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Problem 45 Hard Difficulty

ssm Starting with an initial speed of 5.00 $\mathrm{m} / \mathrm{s}$ at a height of $0.300 \mathrm{m},$ a $1.50-\mathrm{kg}$ ball swings downward and strikes a $4.60-\mathrm{kg}$ ball
that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 1.50 -kg ball just before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision. (c) How high does each ball swing after the collision, ignoring air resistance?


a. 5.56 $\mathrm{m} / \mathrm{s}$
b. $+2.73 \mathrm{m} / \mathrm{s}$$,-2.83 \mathrm{m} / \mathrm{s}$
c. 0.380 $\mathrm{m}$, 0.409 $\mathrm{m}$


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Video Transcript

We have a 1.5 kilogram ball in the end of a string which is already moving with a speed of five meters per second and it swings down from the string and it hits. I don't think it's quite at such a night awesome like that and it hits another ball and the bill sitting there and they have been elastic collision. So there's multiple parts of the problem. First, we need to know what is the speed of the first ball just before it collides with the stationary ball. So for this part, we can use conservation of energy. This is a simple pendulum. We know that at the top, the ball has both potential energy and kinetic energy, and at the bottom is a song. Was we defined? Our reference frame are zero to be the lowest point of the swing, then it only has kinetic energy, so we know that energy is conserved. So we can say that the potential energy at the top, plus the kinetic energy at the top, is equal to the kinetic energy at the bottom. So we know the formula for potential energy is mg times the height. So when I did forget to tell you that the height that we're starting with his 0.3 meters. So we have a massive 1.5 tanks, 9.8 times the height, which is 0.3 plus 1/2 times the mass times. The velocity squared is equal to the kinetic energy at the bottom, 1/2 times the mass times velocity at the bottom, which we don't know squared. So you can see that the 1.5 cancel out we could have done that before plugging our numbers. It's a little bit. It looks messy. It's not so bad, but this is 9.8 times 2.3 plus 1/2 of 0.5 squared is gonna be equal toe half of V squared. So if you saw that using your basic algebra, you get V at the bottom of 5.56 leaders per second. So that's the answer to part A. We're done with the first part of the question. But then the next part of the question says one of the velocities of both balls after impact, and they tell us that the collision is elastic and this is one of those cases where we have an elastic collision where the initial velocity of the second object is zero. So V not object to is zero. So in the text which I've referred to this in other problems, there is a shortcut. It's not an equation that you can use for any other situation. This is just for when the initial velocity is zero for the second object. But the relationship between the final velocity of the first object and the initial velocity of the first object is this. It's the final one is the difference of the two masses divided by the sum of the two masses, times the initial velocity of object one. So this can help us because we're going to use conservation of momentum. But without this extra relationship, we would find that we didn't have enough knowns. We had too many unknowns. So let's save that for a second and we'll do conservation of energy. So I'm sorry. Conservation moment. The initial momentum is equal to the final. In this case, what's initially moving? We have the first ball which strikes the second ball on the second ball is at rest. That one is going to be zero that's gonna be equal to the mass of the first ball times of the final of the first ball times the mass of the second ball times the final of the second ball. So we do have some values we can plug some numbers in, but we're going to get stuck because we don't know enough. So let's start with that. Will save Mass. One is 1.5 and 5.56 was our velocity just before impact, and that's gonna be equal to 1.5 times the final one. I lost 4.6, which is the mass of our second ball times of the final two. So we have two unknowns and that doesn't work. So let's use this special relationship here to find the final one. I'll keep it in sight, but, um, write it down here so the final one is going to be the difference of the masses. So we have 1.5 is our M one 4.6. Is Aaron too? In the denominator, we have the some 1.5 plus 4.6, and then that's gonna be multiplied by the initial velocity of object one which is 5.56 meters per second. So in this circumstance, because the initial velocity of object to zero were allowed to use this equation, Um, it's nice and simple because in one step I confined the initial velocity of object one which comes out if you do your calculations and you calculator negative 2.8 to 6 meters per second. And the negative scientists indicates that is moving in the opposite direction that it originally waas which kind of makes sense if we scroll back and look at our picture, it just means the ball one hit the stationary ball and it bounced back. It bounced back in this direction. So we have an answer for the first all. Now we need an answer for the second ball. So let's, um, plug it back into less equation here, our conservation of momentum equation, which we couldn't solve because we had too many unknowns. Now we can go back and plug in our value for V s one 1.5 kilograms times negative. 2.8 to 6 plus 4.6 tens v Final two. So we need to move this first term on the right hands. I we need to move this over there so it ends up. You add the 2.8 to 6 times 1.5 to the left hand side. What we have is 12.579 and around to the proper number, said kids at the end. So 12.579 is our left hand side and that's equal to 4.6 times the final velocity of object to so our final velocity of object to is positive 2.73 four. So let's summarize our answers and use the right number of significant figures. There's three significant figures in the problem, so the final one is negative 2.83 meters per second and the final two is to positive 2.73 meters per second. So this is the answer to this second part of the problem. Reach that. This is the answer to part B. Okay, the last part of the problem asks us to calculate how high each ball will swing after the impact. So again, looking at that initial set up, the 1.5 kilograms ball hits the 4.6 bounces back up to the left and eventually will come to rest. This ball here, which is on its own string, will swing up in this direction. So how high do they go? In this case, we're disc unused. The conservation of energy again. We know that whatever the kinetic energy has at the bottom will turn into potential energy at the top because we're gonna allow it to swing until it comes to rest. So the 1/2 MV squared in an equal MGH and we're talking about the same ball. So whichever ball it is, the EMS were going to cancel. We want to just developed an expression for H. So if I divide both sides by G, we have hh equals V squared over to G. So the height for object one is going to be 2.83 and there is a negative. Their real that means is it goes to the left. The negative is gonna cancel or go. I would need square it divided by two times 9.8 and that is 0.409 leaders. Words your nine meters. That's for height one and then for high, too. We have 2.73 positive, which just means it's going to the right squared. Divided by two tens 9.8 meters per second squared, that comes out 2.383 0.383 So these two are the heights of the two balls as they swing back up.