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Spheres $A(\operatorname{mas} 0.020 \mathrm{kg}), B(\mathrm{mass} 0.030 \mathrm{kg}),$ and $C$ $(\text { mass } 0.050 \mathrm{kg})$ are approaching the origin as they slide on a frictionless air table (Fig. 8.41$)$ . The initial velocities of $A$ and $B$ are given in the figure. All three spheres arrive at the origin at the same time and stick together. (a) What must the $x$ - and $y$ -components of the initial velocity of $C$ be if all three objects are to end up moving at 0.50 $\mathrm{m} / \mathrm{s}$ in the $+x$ -direction after the collision? $(b)$ If $C$ has the velocity found in part (a), what is the change in the kinetic energy of the system of three spheres as a result of the collision?

(a) $v_{C 1 x}=1.75 \mathrm{m} / \mathrm{s}$

$v_{C 1 y}=0.26 \mathrm{m} / \mathrm{s}$

(b) $\Delta K=-0.092 \mathrm{J}$

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Cornell University

Numerade Educator

University of Washington

Simon Fraser University

{'transcript': "problem. 8.69. So we have these three balls that air colliding with each other and then they're sticking together. And what we want to know is what is the magnitude and direction of velocity that ball see here needs to have In order for the combined three balls, Thio be moving off 1/2 a meter per second in the extraction after they collide. So we're told the masses of these three milligrams zero five kilograms. Now these are moving around on a frictionless table, so we know that the total momentum is going to be conserved. So, actually, let's just call the total mass large M. And then the final velocity will just called V. Since all the other velocities we're going to have subscript on them save us. I was a little bit of writing, and so this is going to be some of the momentum before the collision, and this is what we'd like to know. So let's do X first. This is a colon, so V c X is going to be equal toe one over m received times. Mm, the ex. This is going to be our half meter per second that we're wanting to see at the end by this m a The ex minus m d the ex so be, is coming in at an angle. We can see that this is a 60 degrees, too. The X axis. So this is you. Do you just use the trigonometry to find the components of this? So you co sign, it'll be negative because it's pointing that direction. It's really if you were running to find the actual thing, you'd put me here, and then it would actually be you. 100. Yeah, 180 degrees plus 60. And so the coast, I would be negative. Or you can just save yourself the bother of thinking about that. It's not really that much more bother, but anyway, you can, ah, just put the negative sign in front of the coastline of 60. It works exactly. So then, doing that, we find that this should be one point 75 meters per second, and then this is going to look pretty much the same except with wise everywhere and big ab B Y minus m. A v A. Why now? A is not moving in all of the my direction and the desired. The desire to be y of our desired white component of the velocity of the three balls after the Clyde a zero minus m b the B y. Well, get me negative. So this one's being positive and this would be a negative sign of 60 degrees would be what you want for that. And so then this is zero point two 60 meters per second. So far, so good. So for the kinetic energy, we need to know the magnitude of this vector. And like with all vectors, its magnitude is the sum of the squares of its components. This is 1.7 seven meters per second. It's the knob. Delta K is, of course, final minus the initial. This is 1/2 big Am B squared that minus 1/2 m a the A squared plus M B P V squared plus M. C BC, which we just found squared and this being a completely an elastic collision, or indeed, for any inelastic collision. We expect this to be negative, and fortunately for us, that is exactly what happens. So it's negative. 0.92 Jules"}