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Video lessons › Physics · Unit 4 · Work, Energy & Power · Chapter 8 · Mixed JEE Main-Pattern Problems

Overview

8-minute video lesson in English, narrated with a computer voice.

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How to use the mixed JEE Main-pattern problems on work, energy, power and collisions like a real test: three timed sets of fifteen, 25 minutes a set, g = 10 unless told otherwise, the try-first rule, and two solved problems: the work done when a body speeds up from 2 to 6 m/s, and a bullet that sticks in a pendulum block.

Chapters

  1. 0:00Introduction
  2. 0:41What this chapter covers
  3. 1:39How to use each set
  4. 2:49Example 1: work done as speed rises from 2 to 6 m/s
  5. 4:24Example 2: bullet speed from a pendulum block rising 20 cm
  6. 7:17Quick revision: the three sets, timing, two key habits

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Picture this: a bullet hits a block hanging on strings, and stays stuck inside it. The block swings up by just twenty centimetres. Twenty centimetres, that is nothing! So the bullet must have been quite slow, right? Hmm, hold that thought. We will measure its speed in a few minutes. Questions on work, energy, power and collisions come up often in JEE Main, and this chapter trains you to solve them fast, against the clock. Today I will show you how to use its three timed sets, and then we will solve two of its problems together. So, let's begin.

0:41What this chapter covers

So, first of all, what is in this chapter? Three timed sets, with fifteen original problems in each, written in the style of recent JEE Main papers. Okay? Set one mixes work, kinetic energy and potential energy, all in one set. Right? Set two mixes power, vertical circles, and potential-energy curves. Clear? Now, a quick game. Remember our bullet, stuck in the hanging block? Which set will you find it in? Set three, yes, exactly! Impulse, collisions and explosions. Our bullet sticking in a block is a collision. See, before you solve anything, first spot the idea. That is half the work. Got it? And where are the real past JEE questions? 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?

1:39How to use each set

Now, how should you use a set? Like a real test. Give yourself twenty five minutes per set, okay? So tell me. Fifteen problems in twenty five minutes. How much time does each problem get? Twenty five minutes is twenty five times sixty, that is fifteen hundred seconds; the small s on screen means seconds. Fifteen hundred over fifteen: one hundred seconds a problem. Correct! Less than two minutes each. So speed really matters here, right? Now, here comes the trap. You are on problem five, a little stuck, and the solution is right there. Just one quick peek, what is the harm? No, no, careful! Reading a solution feels like solving, but it is not. So, attempt the whole set first, against the clock, and only then check the solutions, one by one. Clear? And one more thing: 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 another value, use that one. Okay?

2:49Example 1: work done as speed rises from 2 to 6 m/s

Alright, let's solve. First, an easy one from Set one. Read it with me: the speed of a three kilogram body rises from two to six metres per second. On screen, m s to the power minus one means metres per second. Find the work done on it. Pause the video and try it. Less than two minutes! Done? Okay, let's check. The speed rose by four. So the work is half, times three, times four squared. Twenty four joules; the capital J on screen is joules. Easy, right? No, no, careful! That is the trap. Kinetic energy depends on the speed squared, so find the kinetic energy at each speed. The work energy theorem: W net equals K f minus K i. W, with the small word net below it, is the work done by all the forces. K with a small f is the final kinetic energy, and K with a small i, the initial one. With me? So, K f is half, times three, times six squared: fifty four joules. And K i is half, times three, times two squared: six joules. Okay? Fifty four minus six: forty eight joules. Or in one go, as on screen: half, times three, times, in brackets, thirty six minus four. Same forty eight! So the trap answer was just half of the truth. Clear? So remember: square each speed first, then subtract. Never square the change in speed. Got it?

4:24Example 2: bullet speed from a pendulum block rising 20 cm

Now, remember our puzzle? Here it is, as a problem from Set three. Read it with me: a twenty five gram bullet sticks in a two point four seven five kilogram pendulum block, which rises twenty centimetres. On screen, the small g is grams, and c m is centimetres. Find the bullet's speed. Pause the video and try it. Done? Okay, let's check. Look at the picture. Small m is the bullet, and the arrow u is its speed. Capital M is the block, and h is how high it rises. Now, many students write: the bullet's kinetic energy becomes the potential energy of block plus bullet: in brackets, capital M plus small m, then times g h. Energy is conserved, right? No, no, wait! The bullet sticks in the block. That is a perfectly inelastic collision, and in it, kinetic energy is lost, as heat and sound. So, across the impact, use momentum, not energy. The impact is so quick that the strings are still vertical, so no horizontal outside force acts, and momentum is conserved. Okay? After the impact, block and bullet swing up together. Capital V is their common speed just after the impact. During the swing only gravity does work, because the strings pull at right angles to the motion. So, energy is conserved: V equals the square root of two g h; that tick-shaped sign is the square root. Twenty centimetres is zero point two metres. Root of two times ten times zero point two is root four: two metres per second. Right? Now, the impact. We need capital M plus small m. Twenty five grams is zero point zero two five kilograms. So what is the total mass? Two point four seven five plus zero point zero two five: exactly two point five kilograms. Correct! Momentum before the impact is small m times u, the bullet alone. After it, capital M plus small m, in brackets, times capital V, okay? So u equals two point five over zero point zero two five, times two. Block plus bullet is one hundred times heavier than the bullet, so the bullet was one hundred times faster: two hundred metres per second! And on the right of the picture, you see the same two steps in letters: u equals capital M plus small m, over small m, times root of two g h. With me? And the energy-only answer from before? It gives just twenty metres per second. Ten times too small! So, our puzzle is settled: a swing of only twenty centimetres, from a bullet at two hundred metres per second. Most of its kinetic energy became heat and sound. Very good!

7:17Quick revision: the three sets, timing, two key habits

So, quick revision. Three sets: work and energy; power, circles and curves; and collisions. Twenty five minutes a set, and g is ten unless told otherwise. And our two traps: square the speeds first, and across an impact, use momentum, not energy. Now open Set one, start your timer, and go. Attempt all fifteen before you check a single solution. You can do this, I promise. All the best!

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