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How to use the mixed JEE Main-pattern problems on the laws of motion like a real test: three sets on momentum, equilibrium and pulleys, on friction and accelerated frames, and on circular motion, one to two minutes a problem, g = 10, the try-first rule, and two solved problems: a ball rebounding from a wall and two blocks that slip.
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Transcript
0:00Introduction
Hello everyone! Okay, one small puzzle before we start. Picture this: a ball hits a wall at twelve metres per second, and bounces straight back at eight metres per second. The speed changed by just four. So the wall hardly had to push, right? Hmm, hold that thought. We will settle it in a few minutes. Questions on the laws of motion come up often in JEE Main, and each one should take you only one to two minutes. Today I will show you how this chapter of mixed problems trains exactly that speed, and then we will solve two of its problems together. So, let's begin.
0:37What this chapter covers
So, first of all, why this chapter? Questions on the laws of motion come up often in JEE Main, okay? And they come from every corner of this unit. This chapter puts them into three sets. Set one: momentum and impulse, equilibrium, and pulleys. Right? Set two: friction on floors and on inclines, one block sitting on another block, and lifts. Clear? Now, a quick game. Picture a car going round a curved road without skidding. Which set will you find it in? Set three, yes, exactly! Circular motion: cars on roads, and coins on turntables. See, before you solve anything, first spot the idea. That is the real skill here. Got it? And where are the real JEE Main questions, from twenty twenty one to twenty twenty six? At the end of Chapters one to seven. The problems in this chapter are original, written in the same style. So, fresh practice for you. With me?
1:40How to use each set
Now, how should you use this chapter? Each problem should take you about one to two minutes. Not ten minutes, okay? One to two. So tell me. Each set has fifteen solved problems. How long should the whole set take you? Fifteen times one, that is fifteen minutes, up to fifteen times two, thirty minutes. Correct! So one set is a short timed test of fifteen to thirty minutes. Right? Next, g, the acceleration due to gravity. Take g as ten; on screen, m s to the power minus two, which means metres per second squared. But only unless told otherwise! If a question gives a different value, use that one. Clear? Now, here comes the trap. You are on problem five, you feel a little stuck, and the solution is right there. Just one small peek, right? No, no, careful! Reading a solution feels like solving, but it is not. So the rule is: attempt the whole set first, as a test, and only then read the solutions. Then check your answers one by one. Okay?
2:44Example 1: average force on a ball rebounding from a wall
Alright, remember our puzzle? Here it is, as a problem from Set one, an easy one. Read it with me: a zero point five kilogram ball hits a wall at twelve metres per second, and rebounds at eight metres per second, along the same line. On screen, m s to the power minus one means metres per second. The contact lasts zero point zero two seconds. Find the average force of the wall. Pause the video, and try it. One to two minutes! Done? Okay, let's check. The speed goes from twelve to eight. So the change is twelve minus eight, just four, right? No, no, careful! That is the trap from our puzzle. Velocity has a direction. The ball was going into the wall, and now it is coming away from it. So the velocity drops from twelve to zero, and then grows from zero to eight the other way. The total change is twelve plus eight, twenty metres per second. Got it? Now momentum, written p. This little triangle is the Greek letter delta, and it means change in. Delta p equals the mass times the change in velocity: zero point five times, twelve plus eight. That is ten newton seconds; on screen, N s. Okay? And the force is the change in momentum over the time: F equals delta p over delta t, where delta t is the contact time. Ten over zero point zero two: five hundred newtons. Use four instead of twenty, and your answer comes out five times too small! So the wall pushes hard after all, and our puzzle is settled. Clear?
4:26Example 2: two blocks slip, find both accelerations
Now a harder one, from Set two. Read it with me: a three kilogram block lies on a five kilogram block, on a smooth floor. Between the blocks, mu is zero point three; mu is this Greek letter, the coefficient of friction. Forty newtons pull the lower block. Find both accelerations. Pause the video, and try it. Done? Okay, let's check. In the picture, small m is the upper block, three kilograms, and capital M is the lower block, five kilograms. F is the forty newton pull, on the lower block only, and small f is the friction between the blocks. Now, many students write: forty over three plus five, so five metres per second squared, both blocks together. Right? No, no, wait! Check first. Nobody pulls the upper block directly. Only the friction from the lower block drags it forwards, and friction can be at most mu m g. So its acceleration can be at most mu g: zero point three times ten, three metres per second squared. This sign means less than or equal to. With me? So, what is the largest pull for which the two blocks can still move together? Both blocks together, three plus five, that is eight kilograms, at most three metres per second squared. Eight times three: twenty four newtons. Correct! And the text on the right of the picture says the same thing in letters: F at most mu g times, capital M plus small m. Okay? But our pull is forty, and forty is greater than twenty four; that is this sign. So they slip! Now friction takes its full value, mu m g: zero point three times three times ten, nine newtons. Clear? Upper block first: nine newtons forwards on three kilograms. We call its acceleration a one; the small one written below is just a label. Nine over three: three metres per second squared. Now the lower block, a two. Forty newtons forwards, and the same nine newtons of friction backwards. Forty minus nine is thirty one, over five: six point two metres per second squared. So the lower block runs ahead. And that five from before? Wrong for both blocks! Always check before you assume they move together. Very good!
6:46Quick revision: three sets, timing, try first
So, quick revision. Three sets: momentum and pulleys; friction and lifts; and circular motion. One to two minutes a problem, and g is ten unless told otherwise. And attempt the whole set first; read the solutions after. Now open Set one, start your timer, and go. Then try the practice problems on each page. You can do this, I promise. All the best!
More videos in Unit 3 · Laws of Motion
- Force, Inertia and the First Law
- Momentum and the Second Law
- Impulse
- The Third Law and Conservation of Momentum
- Thrust: Forces from a Stream of Mass
- The Common Forces
- Free-body Diagrams
- Equilibrium of Concurrent Forces
- Ropes, Chains and Spring Balances
- Bodies Moving Together
- Pulleys
- Constraint Relations
- Springs and Sudden Changes
- Chains and Ropes over Edges and Pulleys
- Pseudo Forces
- Lifts and Apparent Weight
- Accelerating Vehicles
- Wedges
- Static and Kinetic Friction
- Rough Horizontal Surfaces
- Rough Inclines
- More about Friction
- One Block on Another
- Blocks on Moving Surfaces
- Connected Bodies with Friction
- Friction in Accelerating Frames
- Centripetal Force
- Level Curves, Banked Roads and Railway Tracks
- Conical Pendulums and Rotating Bodies
- Rotating Frames and Speeding Up on a Curve
- Mixed Problems: Set 1 (Momentum, Equilibrium and Pulleys)
- Mixed Problems: Set 2 (Friction and Accelerated Frames)
- Mixed Problems: Set 3 (Circular Motion)
- Overview
- Part A: JEE Main Level
- Part C: Concept Checks