In order to convert a tough split in bowling, it is necessary ...

Question

In order to convert a tough split in bowling, it is necessary to strike the pin a glancing blow as shown in Fig. 9–64. Assume that the bowling ball, traveling at 14.0 m/s just before it strikes the pin, has five times the mass of a pin and that the pin goes off at 75° from the original direction of the ball. Calculate the speed
(a) of the pin and (b) of the ball just after collision, and
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(a) of the pin and (b) of the ball just after collision, and

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with collisions. So in this problem, we have an underwater vehicle that undergoes elastic collision with a buoy. The vehicle which has a mass of three times that buoy approaches the buoy with a velocity of 5 m per second. The buoy is initially at rest and the collision um doesn't add any rotational effects following the collision. The buoy is redirected at an angle of 30 degrees from the vehicle's initial direction. And the question wants us to calculate the post collision speed of the buoy. Below this question, we're given a diagram that shows the initial and final states of the collision. We're given four choices for our answers. Choice A is 6 m per second. Choice B is 7.5 m per second. Choice C is 8 m per second and choice D is 8.4 m per second. The first thing we need to do is label um some things in our diagram. The first thing we need to label is the mass of the buoy. So I'll label that as M buoy or MB for short and the mass of the vehicle is gonna be M vehicle or MV for short. Now, we're told that the mass vehicle is 30 times that of the buoy. So we're gonna set the mass buoy equal to M. That means the massive vehicle is 30 M, I don't know what the value or M is. So I'm gonna leave it as a variable. I also need to label the angle at which the vehicle leaves after the collision. So that's gonna be an angle theta and that's gonna be angle below the X axis that the vehicle leaves after the collision. And the vehicle is initially moving in just the um X direction, we're told that we have an elastic collision. So we're gonna have both conservation of momentum and conservation of connect energy to recall our formulas for momentum. So that is P is equal to MV. The momentum is equal to mass multiplied by velocity and momentum and velocity are both vector quantities. O call our formula for connect energy K is equal to one half MV squared. Dark energy is equal to one half the mass multiplied by the speed squared. So since we're dealing with elastic collision, we're gonna have both conservation of energy and momentum. So we'll start off with the conservation momentum and momentum must be conserved in both the X and the Y direction. Let's start off with the conservation momentum in the X direction. So we'll have the total initial momentum in the X direction is equal to the total momentum in the X direction. So initially in the next direction, I have the momentum of the vehicle. So I'll have the M vehicle multiplied by its initial speed, which is gonna be V vehicle or VV for short. And since that's uh velocity is entirely in the X direction, I don't need to break that up into components, then I'll have plus zero for the initial momentum of the buoy because it's, it's unnecessary at rest. So this is gonna be equal to our final momentum. So I have the final momentum of the vehicle. So we'll have the M vehicle multiplied by the prime vehicle, which is our final speed for the vehicle. But I need the X component of this um momentum vector. So I'm gonna multiply this by the cosine of the, that will give me the X component. We also have the um final momentum of the buoy. So we have the M bui multiplied by V prime buoy. And to get the X component, I'm gonna multiply this by the cosine of 30 degrees, which is the angle which is moving away. So at this point, I can plug in values for the uh mass and the initial speed. So I'll have for the massive vehicle, I have 30 M multiplied by the initial speed which is 5 m per second. And this is going to be equal to 30 M multiplied by V prime vehicle multiplied by cosine of the plus M for the mass of the buoy multiplied by the V prime buoy uh multiplied by cosine of 30 degrees. And I notice that I have the variable M in every single term. So I can cancel it out. That just means that the actual value of the mass of the buoy is not important. What is important is this um factor of 30 between the mass of the buoy and the mass of the vehicle. So I can simplify this a little bit. And what I'm going to do is move the V prime buoy to the left hand side of the equation just to separate out my um final speeds on separate sides of the equation. So I have 100 and 50 m per second minus V prime buoy multiplied by cosine of 30 degrees. That's gonna be equal to 30 B prime vehicle cosine data. So at this point, I have three equ uh three unknowns but only one equation. So I'll just move now to my um moment concentration momentum in the Y direction. So I'll have my total initial momentum in the Y direction is equal to my total final momentum in the Y direction. Now, initially, I have no momentum in the Y direction. Everything is either in the X direction or not moving at all. So that's gonna be zero. So that's gonna be equal. But I do have momentum in the Y direction for the buoy and the vehicle, we'll start off with the buoy. So we'll have the mass of the buoy. So M buoy multiplied by V prime buoy. And I need the Y component of this um momentum vector. So I'm gonna multiply this by the sign of 30 degrees. Then I need to look at the momentum of the vehicle after the collision that's actually going in the negative Y direction. It's gonna come in with a minus sign. So I have minus in vehicle multiplied by the prime vehicle multiplied by the sine of the. So what I'm going to do again is move the V prime bu to the other side of the equation and plug in values for our masses. So I'll have M multiplied by V prime buoy multiplied by the sign of the or sign of 30 degrees. Sorry, that's gonna be equal to the 30 M multiplied by V prime vehicle multiplied by the sign of data. And again, the variable M will cancel out. But I'm still left with, with an equation with three unknowns. I'll need to move to my last condition which is a conservation of K connect energy. So I'll have my total in this energy is equal to our total final connect energy. So initially, I'll have my connect energy for the. So I I'm sorry for the vehicle, I'll have one half vehicle multiplied by V vehicle squared plus zero because our buoy initially has zero, connect energy that's gonna be equal to one half vehicle multiplied by the V prime vehicle. That could be its final connect energy for the vehicle plus the final connect energy of the buoy, which is one half M bui multiplied by the prime bui squared. Um I can cancel a factor of one half since it shows up in every term and I can plug in values for the uh masses and for the initial speed. And I'm also going to just um move my V prime buoy term to the left hand side of the equation. So what I'm going to have is 30 M multiplied by 5 m per second quantity squared minus M multiplied by V prime buoy squared. And that's gonna be equal to 30 M multiplied by B prime vehicle squared. So again, the variable M cancels out and I can simplify things just a little bit. So I will have 750 m squared per second squared minus V prime buoy weird. And that's gonna be equal to 30 the prime vehicle squared. So at this point, I have three equations with three unknowns and so I can solve them simultaneously. There's multiple ways to do this. But the way I'm going to demonstrate here is going to eliminate the theta and the V prime vehicle variables without actually solving for their numerical values. And so way I'm going to do this is I'm first going to start off by taking my um conservation momentum in the X in the Y direction and squaring both of those equations. So for the um conservation momentum in the X direction, I'll square that equation. And so I'll get 100 50 m per second quantity squared plus V prime two squared multiplied by cosine squared of 30 degrees minus two multiplied by the 1 50 m per second, which is multiplied by V prime, which is multiplied by cosine of 30 degrees. And this is going to be equal to the quantity of 30 prime vehicle squared multiplied by cosine squared of data. We'll do the same thing for the conservation momentum in the Y direction equation. So that will give me um V prime buoy quantity squared multiplied by the stein squared of 30 degrees. And that's going to be equal to the quantity of 30 V prime vehicle all squared multiplied by stein squared of data. So one way to do is add these two equations together. And so that will give me 100 50 meters per second quantity squared plus B prime buoy squared. And this could be multiplied by cosine squared of 30 degrees plus sine squared of 30 degrees, which is one, if you recall your trick identities, then I'll have minus 300 m per second multiplied by V prime, which is multiplied by cosine of 30 degrees. And that's gonna be equal to the quantity of 30 B prime vehicle squared. And then that'll be multiplied by cosine squared plus sine squared of theta and that's going to just give me one, I've eliminated the theta term out of this. So now what I need to do is eliminate the V prime vehicle and to do that, I need to take my energy equation and multiply it by negative 30. And in doing so what I will get is negative 30 multiplied by the 750 meters squared per second squared plus 30 V prime buoy word. And then I will have that equal to the quantity of actually negative the quantity of 30 B prime vehicle. Where, so I'm gonna add those two equations out together and lots of things cancel out. The constant term cancels out because 150 is equal to 30 multiplied by sorry, 150 squared is equal to 30 multiplied by 750. And when I, what I'm left with is um 31 B prime B squared minus the 300 meters per second multiplied by V prime E multiplied by cosine of 30 degrees. And that's gonna be equal to zero because I have three, the quantity of 30 V prime vehicle squared minus the quantity of 30 V prime vehicle squared, which gives me zero. So this gives me a quadratic equation. And I could use quadratic formula if this is simple enough that I can just use it uh solve it by factoring. So I'll factor out a V prime buoy. And what I'm left with is um I have B prime buoy equal uh multiplying 31 B prime buoy minus the 300 m per second, multiplied by the cosine of 30 degrees and all that has equals zero. So I'll have two solutions. I'll have V prime buoy is equal to zero or 31 V prime buoy minus the 300 m per second multiplied by cosine of 30 degrees. S equals zero. If us all that for V prime Booy, that equals 8.4 m per second. Now, at this point, I have two answers. And the answer I want to choose is the 8.4 m per second. And that's because my V prime bui equal to zero is just my initial condition. The boo was initially at rest. So one solution is for the con uh collision to not happen. And so that would correspond to a final speed of the V prime buoy equaling to zero. But I do want the collision to happen, but I am going to choose the other speed. And if I look at the choices of answers that actually corresponds to answer D, so just to recap what we've done here, we used the conservation of momentum and conservation of energy for an elastic collision to set up um a system of equations which we then solved to find the final speed of the buoy. So I hope that this has been useful and I'll see you in the next video

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