Mixed Problems: Set 1 (Momentum, Equilibrium and Pulleys)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Here is a three kilogram lamp, hanging from two strings, each at sixty degrees to the ceiling. Its weight is thirty newtons, shared by two strings. So each string carries fifteen newtons, right? Hold on to that answer; we will test it soon. Today is our first set of mixed problems on the laws of motion: momentum, equilibrium and pulleys. Problems like these come up often in JEE Main. So, let's begin.

0:34Idea 1: force is the rate of change of momentum

Idea number one: momentum. Here, a ball of zero point five kilograms hits a wall at twelve metres per second, written m s to the power minus one, and rebounds at eight, along the same line. The contact lasts zero point zero two seconds. Find the average force of the wall, okay? First, the rule. The average force, F, equals delta p over delta t. Delta, this Greek triangle, means change in; p is the momentum, and t is the time. So, the change in momentum, over the time it takes. Clear? Now, the ball came in at twelve and went out at eight. So the change is twelve minus eight, which is four, and delta p is zero point five times four, right? No, no, careful! This is the trap. Momentum has a direction, and the ball turns round: twelve one way, then eight the other way. So the change is twelve plus eight. Delta p equals m times the sum of u and v, where m is the mass, u the speed in, and v the speed out. With me? So delta p is zero point five times twelve plus eight, that is, zero point five times twenty, which is ten. The unit, N s, is the newton second, the same as kilogram metre per second. Okay? Now divide by the time: ten over zero point zero two. Five hundred newtons, and N stands for newtons. So whenever a body turns back, add the speeds. Got it?

2:03Idea 2: equilibrium, resolve the forces

Now, idea number two: our puzzle. Here, a three kilogram lamp hangs from two strings, each at sixty degrees to the horizontal ceiling; the little circle means degrees. Two strings share thirty newtons, so fifteen newtons each, right? Let's check it properly. The lamp is at rest, so it is in equilibrium. The big Greek sigma means the sum of, and F x and F y, with the small x and y written below the F, are the forces along the horizontal x axis and the vertical y axis. Both sums must be zero. Okay? Each string pulls along itself with tension T, at sixty degrees. Its horizontal part is T cos sixty degrees, where cos is the cosine. The two strings lean opposite ways, so these parts cancel each other. Right? Now the vertical. Each string gives T sine sixty degrees upwards, so two T sine sixty in all. This holds up the weight, m g. Taking g, the acceleration due to gravity, as ten, that is three times ten, thirty newtons. With me? So T equals thirty over two sine sixty, and two sine sixty is root three; the tick is the square root sign. Thirty over root three is ten root three. The wavy equals sign means approximately: about seventeen point three newtons. More than fifteen! Only the vertical parts hold the lamp up, so each string must pull harder.

3:29Example 1: Atwood machine acceleration, tension, distance

Clear? Alright, now pulleys, with example one. Here, an Atwood machine carries six kilograms and four kilograms. Find the acceleration, the tension, and the distance each mass moves in two seconds, starting from rest. Pause the video, and try it first. Done? Okay, let's check. On the left of the figure is the Atwood machine: masses m one and m two over one pulley; the small one and two are labels. The heavier, m two, goes down with acceleration a, and m one goes up with the same a; the green arrows show these accelerations. On the right, a block, capital M, on a table, is pulled by a hanging small m; same idea. With a light string and a light frictionless pulley, the tension T is the same on both sides, okay? Now, a quick question. The six kilogram block weighs sixty newtons. So the tension in the string is sixty newtons, right? No, no! If the tension were sixty, the six kilogram block would feel no net force, and could not speed up. So, the Atwood result: a equals m two minus m one, times g, over m one plus m two. Here m two is six and m one is four: six minus four, times ten, over six plus four, which is two metres per second squared, written m s to the power minus two. Clear? Now the tension: two times m one times m two times g, over m one plus m two. Two times six times four times ten, over ten: forty eight newtons. See, less than sixty, just as it must be. Perfect! Last, the distance s: from rest, s equals half a t squared, where t is the time. Half times two, times two squared: four metres. So each mass moves four metres, one up and the other down. Got it?

5:22Example 2: movable pulley accelerations and tension

Now a harder one, JEE level. Here, a six kilogram block hangs from a movable pulley. The string round this pulley has one end fixed to the ceiling; the other end passes over a fixed pulley, and carries a two kilogram block. Find the accelerations and the tension. Pause the video, and try it. Done? Okay. In the Atwood machine, both blocks had the same acceleration. Here also there is one string, so the same acceleration again, right? Oh no, careful! This is the trap. The string's length is fixed, and the six kilogram block hangs on two strands of it. If the block goes down some distance, both strands get longer by that much, and that string comes from the other side. So the two kilogram block rises twice as far; the six kilogram block moves only half as far, okay? So the accelerations are a, down, for the six kilogram block, and two a, up, for the two kilogram block. Clear? Now the six kilogram block, with its pulley, taking the strings and pulleys as light and frictionless. Two strands pull it up, each with tension T, so two T upwards, and its weight, sixty newtons, pulls down. It goes down with a, so sixty minus two T equals six a. Okay? And the two kilogram block: tension up, twenty newtons of weight down, and it rises. So T minus twenty equals its mass, two, times its acceleration. And that acceleration is two a, not a! So T minus twenty equals four a. Right? Now solve. From the second equation, T equals twenty plus four a. Put it into the first: sixty, minus forty, minus eight a, equals six a. The double arrow means implies: fourteen a equals twenty, so a equals ten over seven, about one point four metres per second squared. With me? So the six kilogram block goes down at about one point four, and the two kilogram block goes up at twice that: twenty over seven, about two point nine metres per second squared. And the tension is twenty plus forty over seven: one hundred and eighty over seven, about twenty five point seven newtons. Get that doubled acceleration right, and the rest is easy. Very good!

7:38Quick revision: momentum, equilibrium and pulleys

So, quick revision. Momentum: force is delta p over delta t, and when a body turns back, add the speeds. Equilibrium: resolve the forces, and the sums along each axis are zero. Pulleys: the same tension all along a light string, and a movable pulley moves half as far as the free end. Now go and try the practice problems on this page. All the best!