Take a gun of mass capital M at rest, firing a bullet of mass small m at speed small v. Momentum starts at zero and stays zero, so if the gun moves back at speed capital V, then capital M times capital V equals m times v. Okay? Second, say a body at rest breaks into two pieces. Again the momentum stays zero, so the pieces fly off in opposite directions, with m one v one equal to m two v two; the small one and two below are subscripts for the two pieces. Right? Now think. Both pieces have the same size of momentum. So which one carries more kinetic energy: the lighter piece, or the heavier one? The lighter one, correct! Kinetic energy K is proportional to p squared over m, where p is the size of the momentum; the curly sign between them means proportional to. Same p, so the smaller mass gets more energy: K one is to K two as m two is to m one. See the order? The masses come in the reverse order, so careful there. Clear?

4:49Trap: velocities relative to a moving body

Now, one trap that catches many students. In the momentum equation, every velocity must be relative to the ground. With me? But some questions give a velocity relative to another moving body, like a man jumping off a trolley, or a shell fired from a moving gun, right? So convert that velocity to the ground frame first, and only then write the momentum equation. Example two does exactly this. Okay?

5:15Example 1: recoil speed of a gun firing a bullet

Alright, example one, our puzzle! Here, a four kilogram gun fires a twenty gram bullet at four hundred metres per second; k g is kilogram, g is gram, and m s to the minus one is metres per second. Find the recoil speed of the gun. Pause the video, and try it first. Done? Let's check. Gun and bullet start at rest, so the total momentum before firing is zero, and it stays zero. With me? Now, the masses. The bullet is twenty grams. So, can we put twenty straight into capital M V equals m v? Not with the gun's mass in kilograms! Twenty grams is zero point zero two kilogram. Keep both masses in the same unit, always. Got it? Now, capital M V equals m v: four times capital V equals zero point zero two times four hundred, which is eight. Right? So capital V is eight over four: two metres per second, backwards. Puzzle solved! The gun pushed the bullet forward, so by the third law the bullet pushed the gun back, equally hard. That is the kick. Clear?

6:28Example 2: man runs off a trolley on a smooth track

Now a harder one, with exactly that trap. Here, a fifty kilogram man is on a one hundred and fifty kilogram trolley, at rest on a smooth track. He runs off its end at four metres per second relative to the trolley. Find the speeds of the man and the trolley relative to the ground. Pause the video, and try it. Done? Okay, let's check. The track is smooth, so no horizontal external force acts, and everything starts at rest. So the horizontal momentum is zero, and stays zero. Right? Now, many students write fifty times four equals one hundred and fifty times capital V, straight away. Is that correct? Oh no, careful! This is the trap. Four metres per second is relative to the trolley, and the trolley itself moves. Say the trolley moves back with speed capital V; then the man moves forward at four minus V, the bracket on screen, relative to the ground. Clear? Now the momentum equation, with ground speeds: fifty times, four minus V, equals one hundred and fifty times V, okay? Open the bracket and collect the V terms: two hundred equals two hundred V, so V is one. With me? So the trolley moves back at one metre per second, and the man moves forward at four minus one, three metres per second. Got it? Quick check. Fifty times three is one hundred and fifty, and so is one hundred and fifty times one. Equal and opposite momenta. Perfect!

8:01Quick revision: third law pairs, momentum conservation

So, quick revision. One: the two forces of a third law pair act on different bodies, so they never cancel. Two: zero net external force means the total momentum stays constant. Three: write every velocity relative to the ground first. Now go and try the practice problems on this page. All the best!