Video: Using Conservation of Momentum to Deduce the Final Speeds of Exploding Bodies

A meteor with a mass of 7000 g flies through the air, straight toward the ground, at a constant speed of 1500 m/s. In midair, the meteor explodes into two pieces. The larger piece has a mass of 5000 g and the smaller piece has a mass of 2000 g. After the explosion, the smaller piece has an instantaneous downward velocity of 1200 m/s, as shown in the diagram. What is the instantaneous downward velocity of the larger piece?

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Video Transcript

A meteor with a mass of 7000 grams flies through the air straight toward the ground at a constant speed of 1500 metres per second. In midair, the meteor explodes into two pieces. The larger piece has a mass of 5000 grams and the smaller piece has a mass of 2000 grams. After the explosion, the smaller piece has an instantaneous downward velocity of 1200 metres per second, as shown in the diagram. What is the instantaneous downward velocity of the larger piece?

Okay, so, the diagram shows this meteor that’s falling towards the ground. And initially, it has a velocity of 1500 metres per second. Then, we’ve been told that the meteor explodes into two pieces, the smaller one travelling at 1200 metres per second downward, and the larger one travelling at a velocity 𝑣. We’re trying to find out what this value of 𝑣 is, knowing that the initial mass of the meteor was 7000 grams. And after the explosion, the smaller piece has a mass of 2000 grams, and the larger piece has a mass of 5000 grams.

Now to find the answer to this question, we need to recall what’s known as the law of conservation of momentum. In other words, we need to recall that momentum is conserved in an isolated system. Now, in this case, we’re considering a system that consists of our meteor and basically nothing else. And when the meteor explodes, there are no external influences on the meteor itself. In other words, in this case, we do have an isolated system because the environment around the meteor is not causing it to explode or doing anything to the meteor, for that matter. The meteor is exploding of its own accord. And so, there is no external force acting on the meteor.

Now since we have an isolated system, this must mean that momentum is conserved. In other words, the total momentum of the system, in this case, before the explosion, is equal to the total momentum of the system after the explosion. As well as this, we can recall that the momentum of an object 𝑝 is given by multiplying the mass of the object by the velocity of the object. So, we can use this equation to work out the total momentum of the system before the explosion and after the explosion.

We can say that the total momentum of the entire system before the explosion, which we’ll call 𝑝 subscript bef, is equal to the mass of the meteor before the explosion multiplied by the velocity of the meteor before the explosion. Now we know the mass of the meteor before the explosion is 7000 grams and its velocity is 1500 metres per second. And since there’s nothing else in the system, apart from the meteor itself, we don’t need to add anything here to find the total momentum of the system.

So, we can see that the total momentum of the system before the explosion is given by multiplying the mass of the meteor, 7000 grams, by the velocity, 1500 metres per second. And at this point, we can choose to convert the mass into base units. Because remember, the base unit of mass is kilograms not grams. We don’t actually have to do this, as we’ll see later. But it does make life easier to be working in kilograms than grams.

However, truthfully, the choice is up to us. As long as we stay consistent, then it doesn’t matter whether or not we convert the units from grams to kilograms. Basically, what we mean by staying consistent is that if we choose to go with grams, then we must use grams everywhere. And conversely, if we choose to convert this quantity to kilograms, then we must convert all of the other quantities to kilograms as well. So, let’s do that.

We can recall that one kilogram is equivalent to 1000 grams. And therefore, 7000 grams is equal to seven lots of 1000 grams, or seven kilograms. And hence, we replace the mass of the meteor with seven kilograms. Then, when we evaluate the right-hand side of this equation, we find that the total momentum of the system before the explosion is equal to 10500 kilograms metres per second. Now because of conservation of momentum, this is what the total momentum of the system must be after the collision. So, let’s work out an expression for the total momentum of the system after the collision.

We can say that the total momentum of the system after, which we’ll call 𝑝 subscript aft, is equal to the mass of the smaller piece of the meteor, which we’ll call 𝑚 subscript 𝑆, multiplied by the velocity of the small piece of the meteor, 𝑣 subscript 𝑆. And as well as this, we need to add to it the momentum of the larger piece, which is 𝑚 subscript 𝐿, for the mass of the larger piece, multiplied by 𝑣. We’re not calling it 𝑣 subscript 𝐿 because we’ve been told that the velocity of the larger piece is simply 𝑣. So, let’s use the convention we’ve been given in the diagram.

So, anyway, we know now three of the quantities on the right-hand side of this equation. So, let’s substitute those in. But, of course, we need to remember to convert the masses into kilograms. And so, 2000 grams is gonna be two kilograms and 5000 grams is going to be five kilograms. Hence, we can say that the total momentum of the system after the explosion is equal to two kilograms multiplied by 1200 metres per second plus five kilograms multiplied by 𝑣.

Now the first term on the right-hand side here becomes 2400 kilograms metres per second when we expand out the parentheses. And at this point, we should remember that the total momentum of the system after the explosion must be equal to the total momentum of the system before the explosion. In other words, everything over here on the right-hand side must be equal to 10500 kilograms metres per second. Hence, we can say that 10500 kilograms metres per second is equal to 2400 kilograms metres per second plus five kilograms times 𝑣.

Then, we can try and solve for 𝑣, firstly, by subtracting 2400 kilograms metres per second from both sides of the equation, which means that that term cancels on the right-hand side. And on the left, we’re left with 8100 kilograms metres per second. And then, we can divide both sides of the equation by five kilograms. This way, the five kilograms on the right-hand side cancels. And on the left, we can see that the unit of kilograms in the numerator cancels with the unit of kilograms in the denominator.

Then we simply have 𝑣 on the right-hand side. And on the left, we have 8100 divided by five, metres per second. And this is perfect because we’re trying to calculate a velocity. Therefore, it being in metres per second is ideal. That is the base unit for velocity. Cleaning everything up and evaluating the fraction on the left-hand side of the equation, we find that the velocity 𝑣 of the larger piece of the meteor after the explosion is 1620 metres per second. And that is the final answer to our question.

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