Impulse

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Jump down from a wall one point eight metres high, and land with stiff legs. Ouch! Now jump again, but bend your knees as you land. Much softer, right? But both times you end up at rest, so the change in your momentum is exactly the same. Then why does one landing hurt so much more? Hold that thought. Today's idea, impulse, answers it, and impulse comes up often in JEE problems. So, let's begin.

0:33Impulse: the area under the force–time graph

So, here is the definition. J with an arrow on top is the impulse, a vector. The tall, stretched S is the integral sign, F with an arrow is the force, and d t is a tiny bit of time. So we add up force times time over the whole interval. On the graph, force F goes up and time t goes across, and that sum is the area under the curve, the blue hump. Okay? Now, think. Which quantity of the body does this area change? Its momentum, yes, exactly! By the second law, J equals delta p, the change in momentum. Capital delta, the little triangle, means change in, and p with an arrow is momentum. So J equals m v minus m u: mass m, final velocity v, initial velocity u, each velocity with an arrow. And N s in the bracket is the unit, newton second. Got it? Next, the average force, F with an arrow and the small letters a v below. It equals delta p over delta t, the time for which the force acts. On the graph, it is the red dashed line: the height of a rectangle with the same area as the hump.

1:48Impulsive forces and softer landings

Clear? Look at the blue hump again: a very large force for a very short time. Such forces are called impulsive forces, like a bat hitting a ball, or a collision between two cars. With me? Now, during such an impact, ordinary finite forces give almost no impulse. Take the weight, m g: mass times g, the acceleration due to gravity. It is finite and the impact is brief, so its impulse is negligible. But only for a really short impact; remember this for example two. Okay? Now the big question. Look at the green rectangle on the graph: the same impulse J, spread over a longer time. So, same change in momentum, short contact or long contact: which needs the smaller force? The longer one, correct! Average force is delta p over delta t: keep delta p fixed, make delta t bigger, and the force comes down. The double arrow on screen means implies. That is why the green rectangle is so much lower. Right? And this answers our puzzle! Bent knees stretch the stopping time, so the ground pushes on you with a much smaller force. Same idea when a fielder draws his hands back to catch a fast ball, and in the crumple zones of a car. Longer time, smaller force. Simple, isn't it?

3:09Ball hitting a wall: watch the signs

Now a classic. Take a ball of mass m. It hits a wall along the normal, the line perpendicular to the wall, at speed u, and rebounds at speed v. So the impulse is m times, v minus u. Final minus initial, just as we said. Correct? No, no, wait! This is the trap. Momentum is a vector, and the ball goes back the other way. So take the rebound direction as positive: the initial velocity is minus u, and the change is m v, minus m times minus u. So J equals m times, u plus v. The speeds add! Clear? And if the rebound is elastic, the ball comes back with the same speed: v equals u. Then J equals two m u. Twice the momentum it came in with! Forget the sign, and m times, v minus u gives you zero here. Careful! With me? Last case: a smooth wall, hit at an angle, which we call an oblique impact. Only the component of the velocity, v with an arrow, normal to the wall changes; the component along the wall stays the same. Note the condition: a smooth wall. Okay?

4:17Example 1: ball hit straight back, impulse and force

Alright, example one. Here, a zero point one five kilogram ball moving at twenty metres per second is hit straight back at thirty. On screen, k g is kilogram, m s to the minus one is metres per second, and s is seconds. Find the impulse, and the average force if the contact lasts zero point zero one second. Pause the video, and try it first. Done? Okay, let's check. Many students write: thirty minus twenty, a change of ten metres per second. Is that right? Oh no, careful! The ball is hit straight back, so its direction reverses. Take the final direction as positive: the initial velocity u is minus twenty, and the final velocity v is plus thirty. Okay? So, J equals zero point one five times, thirty minus minus twenty. Two minus signs make a plus: zero point one five times fifty, which is seven point five newton second. Notice, it is just the mass times the sum of the two speeds, twenty plus thirty, our wall result. Got it? Now the average force: J over delta t, seven point five over zero point zero one. Seven hundred and fifty newtons; capital N here is newton, the unit. That is a huge force on a small ball: an impulsive force! Clear?

5:42Example 2: landing with stiff legs vs bent knees

Now example two, our puzzle from the start! Here, a sixty kilogram person jumps down from a one point eight metre wall. Find the average force of the ground on the person if the landing takes, part a, zero point zero five second with stiff legs, and part b, zero point three second with bent knees. Pause the video, and try it. Done? Okay, let's check. After falling through a height h, the landing speed v is the square root of two g h; the tick-shaped sign is the square root. With g equal to ten metres per second squared, that is the root of two times ten times one point eight, root thirty six: six metres per second. With me? From six metres per second, downwards, to rest: the change in momentum is sixty times six, three hundred and sixty newton second, upwards. Okay? Now, think. While the person is landing, which forces act on them? Two forces, yes! Capital N, the normal reaction of the ground, pushing up, and the weight, m g, pulling down. So the net force, N minus m g, equals delta p over delta t. The trap is writing N equals delta p over delta t. The ground must also hold up the weight! Clear? Part a, stiff legs. Three hundred and sixty over zero point zero five is seven thousand two hundred. Add m g, six hundred: N is seven thousand eight hundred newtons. The first capital N is the normal reaction; the last is newton, the unit. Got it? Part b, bent knees. Three hundred and sixty over zero point three is twelve hundred, plus six hundred: eighteen hundred newtons. Same fall, same change in momentum, but less than a quarter of the force! Puzzle solved. And notice: next to twelve hundred, the six hundred from m g is not small, so here we keep gravity. Right?

7:49Quick revision: impulse, wall rebound, softer landing

So, quick revision. One: impulse is the area under the force time graph, and it equals the change in momentum. Two: when a ball rebounds head-on from a wall, along the normal, the speeds add, so J equals m times, u plus v. Three: the same change in momentum over a longer time needs a smaller average force. Now go and try the practice problems on this page. You can do them. All the best!