Collisions Through Springs and on Wedges; Chains of Collisions

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Picture a block sliding up a smooth ramp and back down. If the ramp is fixed to the floor, the block comes back with the same speed, just reversed, right? Now put the ramp on a smooth floor, free to slide. Does the block climb as high? And does it come back with the same speed? Today: collisions spread out in time, through a spring and on a wedge, and then chains of collisions. These come up often in JEE, so let's begin.

0:32A spring between two blocks

So, first, the spring. Two blocks on a smooth floor, a light spring between them, and one block runs into the spring. The blocks never touch; the spring pushes on both, over a longer time. So it is still a collision, just spread out in time. Okay? Now think. The spring gets shorter, and then, for one moment, stops getting shorter. At that moment of greatest compression, what can you say about the two blocks? Yes, exactly! They move with the same velocity. The floor is smooth, so momentum is conserved. So the common velocity, capital V, equals p total over m one plus m two. p total is the total momentum, and m one and m two are the masses; the small one and two written low are just labels. Clear? Now, how much energy does the spring store? All the kinetic energy, right? No, no, careful! Both blocks are still moving, so some kinetic energy is still there. The spring stores only the part a perfectly inelastic collision would lose. So, half k x max squared equals half mu v rel squared. k is the spring constant, x max the greatest compression, mu, the Greek letter, the reduced mass, and v rel the relative velocity of the blocks before the collision. With me? Solve it, and x max equals v rel, times the square root of mu over k. The tick shaped sign is the square root. And mu is m one m two, over m one plus m two, the reduced mass you met in perfectly inelastic collisions. Got it? And after that? The spring pushes the blocks apart until it is relaxed again, and gives back all its energy. So the final result is exactly that of an elastic collision. Right?

2:23A block on a wedge that is free to slide

Now the wedge. Here, a block of mass m runs at speed u onto a smooth wedge of mass capital M, which sits free on a smooth floor. As the block climbs, it pushes the wedge forward. Okay? So tell me. At the greatest height, does the block stop completely? No! Very good if you said no. It stops only relative to the wedge. Block and wedge move together with one common horizontal velocity. Just like the spring: greatest compression there, greatest height here. Clear? The floor is smooth, so horizontal momentum is conserved, and with no friction anywhere, mechanical energy is conserved too. Together they give h equals capital M over capital M plus m, times u squared over two g. The dot means times, and g is the acceleration due to gravity. Now, u squared over two g is the height on a fixed ramp, and capital M over capital M plus m is less than one. So on a free wedge, the block climbs lower. First half of our puzzle, solved! Got it? And when the block slides back down to the floor? No friction anywhere, so no energy is lost. So again, the result is that of an elastic collision. Right?

3:40Chains of collisions

Last idea: chains of collisions. One block hits a second and they stick, then the pair hits a third, and so on, every collision perfectly inelastic. So, must we solve every collision, one by one? No need at all! Momentum is conserved in every impact, so the total never changes, and in the end everything moves together. So the final velocity, capital V, is p total over the total mass, from total momentum alone. One line! Okay?

4:11Example 1: greatest compression of a spring between blocks

Let's try one. Here, a two kilogram block at three metres per second runs into a light spring, fixed to a four kilogram block at rest on a smooth floor. The spring constant k is three hundred newtons per metre. Find the greatest compression. On screen, k g is kilograms, m s to the minus one is metres per second, and N m to the minus one is newtons per metre. Pause the video and try it. Done? Okay, let's check. At greatest compression, the blocks move together. Total momentum: two times three, plus four times zero, that is six. Total mass, six. So capital V is one metre per second. Easy, isn't it? Kinetic energy before: capital K equals half times two times three squared, that is nine joules. The J on screen stands for joules. Okay? At greatest compression, both blocks move at one metre per second, so use the total mass, six kilograms. Half times six times one squared: three joules. Right? So how much energy is stored in the spring now? Nine minus three: six joules, yes! Not the full nine; that is the trap. So half times three hundred, times x squared, equals six, with x the compression. Clear? That gives x squared equals zero point zero four. The double arrow means so. x equals zero point two metre. With me? Quick check with the formula. Mu is two times four over six: four thirds. And v rel is three, since the second block was at rest. So x equals three, times the root of four thirds over three hundred: again zero point two metre. Perfect!

6:01Example 2: block slides up and back down a free wedge

Now a harder one, and it finishes our puzzle. Here, a one kilogram block slides at five metres per second onto a smooth four kilogram wedge, resting on a smooth floor. The block goes up and slides back down to the floor. Find the velocities of block and wedge after that. Pause the video and try it. Done? Okay, let's check. The wedge is smooth, so the block comes back at five metres per second, just reversed, right? No, no, careful! That is true only for a fixed ramp. Here the wedge moves. But everything is smooth, so no energy is lost, and horizontal momentum is conserved. So this is an elastic collision in effect. Okay? So use the elastic collision formulas, with the wedge at rest at the start. v one equals m one minus m two, over m one plus m two, times u one. v two equals two m one, over m one plus m two, times u one. Here m one is the block, m two the wedge, u one the starting speed, and v one and v two the final velocities. Right? Block: one minus four, over five, times five: minus three metres per second. The minus sign means backwards! So the block comes back at three, not five. Second half of our puzzle, solved! With me? Wedge: two times one, over five, times five: two metres per second, forwards. Clear? Quick check. Momentum: one times minus three, plus four times two: five, as at the start. Energy: half times one times three squared, plus half times four times two squared: twelve point five joules, also unchanged. The block comes back slower because the wedge takes some of the energy. Neat!

7:54Quick revision: spring compression, wedge height, chains

So, quick revision. Spring: a common velocity at greatest compression, and x max equals v rel times root mu over k. Free wedge: a common velocity at the greatest height, and h equals capital M over capital M plus m, times u squared over two g. Spring relaxed, or block back down: an elastic collision. For a perfectly inelastic chain, total momentum alone. Now go and try the practice problems on this page. All the best!