Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Put a small block on top of a bigger block, on a smooth floor. Now pull only the top block, gently. Does the bottom block come along with it? Or does the top block simply slide off? Hmm, not so obvious, right? Keep your answer in mind; we will settle it with real numbers at the end. Today is the JEE Main part of the Laws of Motion problem bank: two quick ideas, then two full problems. Problems like these come up often, so let's begin.

0:33Idea 1: what does the machine in a lift read?

Idea number one. Here, a sixty kilogram man stands on a weighing machine inside a lift, and it reads seven hundred and twenty newtons; the capital N after the number is the newton, the unit of force. Find the lift's acceleration. But first, think. What does this machine actually measure? His weight? Not his weight! It measures the normal force N, the push between his feet and the machine. Look at the figure: three lifts, accelerating up, at constant velocity, and accelerating down. The blue arrow N is that push, upwards. The green arrow m g is his weight, downwards. And the dark arrows marked a show the lift's acceleration. Okay? Now Newton's second law, as seen from the ground, with up as positive. The net upward force, N minus m g, equals m times a. So N equals m times, g plus a. That is the formula under the left picture, accelerating up. In the middle, constant velocity, N is just m g; and on the right, accelerating down, it is m times, g minus a. Right? In this problem bank, g is ten metres per second squared. So seven hundred and twenty equals sixty times, ten plus a. Seven hundred and twenty over sixty is twelve, so ten plus a equals twelve. With me? So a equals two metres per second squared; the m s to the power minus two is metres per second squared. And it is upwards, because a came out positive, and we took up as positive. Clear? Now a quick question. The acceleration is upwards. So the lift must be moving up, right? No, no, careful! That is the trap. Only the direction of the acceleration matters, not the direction of motion. The lift could be going up and speeding up, or going down and slowing down. Both give the same reading. Got it?

2:34Idea 2: impulse when the ball turns back

Alright, idea number two. Here, a zero point two kilogram ball reaches a bat at thirty metres per second; the m s to the power minus one means metres per second. It leaves the bat at forty metres per second, in the opposite direction. Contact lasts zero point zero one seconds; the small s is seconds. Find the impulse, and the average force, okay? Impulse, J, is the change of momentum. The little triangle is delta, it means change in, and p is momentum. So J equals m v minus m u, where u is the velocity before the hit, and v is the velocity after. Clear? Now, thirty before, forty after. So the velocity changes by just ten metres per second, right? Oh no, careful! The ball turns back. Velocity has a direction, so we need signs. Take the outgoing direction as positive. Then u is minus thirty, and v is plus forty metres per second. Okay? So J equals zero point two, times forty minus minus thirty. That is zero point two times seventy: fourteen newton seconds. The N s is the newton second, the unit of impulse. If you had used ten, you would get just two newton seconds, seven times too small! So when a ball turns back, add the two speeds. Right? Now the average force, F with av at its foot, for average. It is the impulse over the contact time, delta t. Fourteen over zero point zero one: one thousand four hundred newtons! Such a huge force, for just one hundredth of a second. Forces like this are called impulsive. Got it?

4:17Example 1: distance in 5 s, force removed after 3 s

Now, example one, an easy one. Read it with me: an eight newton force pushes a two kilogram body from rest, on a smooth floor, for three seconds, and then the force is removed. How far does the body go in the first five seconds? Pause the video, and try it first. Done? Okay, let's check. The floor is smooth, so the only horizontal force is the push, capital F, eight newtons. Second law: a equals F over m, eight over two, four metres per second squared. Clear? At t equals three seconds, where t is the time: v equals a times t, four times three, twelve metres per second. And the distance s equals half a t squared: half times four times nine, eighteen metres. With me? Now the force is removed. No force, so the body stops, right? No, no, wait! That is the trap. Zero net force means zero acceleration, not zero velocity. That is Newton's first law. The floor is smooth, so there is no friction either. So v just stays twelve metres per second. Okay? So in the next two seconds, it goes twelve times two, twenty four metres. Total: eighteen plus twenty four, forty two metres. Two stages, two different motions: accelerated, then uniform. Perfect, let's go on.

5:47Example 2: do the two blocks move together

Now, our puzzle from the start, with real numbers, and it is a little harder. Here, a one kilogram block lies on a two kilogram block, on a smooth floor. Between the blocks, mu is zero point two; mu, this Greek letter, is the coefficient of friction. Now a three newton force pulls the upper block. Do they move together? Pause the video and try it. Hint: first assume they move together. Done? Okay, let's check. Small m is the upper block, one kilogram, and capital M is the lower block, two kilograms. If they move together, the only horizontal force on the pair is the pull, capital F, three newtons, because the floor is smooth. So a equals F over, small m plus capital M: three over three, one metre per second squared. Right? Now look at the lower block alone. Nobody pulls it; the three newtons act on the top block, and the floor is smooth. So what makes it accelerate? Friction from the upper block, exactly! It is the only horizontal force on the lower block. So the friction needed, small f, equals M times a: two times one, two newtons. Clear? Now, how much friction can these surfaces give, at most? Mu times the normal reaction between the blocks, and that is the upper block's weight, m g. Zero point two times one times ten: two newtons. But careful: this is only the largest possible value. Static friction is not automatically mu times the normal reaction. Okay? So, two newtons needed, and two newtons available. Together, or slip? Together, yes, but only just! Friction is exactly at its limit. Pull a little harder than three newtons, and the top block slips. So, our puzzle: with these numbers, the bottom block does come along, but only just. Assume together, find the friction needed, compare. That's it!

7:52Quick revision: lift, impulse, two blocks with friction

So, quick revision. In a lift, the machine reads the normal force N, and what matters is the direction of the acceleration, not the direction of motion. When a ball turns back, use signs, so the two speeds add: here, m times, thirty plus forty, and the average force is the impulse over the contact time. When the force is removed on a smooth floor, the velocity stays constant. And for two blocks: assume they move together, then compare the friction needed with mu m g. Now try the practice problems on this page, and then attempt Part A as a test, with a clock on the table. All the best!