Explosions and Energy Release

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. When a rifle fires, the bullet flies forward and the rifle kicks back, with equal and opposite momenta. So, do they also get equal kinetic energy? If they did, your shoulder would be hit as hard as the target! Today: explosions, where momentum is conserved but kinetic energy is not. This comes up often in JEE, so let's begin.

0:29Explosions: internal forces only

First, the explosion itself. Take a bomb at rest. Bang! Pieces fly off in every direction. Now tell me: in an explosion, is momentum conserved? Yes, it is! The pieces push on each other, so the forces of the explosion are internal forces only. Okay? And internal forces cannot change the total momentum. So, p after equals p before. The small arrow on top of p means a vector, and after and before, written low, are just labels. Clear? Now kinetic energy. Before the bang, the bomb was at rest: zero kinetic energy. So, is kinetic energy also conserved? No, no, not at all! It goes up. K after equals K before, plus E, where capital K is kinetic energy and capital E is the energy released by the explosion. So remember: momentum is conserved, but kinetic energy increases. Got it?

1:31A body at rest breaks into two

Now, the common case: a body at rest breaks into two pieces, with masses m one and m two, and speeds v one and v two. The small one and two written low are just labels. Zero momentum before means zero after. So the pieces get equal and opposite momenta: m one v one equals m two v two, in opposite directions. Right? So tell me: which piece moves faster, the heavy one or the light one? The lighter one, of course! v one over v two equals m two over m one. See, the masses come in upside down. Clear? Now, back to our rifle puzzle. Equal momenta. So who gets more kinetic energy? Use K equals p squared over two m: kinetic energy in terms of the momentum, p. Both pieces have the same p, so K goes as one over m. So, K one over K two equals m two over m one. The lighter piece takes more energy! That is why the bullet does the damage, and the rifle only kicks your shoulder. Puzzle solved! With me? To share out the energy released: K one equals m two, over m one plus m two, times E, okay? For K two, swap the masses: m one, over m one plus m two, times E, right? So each piece's share has the other piece's mass on top. So: the lighter piece gets the larger speed, and the larger share of the energy. Giving the bigger share to the heavier piece is a common trap. Got it?

3:01A projectile that explodes in flight

Now, a shell flying along its parabola explodes in mid air, and the pieces fly apart. Okay? So, what happens to the centre of mass of all the pieces? Does it jump off the path? Not at all! The forces of the explosion are internal, and gravity pulls on every piece just as before. So the centre of mass continues on the original parabola, as if nothing had happened. Neat, isn't it? But one condition: only until a piece lands. Then the ground pushes on that piece, an outside force. Clear?

3:32Example 1: body at rest splits in two, energy released

Here, a five kilogram body at rest explodes into pieces of two kilograms and three kilograms. The two kilogram piece moves at fifteen metres per second. Find the velocity of the other piece, and the energy released. On screen, k g is kilograms, and m s to the minus one is metres per second. Pause the video and try it. Done? Okay, let's check. The body was at rest, so the pieces get equal and opposite momenta. Two times fifteen equals three times v. The double arrow means "which gives". v equals ten metres per second, in the opposite direction. Easy, isn't it? Now the energy. Before the explosion: zero kinetic energy. So where does the released energy go? Into the kinetic energy of the pieces, yes! K one, for the two kilogram piece: half times two times fifteen squared, two hundred twenty five joules. The J stands for joules, okay? K two, for the three kilogram piece: half times three times ten squared, one hundred fifty joules. Right? So the energy released, E, is two hundred twenty five plus one hundred fifty: three hundred seventy five joules. See, the lighter piece took more, just like the bullet. Clear? Quick check with the share formula. The lighter piece gets three over two plus three, that is three fifths. Three fifths of three hundred seventy five is two hundred twenty five joules. Perfect!

5:05Example 2: shell explodes at the top, where it lands

Now a harder one, JEE level. Here, a four kilogram shell is fired at forty metres per second, at forty five degrees; the small circle means degrees. At the top of its path, it explodes into pieces of one kilogram and three kilograms, and the one kilogram piece falls straight down from rest. Where does the three kilogram piece land, and how much energy did the explosion release? Take g, the acceleration due to gravity, as ten. Pause the video and try it. Done? Okay, let's check. At the top, the vertical velocity is zero, so only the horizontal part is left: forty times cos forty five degrees, the cosine. About twenty eight point three metres per second. The wavy equals sign means approximately. Okay? Now the explosion. Its forces are internal, so horizontal momentum is conserved. The one kilogram piece is left at rest. So how fast does the three kilogram piece move? Four times twenty eight point three equals one times zero, plus three times v prime. The little dash on v is read prime. So v prime is four thirds of twenty eight point three: about thirty seven point seven metres per second, forward. With me? Now, where does it land? The centre of mass stays on the parabola, so this piece lands where the shell would have landed, right? No, no, careful! That is where the centre of mass lands, not this piece. First, the full range: capital R equals forty squared, times sine ninety degrees, over ten, and that comes to one hundred sixty metres. The top of the path is halfway, at eighty metres. Clear? This piece leaves the top horizontally, so it falls for the same time as the second half of the flight. In that time the shell would go eighty metres; this piece is four thirds as fast. So x, its distance from the gun, is eighty plus four thirds of eighty: about one hundred eighty seven metres. Got it? Let's check with the centre of mass. Both pieces start at the same height with no vertical velocity, so they land together; the one kilogram piece lands right below the top, at eighty metres. One times eighty, plus three times one hundred eighty six point seven, over four: about one hundred sixty metres. Exactly the full range! Beautiful! Last, the energy released: kinetic energy after, minus before. The one kilogram piece is at rest, so it has none. Half times three times v prime squared, minus half times four times v squared. Use exact squares here, not the rounded speeds: v squared is forty squared over two, exactly eight hundred, and v prime squared is sixteen ninths of eight hundred. So, about two thousand one hundred thirty three, minus one thousand six hundred: about five hundred thirty three joules. That's it!

7:59Quick revision: momentum kept, energy shares, centre of mass

So, quick revision. In an explosion, momentum stays unchanged, but kinetic energy rises by the energy released. From rest into two pieces: K one over K two equals m two over m one, so the lighter piece gets more. And in flight, the centre of mass stays on the original parabola, until a piece lands. Now go and try the practice problems on this page. All the best!