9.5 Elastic and Inelastic Collisions

Okay, today's lesson is on elastic and inelastic collisions. It is essentially an add-on to what we have already learned about momentum. We are now considering collisions and whether or not energy is conserved. The first scenario we will examine is when we are dealing with subatomic particles such as atoms and electrons. This is likely the only case where kinetic energy is conserved. By this, I mean that the total kinetic energy of all the particles before the collision must equal the total kinetic energy of all the particles after the collision. This is similar to what we did with momentum, but now we are looking at kinetic energy. Remember, our formula for kinetic energy is 1/2 * MV^2. So, what we are discussing is that the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision. One-half mV^2 for each particle or object before the collision adds up to the same total after the collision. Einstein said that all energy is conserved, it just gets transformed from form to form. So, what we have to be clear on here is that the internal collision energy is conserved, and the total energy remains the same. However, when we're talking about elastic and inelastic collisions, we're only concerned with the kinetic energy. Therefore, most collisions will always be inelastic, as the total and kinetic energy may not be the same before and after the collision. This type of collision is called inelastic and usually involves a loss of kinetic energy, which could be converted into heat due to friction (e.g., car brakes), sound (e.g., car crash), light (e.g., sparks), or other forms. Bending the shapes, like in a car accident, the tin gets bent and twisted and all that. So, that requires energy to do. Because we're using up the energy, it ends up being inelastic. There's a range if you have objects that collide, like pool balls or curling rocks, something like that. Very little energy is lost, so it's still inelastic but quite efficient. If it's a rock being split up or a car accident or something like that, most of the energy ends up being lost, so there's not too much left over at all. Make sure you're clear that in every situation, momentum is still conserved. So, no matter what, we always still have conservation of momentum. So, we can use all of our old formulas. The only thing that's different now is for elastic collisions. Kinetic energy is also conserved. The key thing with kinetic energy when we're doing the calculations is that it doesn't matter if we have vectors because kinetic energy is a scalar. We don't have to worry about the angles and things like that we do with momentum. We just basically do the total kinetic energy of each object and compare them that way. So let's go through this example. This is one of the harder examples that you'd see. It's a ballistic pendulum, which means you shoot a bullet into the wood and the wood and the bullet would swing together. By doing that, we can then calculate how fast the bullet was going before it hit the wood. Part A is asking how we would calculate the velocity of the bullet. The first thing to understand is let's calculate what the energy of the block is before and after. Once the bullet hits the block of wood, it's embedded into the block of wood. So we have the mass of the bullet and the block, and at the bottom, it's going to have some sort of amount of kinetic energy once the bullet] and we can use the conservation of momentum equation to solve for the initial velocity of the bullet. Let's denote the initial velocity of the bullet as v, and the combined mass of the bullet and the block of wood as m. According to the conservation of momentum equation, we have: (m * v) + (m * 2.75) = m * 0Since the block of wood is initially at rest, its velocity is 0. Simplifying the equation, we get:v + 2.75 = 0Therefore, the initial velocity of the bullet is -2.75 meters per second. In conclusion, the bullet has an initial velocity of -2.75 meters per second, and upon impact with the block of wood, the block will move with a final velocity of 2.75 meters per second. block of wood will equal the combined mass and velocity after the collision, so for our bullets, we know the mass was 0.202 velocities. What we're looking for is the block's initial velocity, which we know is zero initially, and we'll have a total mass of 5.77 times that velocity that we just calculated (2.75). We can solve it then for the velocity of the bullet, which is 794 meters per second. So that's part A. We calculated the velocity of the bullet before it hits the block just by using conservation of momentum. Now, for Part B, we want to figure out if the collision was inelastic or elastic. Just erase this part, we don't need it here anymore. Let's calculate our kinetic energy before and after. So, our kinetic energy of the bullet before it hits the wood will be 1/2 * MV^2. Therefore, 1/2 * 0.02 * 794^2 = 60,304 joules. The energy of the block is zero because it wasn't moving. So, we only have our 60,304 joules. After the bullet hits the block of wood, both objects will be stuck together. This means that the collision is completely inelastic. In this case, the total kinetic energy can be calculated by multiplying half of the total mass (5.77) by the square of the total speed. The resulting total kinetic energy is 21.8. It is evident that most of the energy is lost in the collision. The initial total energy before the collision was around sixty-three or four, while the total energy after the collision is 2108. Therefore, it is crucial to understand that when two objects stick together in a collision, such as a bullet and a block of wood, the collision is always completely inelastic and most of the energy is lost. On the other hand, if the objects were to bounce off each other, the collision would be somewhat inelastic but not a total loss of energy. Questions he is now asking are how much is lost and what percentage would it be. So, this one is pretty straightforward. Just do some simple math. If you want to find out how much was lost, subtract 21.8 from 6304. This will give us 6282 joules. So, that would be how much is lost. The questions could be worded differently. They could ask how much energy is lost. In this case, the answer would be 6282 joules. Alternatively, they could ask what percentage is lost or how much percentage is left. In either case, just do the math. In this case, we divide 6282 by 6300, which gives us 99.7. So, that means 99.7% of the energy was lost in the collision. If it was asking how much was left, just subtract the lost energy from the initial amount. Do the reverse, and you would go 21.8 over the total, and that should give you 0.3 percent, right? If you lose 99.7, 0.3 remains. So, depending on how it's worded, you could always answer those questions. And that's it.