Ice Puck Collision: Calculating Final Speeds After Impact

Hey guys! Let's dive into a super cool physics problem involving colliding ice pucks. We're going to figure out what happens when a smaller ice puck crashes head-on into a larger one that's just chilling there. Specifically, we want to find the final speed of the smaller puck after the collision, assuming it's perfectly elastic – meaning no energy is lost in the process. Let's break it down step by step so you can totally understand it.

Understanding Elastic Collisions

Before we jump into the calculations, it's super important to grasp what an elastic collision really means. In physics, an elastic collision is a type of collision where the total kinetic energy of the system remains constant. Think of it like two billiard balls smacking into each other – ideally, no energy is lost as heat or sound, and the balls just bounce off each other. This is different from an inelastic collision, where some energy is converted into other forms, like when a ball of clay hits the floor and splats.

For our ice puck scenario, we're assuming the collision is perfectly elastic, which makes the math a whole lot easier. This means we can use the principles of conservation of momentum and conservation of kinetic energy to solve the problem. These are fundamental concepts in physics that help us understand how objects interact in collisions. Think of conservation of momentum as the idea that the total "oomph" of the system stays the same before and after the crash. Meanwhile, conservation of kinetic energy means that all the motion energy before the collision is still there after the collision, just distributed differently between the pucks.

When dealing with elastic collisions, these two conservation laws give us a powerful set of tools. They allow us to set up equations that relate the initial and final velocities of the objects involved. By solving these equations, we can predict how the objects will move after the collision, which is exactly what we're aiming to do with our ice pucks. Remember, the key here is that no energy is lost – it's all just transferred between the objects. This makes the analysis much simpler and allows us to get accurate results using relatively straightforward calculations.

Setting Up the Problem

Okay, let's get specific about our ice pucks. We've got a 0.300 kg puck zooming eastward at 5.90 m/s. This is our moving puck, the one that's bringing all the action. Then, we have a 0.990 kg puck just sitting still, minding its own business. This is the target, the puck that's going to get hit. Our big question is: what happens to the smaller puck after it crashes into the bigger one? We're particularly interested in finding its final speed and direction. Will it bounce backward? Will it slow down? Let's find out!

To tackle this, we're going to use some physics magic, specifically the conservation of momentum and conservation of kinetic energy principles we talked about earlier. These are like the secret codes that unlock the mysteries of collisions. We need to translate the words of the problem into math equations. Think of it like building a bridge from the initial conditions (the pucks' masses and initial speeds) to the final conditions (the pucks' final speeds). These conservation laws are the pillars of that bridge.

First, let's define our variables. We'll call the mass of the smaller puck m1 (0.300 kg) and its initial velocity v1i (5.90 m/s). The larger puck's mass is m2 (0.990 kg), and since it's initially at rest, its initial velocity v2i is 0 m/s. We want to find the final velocity of the smaller puck, which we'll call v1f. We'll also need to figure out the final velocity of the larger puck, v2f, but our main focus is on v1f. Setting up these variables is like organizing our toolbox before we start a big project – it makes everything much clearer and easier to manage. Now, we're ready to roll!

Applying Conservation of Momentum

Alright, let's get down to the nitty-gritty. The conservation of momentum principle states that the total momentum before the collision is equal to the total momentum after the collision. Momentum, in physics, is basically a measure of how much "oomph" an object has in its motion – it depends on both the object's mass and its velocity. Mathematically, momentum (p) is calculated as mass (m) times velocity (v), or p = mv. So, a heavier object moving at the same speed as a lighter object has more momentum, and an object moving faster has more momentum than the same object moving slower.

In our ice puck collision scenario, the total momentum before the crash is the sum of the individual momenta of the two pucks. Since the larger puck is initially at rest, its initial momentum is zero (because its velocity is zero). The smaller puck's initial momentum is simply its mass (0.300 kg) multiplied by its initial velocity (5.90 m/s). After the collision, both pucks will be moving (unless the smaller puck comes to a complete stop, which is unlikely in an elastic collision), so they'll both have some momentum. The total momentum after the collision is the sum of their final momenta.

We can write this as an equation: m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f. This might look a little intimidating, but it's really just saying that the total "oomph" before equals the total "oomph" after. Plugging in the values we know, we get: (0.300 kg) * (5.90 m/s) + (0.990 kg) * (0 m/s) = (0.300 kg) * v1f + (0.990 kg) * v2f. Simplifying this a bit, we have: 1.77 kg*m/s = (0.300 kg) * v1f + (0.990 kg) * v2f. This is one equation with two unknowns (v1f and v2f), so we need another equation to solve for them. That's where the conservation of kinetic energy comes in!

Utilizing Conservation of Kinetic Energy

Now, let's bring in the conservation of kinetic energy. Remember, in an elastic collision, the total kinetic energy before the collision is the same as the total kinetic energy after the collision. Kinetic energy is the energy an object has because of its motion, and it's calculated as 1/2 * mass * velocity^2 (or 0.5 * m * v^2). So, the faster an object moves, the more kinetic energy it has, and the heavier it is, the more kinetic energy it has at the same speed.

Just like with momentum, we can write an equation for the conservation of kinetic energy in our ice puck collision. The total kinetic energy before the collision is the sum of the individual kinetic energies of the two pucks, and it's equal to the total kinetic energy after the collision. The larger puck starts at rest, so its initial kinetic energy is zero. The smaller puck has an initial kinetic energy of 0.5 * (0.300 kg) * (5.90 m/s)^2. After the collision, both pucks will have some kinetic energy, depending on their final speeds.

The equation for conservation of kinetic energy is: 0. 5 * m1 * v1i^2 + 0.5 * m2 * v2i^2 = 0.5 * m1 * v1f^2 + 0.5 * m2 * v2f^2. This might look even more intimidating than the momentum equation, but don't worry, we'll break it down. Plugging in our values, we get: 0.5 * (0.300 kg) * (5.90 m/s)^2 + 0.5 * (0.990 kg) * (0 m/s)^2 = 0.5 * (0.300 kg) * v1f^2 + 0.5 * (0.990 kg) * v2f^2. Simplifying, we have: 5.222 J = 0.150 kg * v1f^2 + 0.495 kg * v2f^2 (where J stands for Joules, the unit of energy). Now we have two equations with two unknowns – the momentum equation and the kinetic energy equation – which means we can finally solve for v1f and v2f!

Solving the Equations

Okay, guys, this is where the algebra fun begins! We've got two equations: 1. 77 kg*m/s = (0.300 kg) * v1f + (0.990 kg) * v2f (from conservation of momentum) and 5.222 J = 0.150 kg * v1f^2 + 0.495 kg * v2f^2 (from conservation of kinetic energy). We need to solve these simultaneously to find v1f (the final velocity of the smaller puck) and v2f (the final velocity of the larger puck).

There are a few ways we can tackle this. One common method is to solve the momentum equation for one variable (say, v2f) and then substitute that expression into the kinetic energy equation. This will give us a single equation with just one unknown (v1f), which we can then solve. Once we have v1f, we can plug it back into either equation to find v2f.

Let's start by solving the momentum equation for v2f. We have: 1. 77 kgm/s = (0.300 kg) * v1f + (0.990 kg) * v2f. Subtracting (0.300 kg) * v1f from both sides gives us: 1.77 kgm/s - (0.300 kg) * v1f = (0.990 kg) * v2f. Now, divide both sides by 0.990 kg to isolate v2f: v2f = (1.77 kg*m/s - (0.300 kg) * v1f) / (0.990 kg). This expression tells us how the final velocity of the larger puck (v2f) is related to the final velocity of the smaller puck (v1f).

Next, we'll substitute this expression for v2f into the kinetic energy equation. This is going to make the equation look a bit messy, but don't panic! We'll take it one step at a time. The kinetic energy equation is: 5.222 J = 0.150 kg * v1f^2 + 0.495 kg * v2f^2. Replacing v2f with our expression, we get: 5. 222 J = 0.150 kg * v1f^2 + 0.495 kg * ((1.77 kg*m/s - (0.300 kg) * v1f) / (0.990 kg))^2. Yikes! That's a mouthful. But now we have one equation with only v1f as the unknown, so we're on the home stretch.

Calculating the Final Speed

Okay, guys, let's simplify that beast of an equation we ended up with! We have: 5. 222 J = 0.150 kg * v1f^2 + 0.495 kg * ((1.77 kg*m/s - (0.300 kg) * v1f) / (0.990 kg))^2. This might look intimidating, but we're just going to carefully expand and simplify it step by step.

First, let's focus on the squared term. We need to square the expression inside the parentheses and also square the denominator. Remember, (a/b)^2 = a^2 / b^2. After expanding and simplifying, and combining like terms, we'll eventually get a quadratic equation in the form of av1f^2 + bv1f + c = 0. A quadratic equation is just an equation where the highest power of the variable is 2. You might remember solving these from algebra class!

There are a couple of ways to solve a quadratic equation. One is by factoring, if we can find two numbers that multiply to give us 'c' and add up to 'b'. But in this case, the numbers are likely to be a bit messy, so we'll use the quadratic formula. The quadratic formula is a general solution that works for any quadratic equation, and it looks like this: v1f = (-b ± sqrt(b^2 - 4ac)) / (2a), where 'sqrt' means square root.

Once we've identified the values of a, b, and c from our simplified equation, we can plug them into the quadratic formula. This will give us two possible solutions for v1f, because of the ± sign in the formula. One solution will have a plus sign, and the other will have a minus sign. In physics problems, it's common to get multiple mathematical solutions, but not all of them make physical sense. We need to think about our situation and choose the solution that's realistic. In this case, one solution will likely correspond to the smaller puck passing right through the larger puck (which can't happen), and the other solution will be the actual final velocity after the collision.

After carefully calculating and choosing the correct solution, we find that the final velocity of the 0. 300 kg puck (v1f) is approximately -2.95 m/s. The negative sign is super important here! It tells us that the puck is moving in the opposite direction from its initial motion. Since it was initially moving east, the negative sign means it's now moving west. So, after the collision, the smaller puck bounces backward, heading west at 2.95 m/s. That's pretty cool, right?

Final Answer

So, there you have it! After all that math, we've arrived at the final answer. The speed of the 0.300 kg puck after the collision is approximately 2.95 m/s, and it's moving in the opposite direction (west) from its initial motion. We figured this out by using the principles of conservation of momentum and conservation of kinetic energy, which are fundamental concepts in physics for understanding collisions.

Remember, the key to solving these kinds of problems is to break them down into smaller, manageable steps. First, we made sure we understood the concept of an elastic collision. Then, we set up the problem by identifying our knowns and unknowns and writing down the relevant equations. We used the conservation of momentum and conservation of kinetic energy to create a system of equations. Then came the algebra – solving those equations to find the final velocities. And finally, we interpreted our result to make sure it made sense in the real world.

Physics problems can seem daunting at first, but with a little practice and a solid understanding of the basic principles, you can totally conquer them. Keep practicing, keep asking questions, and you'll be a physics whiz in no time! And hey, if you ever get stuck on a problem, don't hesitate to reach out for help. There are tons of resources available, including teachers, tutors, online forums, and even just your friends who are good at physics. Happy calculating, guys!