Set 3 (Impulse, Collisions and Explosions)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle first. Energy is never created or destroyed, right? So when two bodies collide, the kinetic energy just before must equal the kinetic energy just after. True, or false? Hmm, hold that thought. Today, two problems with the very same numbers will settle it. This is Set three: impulse, collisions and explosions, which come up often in JEE Main. First a quick toolkit, then we solve. Let's begin.

0:35One tool behind all of Set 3

So, what is inside Set three? Fifteen problems: impulse, bodies that stick, balls that bounce, springs, wedges, and explosions. Many stories, one tool. Okay? First, impulse. Capital J, the impulse, equals the integral of F d t, where F is the force; this tall curly sign is the integral sign. It equals delta p, the change in momentum; delta is the little triangle. And F av, with small a v below, is the average force: delta p over delta t, the contact time. Right? Now think. Two bodies collide, and gravity keeps pulling on them during the impact. Is their total momentum still conserved? Yes, it is! The contact forces are huge, and act for a very short time. Gravity is a small, finite force, so in that tiny time its impulse is negligible. So across an impact, total momentum is conserved, even with gravity. That is our one tool. Clear? And explosions? The forces that blow the body apart are internal, so the total momentum just before and just after is the same. But the kinetic energy increases, by the energy released. Keep that in mind for our puzzle. Got it?

1:56Toolkit: restitution and the energy lost

Now look at the figure, a head-on collision. Two balls, masses m one and m two; the small one and two written low are labels. Top row, before: velocities u one and u two. Bottom row, after: v one and v two. Middle row, maximum deformation, the greatest squeeze: one common velocity. Remember it. So, momentum: m one u one plus m two u two equals m one v one plus m two v two. Okay? But that is one equation for two unknowns, v one and v two. So we add the restitution rule: v two minus v one equals e, times u one minus u two. The letter e is the coefficient of restitution: the speed of separation over the speed of approach, along the line of impact. Both rules are also on the right of the figure. With me? And e tells you the type of collision. e equals one: elastic. e between zero and one, as the two less-than signs show: inelastic. And e equals zero: the bodies stick together, a perfectly inelastic collision. Right? Now tell me. For which value of e is the most kinetic energy lost? e equals zero, yes, exactly! When the bodies stick. The loss is half, times mu, times one minus e squared, times u one minus u two, whole squared. Mu is the Greek letter like a u with a tail. With e equal to one, the loss is zero; with e equal to zero, it is the biggest. Clear? And mu is the reduced mass: m one m two, over m one plus m two. So, a hint for our puzzle: with e equal to one, nothing is lost; with less, some is. Where it goes, we will see in Example one. Got it?

3:51Example 1: kinetic energy lost when two bodies stick

Alright, an easy one from Set three first. Read it with me: a three kilogram body at four metres per second sticks to a one kilogram body at rest. On screen, k g is kilograms, and m s to the power minus one means metres per second. Find the common speed, and the energy lost. Pause the video and try it. Done? Okay, let's check. Energy is always conserved, so half, times three, times four squared, equals half, times four, times capital V squared. Correct? No, no, careful! That is the trap. The bodies stick, so e is zero, and kinetic energy is lost right there. So across the impact, use momentum only. Three times four, plus one times zero, equals three plus one, times capital V, okay? The double arrow means so: capital V is three metres per second. Clear? Now the energy. Before, only the three kilogram body moves: half, times three, times four squared: twenty four joules. The capital J on screen is joules. Right? After, both move together at three metres per second. So use the total mass, three plus one, four kilograms, not three! Half, times four, times three squared: eighteen joules. Got it? So the energy lost is twenty four minus eighteen: six joules. Not destroyed, okay? It becomes heat, sound and deformation as the bodies squash and stick. Half our puzzle: here, six joules of kinetic energy are gone for good. With me? Quick check. Mu is three times one, over three plus one: three quarters. With e equal to zero, the loss is half, times three quarters, times four squared. Six joules again. Perfect!

5:53Example 2: greatest compression of a spring between blocks

Now the hard one, to finish our puzzle. Read it with me: blocks of one kilogram and three kilograms, joined by a spring, rest on a smooth floor. The spring constant k is three hundred newtons per metre; on screen, N m to the power minus one. The three kilogram block is given four metres per second towards the other. Find the greatest compression. See, the same numbers as Example one! Pause the video and try it. Done? Okay, let's check. The spring gets shorter, and then, for one moment, it stops getting shorter. At that moment, how do the two blocks move? Yes, exactly! With one common velocity, the greatest squeeze from our toolkit. Before it, the heavier block is faster and the gap closes; after it, the gap opens. Clear? The floor is smooth, so no horizontal outside force acts, and momentum is conserved. Capital V equals three times four, plus one times zero, over three plus one: three metres per second. Same as before! Okay? The kinetic energy at the start is twenty four joules, as before. So at the greatest compression, all twenty four joules sit in the spring, right? No, no, careful! That is the trap. Both blocks are still moving at three metres per second, so they keep half, times four, times three squared: eighteen joules. The spring holds only twenty four minus eighteen: six joules. The same six joules that were lost in Example one! Got it? So, half, times three hundred, times x squared, equals six, where x is the compression. x squared is zero point zero four, so x is zero point two metres. With me? Now the topper's way, in one line. x max equals v rel, times the square root of mu over k: max and rel, written small below, mean greatest and relative, and the tick shaped sign is the square root. v rel is four, and mu is zero point seven five, as in Example one. Four times root of zero point seven five over three hundred: four times zero point zero five, zero point two metres. Same answer! Very good! So, our puzzle. When the bodies stick, six joules become heat, sound and deformation, gone for good. With the spring, the same six joules hide in the spring for a moment. When it is relaxed again, it gives them all back: the result of an elastic collision. So kinetic energy is conserved only in elastic collisions; energy as a whole, always. Beautiful, right?

8:39Quick revision: momentum, restitution and energy lost

So, quick revision. Across an impact, conserve momentum, not kinetic energy. Restitution: v two minus v one equals e times u one minus u two, and the loss is half mu, times one minus e squared, times u one minus u two, whole squared. Spring between two blocks: one common velocity at the greatest compression. Now start your timer, and attempt all fifteen problems before you check a single solution. All the best!