(III) Two billiard balls in contact with each other are struck...

Question

(III) Two billiard balls in contact with each other are struck by a third ball moving at 5.5 m/s as shown in Fig. 9–46a. If both of the first two balls move after the collision at an angle of 32° to the initial path of the third ball (Fig. 9–46b), find the velocities of all three balls after the collision. Assume all balls have the same mass, and the collision is elastic.
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Accepted Answer

Hey, everyone in this problem, two identical marbles on a frictionless surface are initially at rest. They are then struck by a third identical marble moving horizontally at 1.5 m per second. As shown in the figure below. We're gonna look at that figure in just a minute after the collision. Both of the first two marbles move at an angle of 45 degrees to the initial path of the third marble. We're told to assume the collision is perfectly elastic and we're asked to find the speeds of all three marbles post collision. Now, we're given the diagram so we can see that marble A and Marble B before the collision are not moving and marble C is moving towards them. Ok, horizontally. And we know that that's a speed of 1.5 m per second. After the collision, we can see that marble A moves 45 degrees above the horizontal at this speed. Va prime marble B moves 45 degrees below the horizontal at a speed of VB prime. And we have that marble C is still going to be moving horizontally. Hm Now we're given four answer choices in this question each of them containing different values for those three speeds. VA prime VB prime and VC prime. And we're gonna come back to these answer choices as we work through the problem. Now, let's start by writing out the information that we're given. So first things first is the mass, OK? And these marbles all have the same mass. So we can say that the mass of marble A MA is equal to the mass of marble BMB is equal to the mass of marble C MC. And we're just gonna call that mass. Maybe these are identical marbles, they have the same gun. Now, we think about what happens before the collision and before the collision, we have no motion in the Y direction at all. OK? Marble C is moving only in the horizontal direction. Marbles A and B are not moving. And so we actually have no motion in the vertical direction at all. So the initial velocity of marble A V not A is gonna be equal to zero. The initial velocity of marble BV, not B is also gonna be equal to zero. They're initially at rest. The initial velocity of marble C in the X direction. Maybe that's where all of its motion is it as a speed of 1.5 m per second will take to the right to be our positive direction. So that's just gonna be 1.5 m per second and we don't know, sorry, we do know that there is no motion in the Y direction. So V not cuy is 0 m per second, there is no initial vertical motion. OK. So that's everything going on before the collision, we can flip to after the collision. And after the collision, there's a little bit more to consider. So after the collision, we're gonna start with marble A, OK. And marble A is gonna have an X velocity after the collision. V fa X that's related to its speed through cosine because we're looking at the adjacent side. So what we have is va prime cosine of 45 degrees. Similarly, in the Y direction, we're gonna have V fa Y is equal to va primes of 45 degrees right now, taking this as positive VA Y tells us that we're taking the upward direction as positive. So when we flip over to looking at marble B, we have to be a little bit careful. It's still going in the positive X direction, but now it's moving downwards. And so that Y velocity for marble B will be negative. OK. So we have VFBX is VB prime cosine of 45 degrees and VFBY is VB prime, sorry, negative VB prime sign of 45 degrees. OK. So don't forget that negative there. And then when we look at C, we know that the final velocity of marble C in the X direction is gonna be its total final velocity VFC. OK? Because it is not moving in the vertical direction according to that diagram. OK. So this is everything we have going on. There's a lot of variables here. Um And so it's a good idea to write them out so that we're very clear and that we don't mix them up when we start substituting things in. Now, the first thing we wanna do is think about a conservation of momentum. We have this collision, we're gonna assume there are no net external forces acting on these marbles. And we're told that the collision is perfectly elastic. So we know we have this conservation of momentum. Now we have motion in both the X direction and the Y direction. So when we talk about conservation of momentum, what it tells us is that the momentum in the horizontal direction will be conserved and the momentum in the vertical direction will be conserved. So we have to consider those two directions separately. So let's start by looking at our conservation of momentum and we're gonna start in the Y direction because there's a little bit less going on there in the Y direction. Marble Sea never has any vertical velocity. Um And so that's already illuminating one of those unknowns for us, we just have to deal with marble A and B. So we're gonna start there. So in the Y direction, we know that the initial momentum in the Y direction P not Y must be equal to the final momentum in the Y direction pfy now initially, right marvels A and B are not moving marble C has only horizontal velocity. And so the initial momentum in the Y direction is just going to be zero. All right. Now, the final momentum in the Y direction is gonna be made up of two components. We have the final momentum for marble A, we have the final momentum for marble B in that direction. Marble C is only moving horizontally even after the collision. And so it is not going to contribute to that momentum in this direction. So what we get is that zero is gonna be equal to now the momentum of marble A after the collision recall that momentum is given by mass multiplied by velocity. So it's gonna be ma multiplied IV FA Y and then we're gonna add the momentum of Marble B after the collision, which is gonna be MBVFBY. And again, looking strictly in that Y direction. Now we know these final velocities in the Y direction, at least in terms of va prime and VB prime, those speeds that we're looking for. So we can substitute those in the mass ma, the mass and B we know that those are equal, we know that that's equal to M. So we can substitute that name as well. And we're actually in a factor that common factor L so we get that zero is equal to the mass M multiplied by va prime sign of 45 degrees minus VB prime sign of 45 degrees. Now, we can divide both sides of this equation by M, we can divide both sides of the equation by sin of 45 degrees. We're gonna be left with zero is equal to va prime minus VB prime, which tells us that va prime must be equal to VB prime. OK. So the speed of marble A and Marl B after the collision is the same, they're moving in different directions, but the speed is going to be the same. If we take a look at our answer choices, we can already eliminate some of these options. So for option A, the speed for A and the speed for B are both 1.60 m, 1.06 sorry meters per second. So that's a possibility. Option B and C have different speeds for A and B, either 1.06 or 2.06 m per second and then they just switch right. So those can't be options. We know these speeds must be the same. So option B and option C can be eliminated. Option D has both of these speeds are 2.06 m per second. So that again is an option. So we've already eliminated two of those answer choices. Now we're gonna keep going with our solution to figure out which of those two remaining choices will be correct. All right. So we've done our conservation of momentum in the Y direction. OK. Now we're gonna flip over and look in the X direction and in the X direction, the same thing goes, we have that the initial momentum in that direction will be equal to the final momentum in that direction. Right now, our momentum, we have three components. Marble ABC. OK. So we have the initial momentum of marble A in the X direction. Who was the initial momentum of marvel B in the X direction was the initial momentum of marble C in the X direction that's equal to the final momentum of A in the X direction. What the final momentum of B in the extraction was the final momentum of C in the extraction. Now, marble A and marble B are initially not moving. So those are not gonna contribute any momentum. So initially, we only have marble C, we're gonna have the mass multiplied by the velocity. So we have the mass of marble C MC multiplied by its initial velocity in the extraction 1.5 m per second. And this is gonna be equal to, we have the mass of a massive B massive C and their corresponding velocity. So we have ma multiplied by va prime multiplied by cosine of 45 degrees was MB VB prime cosine of 45 degrees M CV FC. Now, these masses are all the same. So we can divide by all of those masses. We get that 1.5 m per second. Must be equal to. Now, we found that VA prime was equal to VB prime. OK. So we've divided everything by these masses. We have va prime cosine of 45 degrees plus VB prime cosine of 40 degrees. But VA prime and VB prime are the same. So we can write this as two, the apron multiplied by cosine of 45 degrees plus VFC. Now, we could have done the opposite. We could have written it in terms of VB prime as well, but we went ahead and wrote it in terms of the apron. So now what we can do is write this relationship between this final velocity of marble C and the speed of marble. A after the collision. Hm Now cosine of 45 degrees recall that's gonna be the square root of two divided by two. So we can use that to simplify. We have 1.5 m per second is equal to two va prime multiplied by the square root of two divided by two philosophy FC. And those twos will divide out, we can subtract va prime multiplied by the square root of two from both sides. And we're gonna have that, that final velocity of marble C is gonna be 1.5 m per second minus the square root of two multiplied by va prime. Yeah. So we have two equations. Now we know that VA prime and VB prime are equal. That was equation one. OK. We can label that as equation one. Now we have this second equation, we'll label as equation two that tells us that the final velocity of marble C is 1.5 m per second minus the square root of two multiplied by va prime. Now, this is a perfectly elastic collision. So not only do we have a conservation of momentum, but we have a conservation of kinetic energy as well. So let's look at the conservation of kinetic energy in order to get some more information here. OK. Right now we have two equations with three OMs. We need an additional equation. So conservation of kinetic energy tells us that the initial kinetic energy kno will be equal to the final kinetic energy. KF. Now, initially, we only have kinetic energy from marble C marble A and B are not moving. Recall that kinetic energy is given by one half MV squared. So we're gonna have one half M CV not C squared. This is gonna be equal to that final kinetic energy. And we have to consider all three marbles. So we have one half ma V fa squared plus one half MBVFB squared plus one half M CV FC squared. Right now, these masses are all the same, all right. So we can divide those out. We have one half in every term, we can divide that one half out as well. The initial velocity of marble C 1.5 m per second. So we have 1.5 m per second squared is gonna be equal to the final speed of marvel A we know is VA prime. So we have VA prime squared plus the final speed of BVB prime squared plus the final speed of C squared VFC squared. Hm. Now let's keep going. We know that we can write VFC in terms of VA and we could also write it in terms of VB as well. OK. So we could write VFC is equal to 1.5 m per second minus the square root of two multiplied by VB prime. Those two are equivalent. So let's go ahead and substitute everything in this equation in terms of VB prime. And we wanna leave it with only one unknown. So we can write 2.25 m squared per second squared is equal to VA prime squared, which is just equal to VB prime squared plus VV prime squared plus 1.5 m per second minus the square root of two VB prime all squared going ahead and simplifying on the right hand side, we have 2.25 m squared per second squared is equal to two VB prime squared plus 2.25 m squared per second squared minus 3 m per second, multiplied by the square root of two VB prime plus two VB prime squared. The 2.25 m per second squared that will add to zero. So we have zero is equal to, we can factor out this BB prime term zero is equal to VB prime multiplied by four VB prime minus 3 m per second multiplied by the square root of two. So from the first factor we get that VV prime is equal to zero. OK. Now that corresponds with that initial time point. So that's not the one we want, we want this second solution which is gonna be four VB prime is equal to 3 m per second, multiplied by the square root of two. This tells us that VV prime will be equal to 3 m per second, multiplied by the square root of two divided by four. So three route two divided by four meters per second. And this gives us approximately a speed of 1.061 m per second. OK. So that is our spinal speed for VV. That is VV prime. So that's one of those values we wanted to find. Now we know that VA prime is going to be equal to VV prime. So VA prime is also gonna be 1.061 m per second. The last thing we have to do is figure out VFC. Now VFC, we have this equation two, we have 1.5 m per second minus the square root of two multiplied by VB prime, which we have just found to be three fruit two divided by 4 m per second. Hm What this gives us is VFC is 1.5 m per second minus 1.5 m per second. And so it's just going to be equal to zero. Hm. And so what we have is that marble A and marble B will have the same speed of approximately 1.06 m per second. And marble C will not be moving. It will have a speed of 0 m per second. We take a look at our answer choices. We can see that this corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

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