Calculating Initial Velocity In Vertical Motion A Two-Sphere Collision Problem

Have you ever wondered how physicists calculate the initial velocity of objects in motion? It's a fascinating process that combines the principles of kinematics with a bit of algebraic problem-solving. In this article, we'll break down a classic physics problem step-by-step, making it easy for you to understand the underlying concepts and calculations. So, let's dive in and unravel the mysteries of vertical motion!

Problem Statement: The Two-Sphere Collision

Okay, guys, let's break down the problem we're tackling today. Imagine you're standing on the rooftop of a building, and you decide to throw a sphere vertically downwards with some initial speed, which we'll call v. At the exact same moment, someone on the ground throws another sphere straight up into the air. Our mission, should we choose to accept it, is to figure out the v – that initial downward speed – if the two spheres decide to have a mid-air collision exactly 2 seconds later. Oh, and just to keep things spicy, we know that gravity is pulling everything down at a rate of 10 meters per second squared (g = 10 m/s²). Got it? Awesome, let's get started!

Breaking Down the Problem: Setting the Stage

To solve this problem effectively, we need to dissect it into manageable parts. First, let’s visualize the scenario. We have two spheres moving under the influence of gravity. One is launched downwards from the rooftop, and the other is launched upwards from the ground. They're on a collision course, and we know the exact time of impact: 2 seconds. The key here is understanding that each sphere's motion can be described using the principles of kinematics, specifically the equations of motion for constant acceleration. These equations relate displacement, initial velocity, final velocity, acceleration, and time. By applying these principles separately to each sphere, we can develop a system of equations that will help us solve for the unknown initial velocity, v. Think of it like a puzzle – we have the pieces; we just need to fit them together correctly.

Key Concepts: Understanding the Physics

Before we jump into the calculations, let's make sure we're all on the same page with the physics involved. The most important concept here is uniformly accelerated motion, which simply means that the spheres are moving with a constant acceleration due to gravity. This acceleration, denoted by g, is approximately 9.8 m/s² near the Earth's surface, but for simplicity, we're using 10 m/s² in this problem. This means that for every second that passes, the velocity of the spheres changes by 10 m/s, either increasing in the downward direction or decreasing in the upward direction. Another crucial concept is the independence of vertical and horizontal motion. Since we're dealing with purely vertical motion in this problem, we don't need to worry about any horizontal components. Finally, remember that the displacement of an object is its change in position, and it can be positive or negative depending on the direction of motion. By keeping these concepts in mind, we can approach the problem with a clear understanding of the physical principles at play.

Applying Kinematic Equations: The Math Behind the Motion

Alright, let's get our hands dirty with some equations! This is where the magic happens, guys. We're going to use the kinematic equations, which are like the secret sauce for solving motion problems. These equations help us relate the position, velocity, and time of an object moving with constant acceleration. Now, for this problem, we've got two spheres doing their own thing, so we'll need to set up equations for each of them. Think of it as writing two different stories, one for each sphere, but both stories are happening at the same time and in the same universe (or, you know, the same physics problem).

Setting Up the Equations for Sphere 1 (From the Rooftop)

Let’s start with the sphere thrown downwards from the rooftop. We'll call this Sphere 1. We need to describe its motion using the kinematic equations. The most useful equation here is the one that relates displacement, initial velocity, time, and acceleration: Δy = v₀t + (1/2)at². Here, Δy is the displacement (the change in vertical position), v₀ is the initial velocity (which is what we're trying to find, v), t is the time (2 seconds), and a is the acceleration due to gravity (10 m/s²). Now, let's plug in the values we know. Since Sphere 1 is moving downwards, we'll consider the displacement to be negative. Let's say the height of the building is H. So, the displacement of Sphere 1 is -Δy₁. We get: -Δy₁ = -vt + (1/2)gt². The negative sign in front of v indicates that the initial velocity is in the downward direction, which we've defined as negative. Remember, the direction matters in physics, guys! This equation is our first piece of the puzzle.

Setting Up the Equations for Sphere 2 (From the Ground)

Now, let's turn our attention to Sphere 2, the one launched upwards from the ground. We'll use the same kinematic equation, but this time, the initial velocity is upwards, so we'll consider it positive. The displacement of Sphere 2 is Δy₂, and it's also positive since it's moving upwards. So, the equation for Sphere 2 becomes: Δy₂ = v₀₂t - (1/2)gt². Notice the minus sign in front of the (1/2)gt² term. This is because gravity is acting downwards, opposing the upward motion of the sphere. Now, we need to consider that the initial velocity of Sphere 2 is unknown, let's call it v₀₂. So, our equation becomes: Δy₂ = v₀₂t - (1/2)gt². This is our second piece of the puzzle. We're getting closer, guys!

Solving for the Unknown: Putting It All Together

Okay, we've got our equations set up, and now it's time to put on our detective hats and solve for the unknown velocity, v. This is where the fun begins! We've got two equations, but it might seem like we have more than one unknown. But don't worry, we have a secret weapon: the fact that the two spheres collide. This collision gives us a crucial piece of information – a relationship between the displacements of the two spheres.

The Collision Condition: Connecting the Equations

The key to solving this problem lies in understanding what happens when the spheres collide. At the moment of impact, the sum of their displacements must equal the total height of the building, H. In other words, if Sphere 1 has traveled a distance Δy₁ downwards and Sphere 2 has traveled a distance Δy₂ upwards, then Δy₁ + Δy₂ = H. This is our secret weapon! It connects the two equations we derived earlier. We know from the previous sections that -Δy₁ = -vt + (1/2)gt² for Sphere 1 and Δy₂ = v₀₂t - (1/2)gt² for Sphere 2. We can rewrite the first equation as Δy₁ = vt - (1/2)gt². Now, we can substitute these expressions for Δy₁ and Δy₂ into our collision condition equation: [vt - (1/2)gt²] + [v₀₂t - (1/2)gt²] = H. This equation is a bit messy, but it's progress! We've now got one equation with a few unknowns: v, v₀₂, and H. We need to find a way to eliminate some of these unknowns to solve for v. Stick with me, guys; we're almost there!

Eliminating Variables: Simplifying the Equation

Now, let's simplify things a bit. Notice that we're actually trying to find v, the initial velocity of Sphere 1. We don't really care about the height of the building, H, or the initial velocity of Sphere 2, v₀₂. So, we need to find a way to get rid of them from our equation. Here's a neat trick: we can rearrange the collision condition equation we derived earlier: [vt - (1/2)gt²] + [v₀₂t - (1/2)gt²] = H. Now, let's rearrange the terms to group the unknowns we want to eliminate: vt - gt² + v₀₂t = H. This equation still has H and v₀₂ in it, but we're one step closer. Here's another key insight: we don't actually need to know the height of the building or the initial velocity of Sphere 2 to solve for v. We just need to focus on the relationship between the displacements and the time of collision. Let's go back to our original equations for Δy₁ and Δy₂ and think about what we know. We know the time of collision (t = 2 s) and the acceleration due to gravity (g = 10 m/s²). We can plug these values into our equations and see what happens. This will help us simplify the equation and hopefully eliminate some of the unknowns. Let's do it!

Calculating the Initial Velocity: The Final Showdown

Alright, buckle up, guys! We're in the final stretch. We've set up our equations, we've simplified them, and now it's time for the grand finale: calculating the initial velocity, v. This is where all our hard work pays off. Remember, we're trying to find the speed at which the sphere was thrown downwards from the rooftop. We've got all the pieces of the puzzle; now we just need to fit them together.

Plugging in the Values: Crunching the Numbers

Let's go back to our simplified equation: vt - gt² + v₀₂t = H. We know that t = 2 seconds and g = 10 m/s². Let's plug these values into the equation: v(2) - (10)(2)² + v₀₂(2) = H. This simplifies to: 2v - 40 + 2v₀₂ = H. Now, we need to get rid of the H and v₀₂ terms. Remember, we had the collision condition: Δy₁ + Δy₂ = H. We also had the equations for Δy₁ and Δy₂: Δy₁ = vt - (1/2)gt² and Δy₂ = v₀₂t - (1/2)gt². Let's substitute the values of t and g into these equations: Δy₁ = v(2) - (1/2)(10)(2)² = 2v - 20. Δy₂ = v₀₂(2) - (1/2)(10)(2)² = 2v₀₂ - 20. Now, we can substitute these expressions for Δy₁ and Δy₂ into the collision condition equation: (2v - 20) + (2v₀₂ - 20) = H. This simplifies to: 2v + 2v₀₂ - 40 = H. Now, we have two equations: 2v - 40 + 2v₀₂ = H 2v + 2v₀₂ - 40 = H. Notice anything? Both equations are the same! This means we've essentially gone in a circle, and we haven't eliminated any unknowns. But don't worry, this is a common problem-solving technique. Sometimes, you need to try a few different approaches before you find the right one. Let's go back to our original equations and see if we can find another way to solve for v.

Isolating v: The Final Calculation

Okay, guys, let's try a different approach. We're still aiming to find v, the initial velocity of Sphere 1. Let's go back to the equations for Δy₁ and Δy₂: Δy₁ = vt - (1/2)gt² Δy₂ = v₀₂t - (1/2)gt². We also have the collision condition: Δy₁ + Δy₂ = H. But remember, we don't actually care about H or v₀₂. We just want to find v. Here's a key insight: instead of focusing on the height of the building, let's focus on the relative motion of the two spheres. Since they collide after 2 seconds, the sum of the distances they've traveled must be equal to the initial separation between them. In other words, the distance Sphere 1 travels downwards plus the distance Sphere 2 travels upwards must equal the height of the building. But we don't need to know the height of the building! We just need to relate the distances they travel. Let's think about the displacements again. Δy₁ is the displacement of Sphere 1, and it's negative since it's moving downwards. Δy₂ is the displacement of Sphere 2, and it's positive since it's moving upwards. We can rewrite the collision condition as: -Δy₁ + Δy₂ = 0. This is because the spheres are colliding, so their relative displacement is zero. Now, let's substitute the equations for Δy₁ and Δy₂: -(vt - (1/2)gt²) + (v₀₂t - (1/2)gt²) = 0. Let's simplify this equation: -vt + (1/2)gt² + v₀₂t - (1/2)gt² = 0. Notice that the (1/2)gt² terms cancel out! This leaves us with: -vt + v₀₂t = 0. Now, we can factor out t: t(-v + v₀₂) = 0. Since t is not zero (it's 2 seconds), we must have: -v + v₀₂ = 0. This means that v = v₀₂! This is a crucial piece of information. It tells us that the initial velocity of Sphere 1 downwards is equal to the initial velocity of Sphere 2 upwards. Now, let's go back to our original equations and see if we can use this information to solve for v. We had: Δy₁ = vt - (1/2)gt² Δy₂ = v₀₂t - (1/2)gt². Let's substitute v₀₂ with v in the second equation: Δy₂ = vt - (1/2)gt². Now, we have two identical equations: Δy₁ = vt - (1/2)gt² Δy₂ = vt - (1/2)gt². This means that Δy₁ = Δy₂! But remember, Δy₁ is negative, and Δy₂ is positive. So, we must have: -Δy₁ = Δy₂. Now, let's go back to our collision condition: Δy₁ + Δy₂ = H. We can substitute -Δy₁ for Δy₂: Δy₁ - Δy₁ = H. This simplifies to: 0 = H. This is impossible! The height of the building cannot be zero. This means we've made a mistake somewhere in our logic. Let's go back and review our steps. Ah, I see the mistake! We made an incorrect assumption about the relative displacement of the spheres. The correct condition is that the sum of the magnitudes of their displacements must equal the height of the building. In other words, |Δy₁| + |Δy₂| = H. Let's go back and correct our calculations.

The Correct Solution: Putting It All Together

Okay, guys, we've identified the mistake, and now we're on the right track! Let's recap our steps and make sure we're clear on the correct approach. We have two spheres: Sphere 1 thrown downwards with initial velocity v, and Sphere 2 thrown upwards with initial velocity v₀₂. They collide after 2 seconds. We want to find v. We have the kinematic equations: Δy₁ = vt - (1/2)gt² Δy₂ = v₀₂t - (1/2)gt². We also have the collision condition: |Δy₁| + |Δy₂| = H. We know that t = 2 seconds and g = 10 m/s². Let's plug these values into the kinematic equations: Δy₁ = 2v - (1/2)(10)(2)² = 2v - 20 Δy₂ = 2v₀₂ - (1/2)(10)(2)² = 2v₀₂ - 20. Now, let's consider the collision condition. Since Δy₁ is negative and Δy₂ is positive, we have: |Δy₁| = -(2v - 20) = 20 - 2v |Δy₂| = 2v₀₂ - 20. So, the collision condition becomes: (20 - 2v) + (2v₀₂ - 20) = H. This simplifies to: -2v + 2v₀₂ = H. Now, we need to find another equation to eliminate H or v₀₂. Let's go back to the idea that the sum of the displacements must equal the height of the building. We have: Δy₁ + Δy₂ = H. Substituting the expressions for Δy₁ and Δy₂, we get: (2v - 20) + (2v₀₂ - 20) = H. This simplifies to: 2v + 2v₀₂ - 40 = H. Now, we have two equations: -2v + 2v₀₂ = H 2v + 2v₀₂ - 40 = H. Let's set these equations equal to each other: -2v + 2v₀₂ = 2v + 2v₀₂ - 40. Now, we can simplify this equation: -2v = 2v - 40. Adding 2v to both sides, we get: 0 = 4v - 40. Adding 40 to both sides, we get: 40 = 4v. Dividing both sides by 4, we get: v = 10 m/s. So, the initial velocity of Sphere 1 downwards is 10 m/s. Congratulations, guys! We've solved the problem!

Final Answer: 10 m/s

So, there you have it! The initial velocity (v) at which the sphere was launched downwards from the rooftop is 10 m/s. We navigated through the problem step by step, breaking it down into manageable chunks, applying the principles of kinematics, setting up equations, and finally, solving for the unknown. Physics problems can be challenging, but with a systematic approach and a clear understanding of the underlying concepts, you can conquer them. Remember, the key is to break down the problem, identify the relevant information, apply the appropriate equations, and solve for the unknowns. And most importantly, don't be afraid to make mistakes! Mistakes are a natural part of the learning process, and they often lead to deeper understanding. So, keep practicing, keep exploring, and keep solving! You got this, guys!