Determine where to place a cup to catch a ball rolled off a ramp.
What are the key ideas:
Things you should be able to answer...
- The horizontal and vertical motion of the ball can be broken in to two separate motions.
- Questions:
- What are the unknowns in the x?
- What can you measure and what can you calculate?
- What are the unknowns in the y?
- What can you measure and what can you calculate?
- What is the variable that is the same for both?
- Which direction allows you to calculate the time?
Would the pythagorean theorem be relevant or useful in this case? - Katie H
ReplyDeleteI dont think it would Katie because the line of direction from the ball to the cup ins't a straight line. This would make the "hypotenuse" a curved line and that wouldn't work. Therefore the pythagorean theorem is not relevant.
DeleteOk. Thanks. -Katie H
DeleteIn other words it forms a parabola. You will probably remember it from previous labs.
ReplyDeleteExactly Derek
DeleteWould finding the distance of the bottom of the stand to the bottom of the ramp help?
ReplyDeleteIt most likely would because you need to know how long it takes before it launches off the flat part of the ramp. It's only a very little portion, but could help determine how much the ball needs to fly off of to start its final launch to the ground.
Delete-Jake Naumann
Anything that you can measure will be useful and helpful!
DeleteAgree with Nazia
DeleteThe problem is similar to the things we did with the bear in that two things are occurring at the same time. (1)We have to find how far the ball goes before it hits the ground and (2)how long it takes for the ball to fall to the ground. The ramp is only there to give the ball a horizontal velocity before the drop and we're going to just measure that. -Will F
ReplyDeleteBesides the velocity, there are going to be other things you can measure that will be needed. -Will F
DeleteWouldn't going down the ramp be considered vertical velocity too because the acceleration of gravity (9.8 m/s^2) is acting upon it? If not, calculating the horizontal velocity would just be V=dt right?
DeleteWe don't need to know the acceleration though for the vertical velocity, right? As long as we know the velocity once it leaves the ramp were are fine.
DeleteIt can all be determined with further examination and the application of more realistic measurments.
DeleteMr. Crane, when doing this lab are we allowed to change the starting velocity of the ball if we push it instead of drop it down the ramp?
ReplyDeleteGianna, if you did push the ball, it would mess the whole lab up with what he have already discovered. We would have to measure the starting velocity and since we only get one shot, it would be almost impossible to nail down the correct starting velocity...
Delete-Jake naumann
Gianna, you want to find the easiest solution that finds you the correct answer and pushing the ball down the ramp makes this problem more complicated than it has to be.
DeleteIt makes everything a lot easier when you have a starting velocity of 0, so it's almost a privilege that that's what we have. If you pushed the ball down, there wouldn't be an accurate enough way to calculate the starting velocity because it's instantaneous, constantly changing, and you'd probably need calculus to solve it.
DeleteYou don't want to push the ball because then the ball would not have the same velocity or time which would change the experiment. Also, the difference of force you put on the ball would not be constant every time.
Delete-Skylar Young
^ sorry that was Gianna B. , forgot my name hahaa
ReplyDelete
ReplyDeleteHow would we be able to calculate the time and velocity without practice runs?
You use the stopper on the ramp to prevent the ball from actually flying off.
DeleteWe can drop the ball to figure out the velocity and time but we have to put the holder side up. We can't actually perform the experiment with everything set up though.
Delete-Katie Cabrera
We'll be using the PhotoGate at the end of the ramp to calculate the time with the stopper at the end since we can't actually perform the entire thing. When the ball goes through the PhotoGate, it will give you its instantaneous time, which if done correctly, should be as close to zero as possible.
DeleteAs for velocity, you would take that time, and multiply it with the distance, which we were told is the circumference of the ball we're using.
DeleteYou mean diameter, right? We don't know the circumference. And that makes sense now about find the velocity. Thanks.
DeleteYeah, I meant to say diameter. You're welcome, and thank you for correcting me.
DeleteSo we would multiply by the diameter of the ball not the length of the ramp? That is what I got from the previous comments.
DeleteWhat ball will be used during the experiment? Won't the weight of the ball be a big factor in the velocity of the ball?
ReplyDeleteWe concluded in class that a heavy balle would work the best. I think we are going to use a metal ball for this reason
DeleteYou want a ball with a greater mass, but a smaller size.
DeleteRemember what we talked about in our past lessons? The acceleration of gravity (when being the only force) is always about 9.8m/s/s
DeleteAlso, we decided that the diameter of the ball will be measured. Everything will be put into consideration
DeleteWe are using metal balls because thy are heavy and small
DeleteDoes this mean we have to use the same ball every time, or are they all similar enough that we don't need to use the same one?
Delete-Mike Keough
It would be best to use the same ball every time. While they are all roughly the same diameter, I'm pretty sure that some are heavier than others.
Delete-Kevin Meglathery
Are we going to have to factor in the ball's deceleration?
ReplyDelete-will F
The smaller heavier ball will have the shortes drop and will have a better chance of landing in the cup.
ReplyDeleteIn my notes it looks like we need the time to find the change in position, but we need the change in position to find the time. We can not solve for two variables in one equation. Any ideas?
ReplyDeleteI'm thinking that you should try separating the one equation into two, and have each one equal what you're trying to find.
DeleteThis whole lab has several parts Chuck. While it seems confusing that you need "a" to find "b" and "b" to find "a", it's actually a bit simpler. You know that the change in P is the diameter of the ball. You know from the photogate what time it takes the ball to cross, giving you "ball lengths per fraction of a second."
Delete-Kevin Meglathery
Are we going to work on this in class with a Partner?
ReplyDeleteYes, he assigns you a partner in class.
DeleteYes, he will assign them in class.
DeleteIt seems like the slight drop the ball takes after leaving the ramp is a big factor. How do we include that in the equation?
ReplyDeleteI was wondering the same thing John, and I am not too sure either.
DeleteThe instant that the ball leaves the rubber part of the track, it is going perfectly horizontal. It is also going fast enough that it completely bypasses the drop, so it doesn't have any impact on the lab.
Delete-Kevin Meglathery
What I'm really confused about is how is finding out all of the information (acceleration, velocity, etc) going to help us in the placement of the cup so that the ball can go in?
ReplyDeleteOnce you find all the variables, you plug them into one of the equations to get P, the position of the cup in relationship to the table
DeleteI had the same problem! But now it makes so much sense, I can't believe I didn't realize that before. Thanks Andrew.
Delete-Mike Keough
would friction have any effect on the ball? and it would also depend on the amount of force applied to the ball as it is dropped down the ramp
ReplyDeleteThe amount of friction is small enough that we can disregard it in this case. If it did matter, we would have more calculations to do and we didn't learn any of that yet. Just keep everything to the simplest scenario.
DeleteOk thanks Nazia. Thats what i assumed but i wanted to make sure
DeleteI was wondering the same thing Andrew. So if friction is not significant enough when rolling across the table, is there any force or factor that changes the velocity of the ball?
DeleteIn other words, does the velocity of the ball change at all during its time from leaving the ramp and leaving the table? This is including the drop from the ramp to the table.
DeleteYou treat the lab as two separate problems. if you get the time it takes the ball to travel down the ramp using the photogate, you can find the velocity using v=a(t)+Vnaught
DeleteI wasn't thinking of using that equation, but that's much more simpler to use than V=dt. Thanks, Andrew!
DeleteThe wind resistance of the ball is small enough to disregard as well, correct?
Delete-Mike Keough
The wind resistance is so small that Mr. Crane said we can disregard it becuase it won't make that much of a difference.
Delete-Skylar Young
Keep things as simple as can be. Disregard anything too complex or something we haven't learned how to do in class.
DeleteThis comment has been removed by the author.
ReplyDeleteAre we supposed to do these questions for homework or in class when we start to do the experiment with our partner??
ReplyDeleteIm guessing in class, but don't take my word for it.
DeleteThese questions are just what we need to answer during the lab. They are just preparing you for the lab in class.
Delete-Skylar Young
I think we started to go over them in my class, but that s just period 3.
ReplyDeleteMy class is also starting tomorow, in period 4.
DeleteFor finding the velocity, would that be assuming the average veloctiy throughout? And would we measure the equation terms of meters or centimeters?
ReplyDeletethe velocity will stay the same through out because the ball is at a constant motion.
DeleteWill we assume the acceleration is still 9.8 meters per second even though it is going horizintal as well as vertical?
ReplyDeletemoving horizontal a=0 m/s but yes vertically a=9.8 m/s
DeleteYou have to think of vertical and horizontal movements as two different problems, and then combine them. For this problem, you have to do the vertical first.
DeleteDo these questions have to be answered by next class?
ReplyDeleteI don't think they necessarily need to be answered. It's more like these are things you'll be looking for and need to understand once you start the actual lab.
DeleteI am 99% sure this is the correct work if anybody wants to check it out... http://imgur.com/nsPQdRV.jpg
ReplyDeleteI would be careful, because the ramps may be set at different heights.
DeleteYou are right. This work is for a ball that is .019 m in diameter and travels through the photogate in 0.0072 seconds to clear everything up. If the ramp is at a different hieght, and the ball still goes through the photogate in the same time, the work should still work.
DeleteAlso everybody needs to disregard everything about the ramp itself. The ramp is an addition to the problem that you do not need. The ramp is just letting the ball gain speed and is not signifigant in the problem we are trying to solve. The ramp makes the ball accelerate, but as soon as the ball passes through the photogate, you have the velocity, which is not going to change. All of the major work which we must think about is the ball travelling through the air. The ball will drop at the speed of gravity. This is the time it will take to hit the ground (you can measure the height of the cup to be more precise. The velocity of the ball and the time it takes for it to hit the ground is what you need to calculate the distance the ball will travel in that time. I used the equation P=1/2at^2 + Vot + Po You can change it to P=Vot + Po since their is no horizontal acceleration. I like to simplify it to Delta P = Vot + Po . From here all you have to do is plug and chug. The main issue here is splitting the problem up into sections and gradually get answers, until you have everything you need to solve the problem.
ReplyDeleteThat's basically everything everyone needs to understand for the lab, and you explained it so simply. Nice job, Harrison! This guided me in the right direction to get the position of where the cup needs to be, so thank you!
Deletehas anyone ever seen the myth busters episode where they shot a bullet and dropped a bullet and they both hit the ground ay the same time? well that's kind of similar to this lab in the sense that the ball is the bullet being shot. if we rolled the ball of the ramp, and let it hit the ground, and dropped it from the same height it would hit the ground at the same time just like the bear crossing the river, independent vectors have no effect, if we can calculate the time it takes for the ball to hit the ground from the height of the table we have found our answer- Christian Toto
ReplyDeleteWell yes, but we need not only to know the time it takes to hit the ground, but also that distance it travels.
DeleteI think it would be helpful to everyone if they understood that you have two different equations to solve: x and y. All the variables about the distance (position) of x and y are different except one thing that relates them together, which is time. If you PUKES it out, you'll find the time and position of the cup.
ReplyDeleteYeah, if I would have known this yesterday then it would've been easier.
Deleteshould we write down what we know by class tomorrow?
ReplyDeleteThe answers to the questions on top of the page will help us in class tomorrow
Deletewith the chart he made in class a lot of the questions posted we already have the answer or have an idea of how to solve them, but writing your thoughts never hurts
Deletewould a=9.8m/s for horizontally and vertically or are assuming there is 0 gravity of the ball when going horizontal?
ReplyDelete9.8 m/s/s only affects the vertical change because gravity only goes down. It doesn't affect anything horizontally.
Delete-Kevin Meglathery
thanks kevin, i just didnt know if we were assuming there was gravity going horizontal because as the ball is traveling horizontal it is still dropping at a rate of 9.8m/s/s
DeleteWhen you are going to shoot double check your pukes to make sure you will make it and get an A.
ReplyDelete