A 500 g wood block on a frictionless table is attached to a horizontal spring. A 50 g dart is shot into the face of the block opposite the spring, where it sticks. Afterward, the spring oscillates with a period of 1.5 s and an amplitude of 20 cm. How fast was the dart moving when it hit the block?

Question

Accepted Answer

Hey, everyone. So this is a conservation of momentum and conservation of energy problem using a spray. We've got a lot going on here, but let's see what they're asking us and we'll work through it. A metallic block of mass, 650 g has a cavity with a spongy lining that holds balls shot into the block. The block is fixed to a horizontal spring. A 75 g ball is shot to the into the block in a direction parallel to the spring on the face fixed to the spring stretching the spring. After that, the block is observed to oscillate at a frequency of 0.8 Hertz. If the block has an amplitude of 30 centimeters, determine the speed of the ball when it enters the cavity. OK. Our multiple choice answers here are a 14.6 m per second. B 0.156 m per second. C 13.1 m per second or D 13.8 m per second. So the first thing I'm going to do here is draw a diagram to kind of help myself think about what's going on here. So we have this block, it's a, it's a fix to a spring, right? And then there is a ball that is shot into it and then it's squishy. So the ball um is kind of absorbed by the block. So we have three moments in time where we have to think about what's happening uh with the physics of this problem. So I'm going to use zero to denote right before the collision. I'm going to use one to denote right after the collision. And I'm going to use two to denote at the time of maximum compression of the spring. And so as we walk through all of the different steps, there are a number of steps here to, to get to our final answer. Um But as we walk through those steps, you'll understand why we are delineating what is happening this problem into those three distinct more discrete moments in time. OK. So the first part of this problem, we're going to recall from our conservation of momentum that our initial momentum is equal to our final momentum or P nought is equal to P F and P or momentum is given by mass multiplied by velocity. So initially, we've got, so we're, we are asked to find the speed of the ball. So initially, we were only worried about that ball. So we have the mass of the ball multiplied by the speed of the ball. The initial speed of the ball is equal to the mass of the ball multiplied by the speed of the ball at 0.1. Right. So right after the collision plus the mass of the block, I'm gonna call that because it's a metallic block. I'm gonna use the subs for m there multiplied by the speed of the metallic block at time one. And so from there, we can look at all of these terms, we know the mass of the ball we are solving for that initial speed. We know the mass of the me metallic block. And we also know that directly after the collision happens because of the conservation of momentum, the speed of the ball and the speed of the block are the same. And so we're just gonna call that V one so that simplifies it just a little bit. And so again, we're solving for this initial, um we're at at time zero here, initial speed of the ball. And so when we simplify this out, we get the mass of the ball. Actually, let me just take this one step further just so we can write it just a little bit easier. The mass of the ball plus the mass of the block. And you're just going to call mass total, that'll come up a couple of times here. So I think that's a little bit easier. So our total mass multiplied by R V one that speed immediately after the collision, all divided by the mass of the ball is going to be our initial speed. Of the ball. So we are solving four this term. Now we need to figure out the one because we don't have that. So we can think about right after the collision, we have conservation of energy as the spring is compressing. So that equation we can recall is given by one half and B squared equals one half K X squared where our mass is the total mass because this is after the collision. So we have the ball and the block together. And then that V here is this V one that we are trying to solve for and that equals one half K X squared. OK. So from our simple harmonic motion equations, we can recall that the frequency is given by one divided by two pi multiplied by the square root of K divided by M. And so we have frequency so we have everything we need to solve four K. So we'll rearrange this equation. We have two pi f that quantity squared loops multiplied by. Um And so when we plug in those values, the problem told us that the frequency was 0.8 Hertz multiplied by the mass. And this mass is that total mass. So let's actually solve for that mass of the block was 0. five, no, no, sorry 0. kg because we are going to rewrite these in standard units. So 75 g is 750.75 point kg plus mass. The metallic block g, which is 6500.650 kg. And that equals our total mass. And that is 0. programs is our total mass there. So we can plug that in here, 0.7 to five kg and solve for our spring constant k, 18.3 newtons per meter. And now we have everything we need to solve for V one which you will then plug into this first equation to solve for V B knot. So let's see what that looks like over here have cancel V one when we isolate, it is going to be the square root of K X squared divided by total. And so we've sol for K that's 18.3 newtons per meter, our or our displacement was given to us in the problem uh 0.30 centimeters or 0. m. So that's gonna be the maximum compression of the spring. So that's the um the kind of the definition of of that maximum compression is that amplitude? OK. So 18.3 newtons per meter multiplied by 0.3 m. That guy sweared divided by 0.7 to five kg and we get 1.5, one m per second. And then our last step plugging that in for that V B O equation V B zero of and total multiplied by V one V one divided by the mass of just the ball. So in total 0.725 kg V one, we just saw 41.51 m per second divided by the mass of the ball, which was point 075 kg. And that leaves us with 14.6 m per second. Finally, lots of steps to this one. But you can see that uh we're calling those just a few of those key equations helps us get to that final answer. And so when we come back up here, we look at our multiple choice answers that aligns with answer choice. A so that's all we have for this one. We'll see you in the next video.

A 500 g wood block on a frictionless table is attached to a horizontal spring. A 50 g dart is shot into the face of the block opposite the spring, where it sticks. Afterward, the spring oscillates with a period of 1.5 s and an amplitude of 20 cm. How fast was the dart moving when it hit the block?

Verified Solution

Key Concepts

Conservation of Momentum

Simple Harmonic Motion (SHM)

Spring Constant and Period of Oscillation