Two blocks with masses mA and mB are connected by a massless
string over a massless, frictionless pulley. Block B, which is
more massive than block A, is released from height h and falls. b. A 1.0 kg block and a 2.0 kg block are connected by a massless
string over a massless, frictionless pulley. The impact speed
of the heavier block, after falling, is 1.8 m/s. From how high
did it fall?

Question

Accepted Answer

Hey, everyone. So this is a conservation of energy problem. Let's see what they're asking us. A simple Atwood machine is formed by two masses. M one and M two hanging over a massless pully. A light cord connects the two masses from a laboratory experiment. A student puts mask one on the floor and holds with her hand mass two at a height of H. The mass of M two is double that of M one. The student releases the mass M two and measures with a motion detector, the impact speed to be 2.6 m per second for M two. From what height did the student release M two? Her multiple choice answers here are a 0. m. B 0.5 m C one m or D 1.5 m. OK. So I'm going to draw a diagram here. We can recall that an Atwood machine is an ideal ideal pulley system, which means we don't have any friction losses. So we have our connected by a light chord that in physics speak means you know a massless chord. We have M one on the floor And then M at a distance of H, we can excuse me. So we can recall that the conservation of energy gives U I plus K I equals U F plus K F. In other words, the sum of your potential and kinetic energy initially is equal to the sum of your potential and kinetic energy. At the final point that we're looking at here, which would be um after mass two is released and it impacts the floor. So if we recall that potential energy is given as M G H and kinetic energy is given as one half M V squared, we can expand out this first equation and then determine what information we have, what information we need to solve for this height. So you, I is going to be M one G H one I plus M two G H two I plus one half M one V one I squared plus one half M two V two I squared. So this might look a little bit more complicated or a little bit different than other conservation of energy problems that you've seen before. And what we need to re recall here is that this entire system is where we have the conservation of energy. So we have two masses in that system. And so it's those masses together uh even though they may be they, even though they are separate here, um looking at them together as a system that is what is conserved though that energy is what is conserved. So that's why we have to break it out between mass one and mass two, both the potential energy of mass one and the kinetic energy of mass one. And then the potential energy of mass two and the kinetic energy of mass two initially. And that equals we'll do the same thing for the second, for the other side of the equation for that final um point. So we have M one G H one F plus M two G H two oops two F plus one half M one V I F squared plus one half, M two B two F squared. I'm sorry, let's go back here. That's the third term is for kinetic energy is one half M one V one F squared. OK. So lots of terms here, let's take each one, we will break it down and see what we have And what we need to figure out. So M one was given to us and the problem, it's just M1 G gravitational constant. We can recall that that's 9.8 m/s squared. H one initial is zero because it's on the floor and two. So the mass of M two is double that of M one. So M two equals two M one And then H two initial is that capital H that we are solving for? Mhm We've already, so one half M one we've established. So V one initial is zero because we are from where we are starting, the student is holding mass two, it's not moving. So V one initial and V two initial are both going to equal zero m per second. And those are all of our variables in the first half of the equation. So let's look at the second half, We have M one. So H one final yes is going to be, is going to also be H because if mass two moves a distance of H because they are connected by this string, mass one will also move up the same distance. So that will also be this H pot, this capital H that we're solving for. So M two, we have already established and then H two F is going to be zero because it Um hits the floor, right. That impact speed means H two goes to the floor V one F is going to equal V two F because they are connected by that string. So they're moving at the same speed. And that's helpful to recognize because we are given V two F as in the problem is 2.6 m per second. OK. And so we have now everything we need to solve for H I know there's a lot going on here. We this will simplify out will get a little bit easier. So let's look at our zero terms, our zero values because that those, that uh zeros out those terms uh and simplifies things pretty quickly. So H1 I is zero. So this term goes to zero V I one goes to zero V sorry, V I V one, I and V two, I are both zero And then H two F is also zero. OK. So let's see what this equation looks like. Now we have M two G H two, I equals M one G H one F plus one half, M one, V one F squared plus one half M two V two F squared. So we've established V one F and V two F are the same. We can just call that V F and then M two, we will substitute in um that two M one G and then H two I is that a capital H is equal to M one G H I F is also H plus one half and one V F squared plus one half. And M two is two M one V F squared. OK. So this is looking more manageable. So let's get the capital H is what we're solving for on the same side of the equation. So we'll have two M one G H minus M one G H Which simplifies to just M1GH not equals. So we have one half M one V F squared and then this one half and the two cancel. So we have one half M one V F squared plus M one V F squared. And that simplifies to three halves or 1.5 M one V F squared from here, our mass is canceled and we have everything we need to solve for H. So we'll divide both sides by G and then we will have three multiplied by V F which is 2.6 m per second squared, divided by two multiplied by gravity which is nine .8 m/s. And when we plug that into our calculators, we get 1.0 m. So that is the answer to this problem. It was a lot of steps. But we take everything methodically, we can see how the problem gave us all of the information we needed to solve it. And that answer aligns with answer choice C so C is the correct answer for this problem. That's all we have for this one, we'll see you in the next video.

Verified Solution

Key Concepts

Newton's Second Law of Motion

Conservation of Energy

Kinematic Equations