Perfectly Inelastic Collisions

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Bullets fly far too fast for any stopwatch. Now, I give you a block of wood, two strings, a weighing scale and a ruler. Can you measure the speed of a bullet with these? Sounds impossible, right? By the end of this lesson, you will do it in two lines. The idea is the perfectly inelastic collision, which comes up often in JEE, so let's begin.

0:28Bodies that stick: one common velocity

So, first, what does perfectly inelastic mean? The two bodies stick together, and then move as one body, okay? Like a bullet that stays inside a block. So after the collision, there is just one velocity for both. Now think. Across the impact, which quantity can we trust to stay the same? Momentum, yes, exactly! The forces between the two bodies are equal and opposite, so they cancel. And outside forces like gravity or friction give almost no impulse in such a short time. So the total momentum just before equals the total just after. On screen, m one and m two are the masses. The small one and two written low are subscripts, just labels. u one and u two are the velocities before, and capital V is the common velocity after. So, m one u one plus m two u two equals m one plus m two, in brackets, times capital V, right? Divide by the total mass, and you get the common velocity. Capital V equals m one u one plus m two u two, over m one plus m two. Total momentum over total mass. Clear? But wait. What if the two bodies come towards each other, head-on? Do we just add the two momenta? No, no, careful! Velocity has a direction. Pick one direction as plus, and a velocity the other way gets a minus sign. So head-on, the two momenta partly cancel. Got it?

2:00How much kinetic energy is lost?

Now, momentum is conserved. But what about kinetic energy, K E? Is that conserved too, when the bodies stick? No, it is not! Some kinetic energy is lost. How much? Delta K equals half, times mu, times u one minus u two, whole squared. Delta is the capital Greek letter that looks like a little triangle. Delta K is the loss of kinetic energy, and mu is another Greek letter. Okay? So what is mu? It is called the reduced mass: m one times m two, over m one plus m two. And u one minus u two is the relative velocity, so the signs matter here too. Head-on, the two speeds add. Right? For given starting velocities, this is the largest loss of kinetic energy possible. Clear? And one special case. If m two was at rest, the fraction of the kinetic energy lost is simply m two over m one plus m two. No velocities needed at all! Okay? So tell me. Suppose a light body hits a very heavy body at rest, and sticks. Is the loss big or small? Big! Very good! When m two is much bigger than m one, this fraction is close to one. So almost all the kinetic energy is lost. With me?

3:20Ballistic pendulum: momentum, then energy

Now, back to our puzzle. This is the ballistic pendulum, in the figure. On the left, a bullet of small mass m, flying with speed u, the red arrow. It hits a block of mass capital M, hanging on two strings, and stays inside. Then the block swings up and rises a height h, the small double arrow. The dashed box shows where the block started. On the right of the figure are the three equations we are going to build now, one by one. Okay? Quick idea. The bullet's kinetic energy becomes the potential energy of the block. Half m u squared equals capital M plus m, times g h. Done, right? No, no, careful! This is the trap. The bullet sticks, so the impact is perfectly inelastic, and kinetic energy is lost right there. So across the impact, use only momentum. m u equals capital M plus m, times capital V, where capital V is the speed of block and bullet together, just after the impact. Clear? After the impact, the block swings up, and now energy conservation is fine. The kinetic energy just after becomes potential energy at the top, the mass cancels, and capital V equals root two g h. The tick shaped sign is the square root, and g is the acceleration due to gravity. With me? Put the two together. u equals capital M plus m, over m, times root two g h. Weigh the bullet and the block, measure h with a ruler, and you have the bullet's speed! Got it? Never use energy conservation across an inelastic impact. Momentum for the impact, and energy only for the motion before or after it. Right?

5:01Example 1: 2 kg body sticks to a 4 kg body, KE lost

Alright, let's try one from the course. Here, a two kilogram body moving at six metres per second hits a four kilogram body at rest, and they stick. Find their common speed, and the kinetic energy lost. On screen, k g is kilograms, and m s to the minus one means metres per second. Pause the video and try it. Done? Okay, let's check. Total momentum: two times six, plus four times zero, since the second body is at rest: twelve. Total mass, six. So capital V is twelve over six: two metres per second. Easy, isn't it? Now the kinetic energy. Before the impact, which body has kinetic energy? Only the moving one, correct! K, the kinetic energy, is half times two times six squared: thirty six joules. The J on screen stands for joules. Okay? After the impact, both bodies move together. Oh no, careful: use the total mass, six kilograms, not four! Half times six times two squared: twelve joules. Right? So the kinetic energy lost is thirty six minus twelve: twenty four joules. Clear? Now a quick check. The second body was at rest, so the fraction lost is m two over m one plus m two: four over six, that is two thirds. Two thirds of thirty six is twenty four joules. It matches! Perfect!

6:34Example 2: bullet speed from a ballistic pendulum

Now, let's finish our puzzle with real numbers, using the same ballistic pendulum in the figure. Here, a ten gram bullet sticks in a block of one point nine nine kilograms, hanging on strings, and the block then rises five centimetres. Find the bullet's speed. Careful with the letter g: after the ten it means grams, and in the brackets, g equals ten is the acceleration due to gravity. Pause the video and try it. Done? Okay, let's check. We know the height, and we want the bullet's speed. So where do we start: from the impact, or from the swing? From the swing, yes! But first, the units. Grams and centimetres must become kilograms and metres. So m is zero point zero one kilogram. Capital M plus m is one point nine nine plus zero point zero one, that is two kilograms. And h is zero point zero five metre. Okay? Now the swing, with energy. Capital V equals the root of two times ten times zero point zero five. That is root one: one metre per second, just after the impact. With me? Now the impact, with momentum. u equals capital M plus m, over m, times capital V, right? That is two over zero point zero one, times one. Two hundred metres per second! Just a block, two strings and a five centimetre rise, and we have measured the speed of a bullet. Puzzle solved!

8:07Quick revision: common velocity, KE lost, ballistic pendulum

So, quick revision. When bodies stick, momentum gives the common velocity: total momentum over total mass, with signs for directions. The kinetic energy lost is half mu times u one minus u two, whole squared, the largest loss possible. And across the impact, only momentum; energy only before or after it. Now go and try the practice problems on this page. All the best!