Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Picture this: a tiny bullet hits a wooden block hanging from long strings, and stays stuck inside. The bullet carries six hundred and seventy five joules of kinetic energy. If all of that became height, the block would rise forty five metres! But it rises less than half a metre. So where did the energy go? We will settle that at the end. Today, the JEE Main part of this unit's problem bank: two quick ideas, then two full problems, of a kind that comes up often. Let's begin.

0:37Idea 1: a constant force ignores the path

Idea number one. Here, a particle moves from zero comma zero to three comma two, in metres, along the curve y equals two ninths x squared. The force F has a small arrow on top, since it is a vector: two i hat plus four j hat newtons. The letters i and j with little caps are the unit vectors along x and y; capital N is newtons. The force is constant. Find the work done. Now, the path is a curve. Must we integrate along it? No, not at all! The key word is constant. For a constant force, the work W is F dot s, the dot product of the force with the displacement, s with an arrow. The displacement goes straight from start to end, so the shape of the path does not matter. Okay? So s is three i hat plus two j hat metres: the end point minus the start point. With me? Now multiply the matching parts and add: two times three, plus four times two, six plus eight. So W is fourteen joules; capital J is for joules. The curve was only there to scare you! But this works only for a constant force. Clear?

1:48Idea 2: power of a car climbing a slope

Alright, idea number two: power. Here, an eight hundred kilogram car climbs a slope of one in twenty five, along the road, at ten metres per second, against a resistance of six hundred newtons. Find its power. On screen, k g is kilograms, and m s to the minus one is metres per second, okay? In the figure, the red arrow, capital F, is the engine's force, up the slope. R is the resistance, down the slope, and so is m g sine theta, the part of the weight along the slope; theta, this Greek letter, is the angle of the slope. With me? Now, the speed is steady. So the engine just has to beat the six hundred newtons, right? No, no, careful! Gravity also pulls the car back, with m g sine theta. Steady speed means no acceleration, so the forces along the slope balance: F equals m g sine theta plus R, as on the right of the figure. Clear? One in twenty five along the road means you rise one metre for every twenty five metres of road. So sine theta is one over twenty five. Okay? In this problem bank, g is ten metres per second squared. So m g sine theta is eight hundred times ten times one over twenty five: three hundred and twenty newtons. Got it? So the engine's force F is three hundred and twenty plus six hundred: nine hundred and twenty newtons. Right? And the power P is F times v, force times speed, the last line in the figure: nine hundred and twenty times ten, nine thousand two hundred watts, or nine point two kilowatts. Here capital W means watts, not work, and k W is kilowatts. Forget the slope part, and you get only six hundred times ten, six kilowatts. That is the trap. Clear?

3:31Example 1: block pushed on a rough floor, then released

Now, example one. Here, a five kilogram block at rest is pushed four metres along a floor by a horizontal twenty five newton force, and then released. Mu, this Greek letter, is the coefficient of friction, zero point three. Find the total distance it moves. Pause the video, and try it first. Done? Okay, let's check. On a horizontal floor, the normal reaction is the weight, m g. So friction, small f, is mu m g: zero point three times five times ten, fifteen newtons, and it acts the whole time the block slides. Okay? Now the push: twenty five times four, so one hundred joules of kinetic energy, right? Oh no, careful! Friction acts during the push too. By the work energy theorem, the net work equals capital K, the kinetic energy gained. Net force, twenty five minus fifteen, ten newtons, over four metres: forty joules, not one hundred. Got it? After the release, only friction acts, and it eats up those forty joules. Fifteen times x equals forty, where x is the extra distance, so x is forty over fifteen: about two point six seven metres. The wavy equals sign means approximately. Many students stop here, but the question asks for the total distance. With me? So the total is four plus two point six seven: about six point six seven metres. Friction works in both stages. Perfect!

5:02Example 2: ballistic pendulum, bullet sticks in a block

Now, our puzzle, with real numbers. Here, a fifteen gram bullet at three hundred metres per second sticks in a one point four eight five kilogram block, hanging from long strings. The small g means grams. How high does the block rise? Pause the video and try it. Hint: two stages. Done? Okay, let's check. Quick idea: the bullet's kinetic energy, half times zero point zero one five times three hundred squared, is six hundred and seventy five joules. Put it all into height, and h is forty five metres. Done, right? No, no, wait! That is the trap. The bullet sticks, so the impact is perfectly inelastic, and kinetic energy is lost right there. But the impact is very short and the strings are vertical, so nothing pushes sideways: horizontal momentum is conserved. In the figure, the bullet, small m, comes in with speed u, the red arrow; capital M is the block. So m u equals capital M plus m, times capital V, the common speed just after. Clear? Fifteen grams is zero point zero one five kilograms, so m u, zero point zero one five times three hundred, is four point five. The total mass is exactly one point five kilograms. So capital V is four point five over one point five: three metres per second. Okay? Now the swing. The string tension is perpendicular to the motion, so it does no work: energy is conserved. The mass cancels, and half V squared equals g h, where h is the rise, the small double arrow; the dashed outline is where the block started. Beside the figure, this reads V equals root two g h; the tick shaped sign is the square root. With me? So h is V squared over two g: nine over twenty, zero point four five metres. Forty five centimetres, not forty five metres! And the last line beside the figure joins both stages into one formula for the bullet's speed u. Right? Now, the missing energy. Before the impact, six hundred and seventy five joules; just after, half times one point five times three squared, only six point seven five joules. The arrow means becomes. So ninety nine percent is lost in the impact, as heat, sound and deformation. Puzzle solved! Momentum across the impact, energy only for the swing. That's it!

7:28Quick revision: slope power, friction and the bullet trap

So, quick revision. Constant force: W is F dot s, and the path does not matter. Steady speed up a slope: power is m g sine theta plus R, times v. Friction acts during the push too, and the total distance includes the push. Bullet sticks: momentum across the impact, energy only during the swing. Now try the practice problems on this page, then take Part A as a timed test. All the best!