Using series in a 2D kinematics problem

[CherryWine]

Homework Statement

A ball is rolling towards a rectangular hole which is 40cm deep and 2cm wide with a velocity 1m/s. It falls through the hole, bounces off the walls a couple of times and falls down. The direction of balls motion is perpendicular to the hole (falling in it from one side). Diameter of the ball is 0.6cm. How many times will the ball bounce off the walls until reaching the bottom? Suppose ideal elastic collisions with the walls of the hole that last for negligible time.

Homework Equations

All 2D equations.

The Attempt at a Solution

I solved the problem by using series. Firstly, using y(t) equation for 2D, I found expressions for y after one, two and three bounces from the walls. Then I checked for a pattern and saw that the general term for y is (n-1/2)*gt^2. Then I formed a series ∑(n-1/2)*gt^2 (n going from 1 to i, where i is the number of bounces needed) and then equated that series with 0,4 m which is the height of the hole. By trial and error I found the correct answer which is i=20 (the ball bounces 20 times), but I would like to know some other method because I doubt that this problem should have been solved by using series. And finally, in the series I found i by trial and error with a calculator, but I am also interested in a purely mathematical method that would give me that i. Also I found t, by using this equation D-d=V*t, where D is the width of the hole and d is the diameter of the ball, t is equal to 0,014s, that is the time needed for one bounce, and it is constant.

 

[Fightfish]

Well, because the collisions are elastic, the y-component of the velocity of the ball is unaffected by the bounces with the wall! That is to say, the ball takes the same amount of time to fall to the bottom in this case as one in which the ball simply falls to the bottom without bouncing.
Because of the elastic nature of the bounces, the magnitude of the x-component of the velocity remains the same; only its direction is reversed.

 

[CWatters]

Nice problem :-)

 

[CherryWine]

Well, because the collisions are elastic, the y-component of the velocity of the ball is unaffected by the bounces with the wall! That is to say, the ball takes the same amount of time to fall to the bottom in this case as one in which the ball simply falls to the bottom without bouncing.
Because of the elastic nature of the bounces, the magnitude of the x-component of the velocity remains the same; only its direction is reversed.

I don't think that is relevant to the questions asked here:

1. Is there another method for solving this problem without using series?
2. Is there a mathematical method for finding i in this problem without using trial and error method?

 

[Fightfish]

1. Is there another method for solving this problem without using series?
2. Is there a mathematical method for finding i in this problem without using trial and error method?

Umm, I was under the impression that I just told you such a method. You have the following facts:
(1) The time taken for the ball to fall to the bottom is the same as though there were no bounces (think about it, if you're not convinced). So, you can calculate the time taken for the ball to fall to the bottom of the well by simply using your kinematic equation [itex]\Delta y = u_{y}t + \frac{1}{2}g t^{2}[/itex] (treating downwards as positive).
(2) The magnitude of the x-component of the velocity is constant. So, the total distance traveled by the ball in the x-direction is just [itex]d_{x} = u_{x}t[/itex]. The ball hits the wall once every [itex]2 - 0.6 = 1.4[/itex]cm. So, ...hopefully you can carry on from here?

Edit: Maybe this picture will make it clearer to you. The yellow curve denotes the trajectory if there were no walls, while the blue curve denotes the trajectory if there were walls. You can see that the effect of the bounce is just to "fold" the curve horizontally.

 

[CherryWine]

Umm, I was under the impression that I just told you such a method. You have the following facts:
(1) The time taken for the ball to fall to the bottom is the same as though there were no bounces (think about it, if you're not convinced). So, you can calculate the time taken for the ball to fall to the bottom of the well by simply using your kinematic equation [itex]\Delta y = u_{y}t + \frac{1}{2}g t^{2}[/itex] (treating downwards as positive).
(2) The magnitude of the x-component of the velocity is constant. So, the total distance traveled by the ball in the x-direction is just [itex]d_{x} = u_{x}t[/itex]. The ball hits the wall once every [itex]2 - 0.6 = 1.4[/itex]cm. So, ...hopefully you can carry on from here?

Edit: Maybe this picture will make it clearer to you. The yellow curve denotes the trajectory if there were no walls, while the blue curve denotes the trajectory if there were walls. You can see that the effect of the bounce is just to "fold" the curve horizontally. View attachment 93816

I understand now. It was just counterintuitive, but now I understand it completely. Thank you, once again!