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A perfectly elastic ping-pong ball is dropped in vacuum from a height equal to ten times its radius onto a perfectly elastic fixed sphere of the same radius. Neglec...

Question

A perfectly elastic ping-pong ball is dropped in vacuum from a height equal to ten times its radius onto a perfectly elastic fixed sphere of the same radius. Neglecting effects due to earth motion, estimate the largest number of bounces against the fixed sphere that the ball can be expected to make under optimum conditions of release.

A perfectly elastic ping-pong ball is dropped in vacuum from a height equal to ten times its radius onto a perfectly elastic fixed sphere of the same radius. Neglecting effects due to earth motion, estimate the largest number of bounces against the fixed sphere that the ball can be expected to make under optimum conditions of release.



Answers

A small ball of mass $3.0 \times 10^{6} \mathrm{kg}$ is dropped on a table from a height of 2.0 $\mathrm{m}$ . After each bounce the ball rises to 65$\%$ of its height hefore the bounce because of its
inelastic collision with the table. Estimate how many bounces occur before the uncertainty principle plays a role in the problem [Hint: Determine when the uncertainty in the ball's speed is comparable to its speed of impact on the table.

Things problem. We're looking at a bell that's bouncing, and every time it bounces, it bounces it. Half of its hide from last bounce so far. Just draw a rough sketch. If I dropped it from here comes down. It comes up to half of its height for the next one, so it would come up here, bounce back again. It goes to 1/2 of its height, so it comes up here again. Half of its height. The street comes up here half of its hide, and it continues this process until eventually stopping. We need to use a infinite geometric approximation for the total distance of the ball travels after being dropped from one meter until it comes to breast. So our a sub one is one meter, because that is the height that our ball is dropped from. That's its initial starting plane and we know are Are is 1/2 because they told us that it is cut in half every time. Every time the ball drops, it has a tight. Now that we know that we can put at, um, all this information into our formula are a sub n equals a one divided by one minus our art. And we know that we have a sum, because if we take the absolute value, what's changeless? Two different color like I thought I did. Absolute value is less than one. So let's go ahead and plug button one divided by one minus 1/2. Give me one divided by 1/2 and not equals two. So how far does the ball ball travels? It travels to beaters.

In order to solve this question. You know the equation for volume of a sphere and the equation for service area sphere. I have provided both of those or you. So aside from that, we know are starting radius is 25 centimeters and wants to know what the estimate maximum error is if our radius is within half a centimeter. All right, so we're gonna do this twice. We need to find me prime, which is equal. Thio or I r squared every defined rs a prime which is equal to a high art, of course, but that doesn't obtain using our power l a. To find our dumps and b we simply plugging over our most by fire change Another So four times 25 I swear I'm supporting it. Type this directly into a calculator and we get a delta V are changing volume rocks, family evil too. 3927 centimeters. And I'll do the same thing for a service area. So we need to play an alarm off by adults arm. So don't s a When people to pigs high times 25 times point Father, that is equal. Thio. Approximately 314.2 centimeters squared

From this question, we're gonna need to break it into a couple of parts, and it's gonna include a few different concepts. So we have a ball that's attached to a spring, and the unstructured spring length, which I call l, is 0.2 meters. We didn't start to whirl it in a circle where the circle is parallel to the ground. So just think of a thinking about holding it and then whirling it above your head in a circle. Um, and the velocity of it moving around 10 gentle in the circle V is three meters per second. When we're doing this, the supplies a centripetal force to the ball and that causes it to stretch the spring and the string stretches and amount x one of 0.1 meters. So the radius of orbit is going to be the length. The strength are the spring unstrap etched plus the length of the stretching of the springs. That's gonna be the total radius of orbit of this ball moving in a circle. That's gonna be X one plus out. So as you're doing this, if the spring isn't stretching anymore than that x one amount, the force from the centripetal acceleration f sub c. It's going to be an equilibrium with the force from the spring. And since it's a situation one, which is why I called the X one for displacement will call this s as someone for the force of the spring and situation. One quill centripetal force is equal to the mass times the velocity squared, divided by our and the force of the spring and situation one is the spring constant K times displacement X one from hooks law So we can rearrange this to solve for Kay. We find that kay here is equal to ah in the squared divided by X one times are where are is equal to X one plus l Now we don't know him, but that's okay because we can use this in the second part of the situation. Since it's the same spring, the spring constant K is gonna be the same, and our mass values are gonna end up canceling because really, what we want to find it. This is the displacement in the next situation. So now we're gonna take the same ball and the same spring, and we're going to hang it in gravitational equilibrium. So we're just gonna hang it horizontally. And so now the force of gravity s a G is an equilibrium with the force from the spring of suggest. But this this situation to something called f sub s too. So what we have here force of gravity is mass times acceleration due to gravity G 9.8 meters per second squared Mrs Equal to K. It's the same K as previously because it's the same spring times the displacement next to so If we saw for the displacement x two, we find that it's equal to M g over K. But if we go back, we just found an expression for Kay. We thought it was n v squared over x one, plus her ex one times x one plus l Sophie plugged that in. We find that this is equal to m times g times X one times x one close l divided by n v squared. So now conveniently the masses cancel. So it's okay that we didn't know that mass term because it's the same ball. So that same mass So this cancel their And so we're left with, uh, all values that we know since G is just the acceleration due to gravity and that's a constant. So we plugged the values into this expression and we find the displacement here is equal to 2.29 times 10 to the minus three meters. We can go ahead and box this and as our solution to the question.

So when you dropped from a height age starting from rest. When you reach the ground, you have velocity ik westworld off to G H. Now, if you bounce awfully high velocity of the prime, then you reach a height of age prime that say on they are related by the prime equals spoiled off to G H crime. So coefficient of restitution is given by the prime of RV equals a crime of what age Now Age is given by 100 inch. So we have a prime is between 53 inch and 58 INGE. So squared it off H prime over 100 inch is given by squared off 50 over 100 um 58 over 100 because the ball bounces begin to pity and 58 so saucing this We have the coefficient of restitution between zero born, 73 and 0.76


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