INT An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 8.2 m before stopping. How far does the lighter fragment slide? Assume that both fragments have the same coefficient of kinetic friction.

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Hey, everyone in this problem, we have an experiment where a physicist uses a firecracker to break apart a cube, the cube is placed on a rough horizontal floor. And when the explosion happens, it's gonna break into two pieces that move in opposite directions. So the first piece has a mass M one, it's gonna move 65 centimeters on the floor before it comes to rest. And the second piece has a mass M two and it's gonna move some distance D and that distance D is what we're trying to find. And we're trying to find it. If the mass M one is equal to five times the mass M two, we're told to consider that the coefficient of kinetic friction for both pieces is the same. We have four answer choices all in meters. Option A 3.3 option B 6.5 option C 9.5 and option D 16.2 right? So if we think about the movement here and what we're really trying to find is how far this second piece moves. The problem is we don't know the initial speed and we don't know once this cube breaks apart how fast it's going. OK. We can figure out the acceleration using some information about the forces. We don't know how long it takes to travel. And so there's just not enough information yet in order to go straight to our kinematic equations. So what we're gonna do is consider this explosion first try to find the speed of the cube immediately after the collision or sorry after the explosion. Now this explosion, we're gonna assume that friction does not impact the explosion. OK. So the explosion of this cube that breaking apart of the cube, the friction does not affect them. The friction only acts once these cubes have started moving and sliding outward from the explosion. OK. So if we assume that the friction doesn't actually affect the breaking apart of the cubes and we have a conservation of moment and the conservation of momentum tells us that the initial momentum P I is gonna be equal to the momentum after the collision P A. And we know that after the collision, our momentum is gonna be made up of two components, the momentum of piece one after the collision plus the momentum of piece two after the collision. OK. So we have this conservation of momentum. Recall that momentum is given by mass multiplied by velocity. So we have M I V I, it is equal to M one V one A plus M two V two A. Now initially this cube is at rest, it's placed on the floor. So the initial velocity is zero, which means the initial momentum is zero. OK. Our mass M one, well, that's gonna be the mass of that first block. OK. We know that that's M one M two, that's our mass M two. And we know the relationship between M one and M two. So we're gonna wanna use that. Now, the problem is we don't know V one A and we don't know V two A, right? So let's think about finding these and coming back to our conservation of momentum. So let's think about mass one immediately after the collision. OK? Or after the explosion. Sorry. So we have mass one immediately after the explosion and its initial velocity V not one is going to be V one A A that velocity we just wrote down in our momentum equation, the velocity immediately after the collision, we know that its final speed is gonna be 0 m per second because it's going to come to rest. Delta X one is going to be equal to 0.65 m. We know it comes to rest within 665 centimeters. So 0.65 m, we don't know how long this takes, but like I mentioned earlier, we can figure out some information about the acceleration. OK. So this acceleration, let's think about drawing a free body diagram for each block. Hey, we have a normal force pointing upwards, a force of gravity pointing downwards and we have a force of kinetic friction, it's kinetic friction because of the pieces are moving and that kinetic friction is going to oppose the motion. OK. So if we say that block one is moving to the right, this would be the free body diagram for block one, block two is going in the opposite direction. So the force of kinetic friction would be on the other side, it would be pointing right. But the magnitude of the forces in the acceleration acting is going to be the same for both blocks. OK? Just in opposite directions. So if we think about Newton's second law, the sum of the forces acting in this extraction is gonna be the mass multiplied by the acceleration at the pieces experience, the force of kinetic friction is acting against the motion. So we're gonna take the direction of motion to be positive, then the force of kinetic friction is going to be negative. So we have negative FK is equal to the mass multiplied by the acceleration. The force of kinetic friction recall is given by UK multiplied by the normal force and this is equal to the mass multiplied by the acceleration. Now, what is our normal force? OK. Well, in this case, our normal force must be equal to the force of gravity because those are the only two forces we have acting in the y direction. Some of the forces in the Y direction must be zero because these pieces are not moving up or down just side to side. So those two can two forces must cancel each other a equal in magnitude opposite in direction. So N must be equal to MG. So what we get is that negative mu K multiplied by M multiplied by G is equal to ma and so the acceleration is going to be equal to negative mu K multiplied by G. OK. We can go back to the information we had for M one A one is gonna be negative mu K multiplied by G. Yeah, we don't know what Mu K is but we know that it's equal or block one or piece, one and piece two. So we're gonna leave it as mu K. We're gonna go ahead with our kinematic equation. We're gonna see where that gets us. Now, for the kinematic equation, we want to include the four variables we've written here. We're excluding time because we don't have any information about it at all. And that equation is gonna be VF one squared is equal to V, not one squared plus two A one delta X one. OK. So we have zero is equal to V one A squared. Who was to multiplied by negative mu kg multiplied by 0.65 m. We can rearrange, we're gonna move everything in the second term over. We get V one A squared. It's gonna be equal to. Now G is 9.8 m per second squared. So if we have two multiplied by 9.8 multiplied by 0.65 we get 12.74 meter squared per second squared multiplied by UK. And we can take the square root to get that V one A. It is going to be equal to polo or minus a 3.56 9 3 m per second multiplied by the square root of M. All right. So we have now the speed of block one immediately after the collision, at least in terms of the coefficient of kinetic version. Now we're gonna do the same for block two, right? And block two, we have similar information. A mass and two, we know that the initial speed is gonna be some speed V two A, we know the final speed will be zero. It's gonna come to rest and it's gonna do so in a distance delta X two that is equal to D A value we're looking for our acceleration is gonna be the exact same because we have the exact same friction acting on this piece. And that's the only force we have acting in that direction of motion. So we have negative mu kg. We use the exact same equation. The F two squared is equal to V not two squared plus two A two multiplied by delta X two. OK. Zero is gonna be equal to V two A squared plus two multiplied by negative mu K multiplied by 9.8 m per second squared multiplied by the distance D. And in this case, if we rearrange moving the second term to the left hand side, and we take the square root, we get that V two A is going to be equal to plus or minus the square root of 19.6 m per second squared multiplied by mu K multiplied by the distance D OK. So this looks a little messy right now. It doesn't look super helpful. So let's take a look at the equation we had and you'll see why it is helpful. We have a relationship between M one and M two. That's gonna allow us to simplify those terms. We now have V one A and V two A written in terms of only UK and the distance D we're looking for. And so we're gonna substitute those in. And what we're gonna see is that those UK terms are going to cancel out. And when they do, we're gonna be left with just an equation with one unknown that devalue we're looking for. So what we have now and we're gonna continue from our conservation of momentum equation. We're gonna do it down below where we have a little bit more room. Yeah. So we were at the stage where we had the initial mass multiplied by the initial velocity is equal to M one multiplied by V one A plus M two multiplied by V two A. Let's substitute everything we know. So on the left hand side, we already said this would be equal to zero because initially this cube is not moving, it's at rest, the speed is zero. The mass M one. OK. We know that this is equal two M one. OK. The initial speed of block one, immediately after the collision, we found to be 3.56 93 meters per second multiplied by the square root of MK. What you'll notice is I've chosen the positive route for V one A. So we're taking the positive velocity either direction would have been OK. As long as we keep it consistent with the second piece. OK. So in this case, we're saying that mass M one has a positive velocity. We're told in the problem that they move in opposite directions. So that means when we get to mass two, we have to choose the negative, it's gonna have a negative velocity. OK? All right. So we're done with mass M one. Now we go to mass M two. So we have plus M two multiplied by the negative square root of 19.6 m per second squared, multiplied by mu K multiplied by D right. Let's move this negative term to the left hand side. On the left hand side, we're gonna get M two multiplied by the square root of 19.6 m per second squared UK D that's gonna be equal to on the right hand side. And one we know is equal to five and two. So we get five M two multiplied by 3.5693 m per second, multiplied by the square root of UK. Now, on both sides, we have M two. So that will divide out. OK. So now we don't have to worry about the mass. All we need to just know was the relationship between the two masses. We also have this square root of UK that will divide it. OK. So on the left hand side, we're gonna be left with the square root of 19.6 m per second squared, multiplied by D. That's gonna be equal to five multiplied by 3.5693 m per second, we can square both sides. So 19.6 m per second squared multiplied by D is going to be equal to five multiplied by 3.5693 m per second, all squared. We're gonna divide both sides by 19.6 and simplify and what we get is that this distance D is going to be equal to 16.2499 m. OK. And that is the distance we were looking for. That's the distance that the second piece is going to travel before stopping on that surface. Let's go back up to the answer choices. We're gonna route to one decimal place. And what we can see is that the correct answer is going to be option D 16.2 m. Thanks everyone for watching. I hope this video helped see you in the next one.

INT An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 8.2 m before stopping. How far does the lighter fragment slide? Assume that both fragments have the same coefficient of kinetic friction.

Verified Solution

Key Concepts

Conservation of Momentum

Friction and Kinetic Friction Coefficient

Newton's Second Law of Motion