Impulse and Momentum

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Imagine a fast cricket ball coming straight at you. You can catch it with stiff hands, or draw your hands back as you catch it. Either way, the ball comes to rest, so it loses exactly the same momentum, right? Then why do stiff hands sting so much more? Today we will answer this properly, with numbers. And this idea, impulse, comes up often in JEE, in collisions and in force time graphs. So, let's begin.

0:31Impulse: the area under the force–time graph

Look at this graph. Up the side is the force F during a collision, and along the bottom, the time t. The force shoots up to a huge peak and falls back to zero, all within the contact time, delta t. This little triangle is the Greek letter delta, and delta t is the short time the contact lasts, okay? Now, the shaded area under this curve has a name: impulse. J equals the integral of F d t. The small arrows on top of J and F mean vectors, quantities with a direction. And this long S is the integral sign: it adds up F times d t, tiny slices of time, over the whole contact. With me? Now recall Newton's second law. F equals d p over d t, the rate of change of momentum, p, okay? So F times d t is a tiny change in momentum. Add all of them up over the contact. What do you get? Yes, exactly! The total change in momentum, delta p. J equals delta p. This is the impulse momentum theorem, and it is the heart of today's lesson. Clear? Now see this dashed red line on the graph, F av, the small letters av meaning average. It is the steady force that gives the same area in the same contact time. So F av equals delta p over delta t. Got it? And the unit of impulse is the newton second, N s, the same as kilogram metre per second, k g m s to the power minus one. The unit of momentum, of course! Right?

2:07Same change in momentum, different contact times

Now, suppose the change in momentum is fixed, but the contact lasts ten times longer. What happens to the average force? Yes, exactly! It becomes one tenth! Same delta p, longer delta t, so a smaller average force. This double arrow means so. Okay? That is why a fielder's hands move back, why we land on sand, and why cars have crumple zones. Each one makes the stopping time longer. Clear? Now, our puzzle in numbers. You catch a one hundred and fifty gram ball, g for grams, moving at twenty metres per second. Delta p is zero point one five times twenty, the cross meaning times: three newton seconds, however you catch it. Right? Hands drawn back, the ball stops in zero point one second. F av is three over zero point one: thirty newtons, N for newtons. With me? Stiff hands, it stops in zero point zero one second. Three over zero point zero one: three hundred newtons! Ten times the force on your hands. That is why stiff hands sting. Puzzle solved!

3:16Impulsive forces in a collision

Now, the force in a collision is huge, and it acts for a tiny time. We call such a force impulsive. Okay? Now, a trap. Two balls collide in mid air, and gravity keeps acting on both of them even during the collision, right? So, is their total momentum conserved across the collision, or not? Many students say no, because gravity is an outside force. Careful here! Gravity and friction are finite forces. In that tiny delta t, their impulse, force times time, is negligible. Got it? So yes, the total momentum of the colliding bodies is conserved across the collision, even when gravity acts. You will see how tiny gravity's share is, in numbers, in Example two. Clear?

4:02Ball striking a wall at an angle

One more result that comes up often. Picture a ball hitting a wall with speed v, at an angle theta to the normal, the line perpendicular to the wall. Theta is this Greek letter. It rebounds elastically, so it leaves with the same speed, at the same angle. Okay? So tell me, which part of the velocity does the wall actually change? Only the normal part! The component along the wall is unchanged, but v cos theta reverses: in towards the wall, then out, away from it. With me? So, with m the mass, and away from the wall taken as positive, the normal momentum goes from minus m v cos theta to plus m v cos theta. The change is twice that: J equals two m v cos theta, along the normal, away from the wall.

4:50Example 1: impulse and force on a ball hit straight back

Got it? Now, Example one, an easy one. There is a zero point one six kilogram ball, moving at twenty five metres per second, and it is hit straight back at thirty five metres per second. Find the impulse, and the average force if the contact lasts eight milliseconds, written m s. Pause the video and try it. Done? Okay, let's check. The change in velocity is thirty five minus twenty five, that is ten, right? No, no, careful! This is the classic trap. The ball comes in at twenty five, and goes back at thirty five, in opposite directions. Take the return direction as positive, and the velocity goes from minus twenty five to plus thirty five. Clear? So the speeds add: delta v is thirty five plus twenty five, which is sixty. J equals m times delta v: zero point one six times sixty, which is nine point six newton seconds. Okay? Now, eight milliseconds is zero point zero zero eight seconds. F av equals nine point six over zero point zero zero eight: one thousand two hundred newtons! That is a huge force, from one short hit. That's it!

6:07Example 2: average force of the floor on a bouncing ball

Now a harder one, JEE level. There is a zero point one kilogram ball, dropped from five metres, and it rebounds to one point eight metres. The contact lasts zero point zero one second. Find the average force of the floor on the ball, with g as ten. Pause and try it. Done? Let's check. First, the speed just before it hits the floor. It falls five metres from rest: v equals root two g h, with h the height; this is the root sign. Root of two times ten times five: ten metres per second, downwards. Okay? It rises to one point eight metres, so it must leave the floor at root of two times ten times one point eight: six metres per second, upwards. Right? Now, ten metres per second down, then six up. Tell me, what is the change in momentum? Zero point one times ten plus six: one point six newton seconds, upwards. Yes, the speeds add again, because the direction reverses, just like Example one. Clear? Now, the second trap. Delta p over delta t: is that the force of the floor alone? Not quite! It is the net force. During contact, the floor pushes up with N, here the normal force, not newtons, and the weight m g pulls down. So N minus m g equals one point six over zero point zero one: one hundred and sixty newtons. With me? So N equals one hundred and sixty, plus zero point one times ten: one hundred and sixty one newtons. And look, gravity's share is just one newton out of one hundred and sixty one! That is the impulsive force idea, in numbers. Perfect!

8:02Quick revision: impulse, average force and rebounds

So, quick revision. Impulse is the integral of F d t, the area under the force time graph, and it equals delta p. F av is delta p over delta t: same delta p, longer contact, smaller force. When a ball bounces straight back, the speeds add, and in a collision, the impulse of gravity is negligible. Now go and try the practice problems on this page. You did really well today. All the best!