Author |
Topic: Falling Ball (some physics required) (Read 4217 times) |
|
NickH
Senior Riddler
Gender: 
Posts: 341
|
 |
Falling Ball (some physics required)
« on: Feb 8th, 2003, 4:24pm » |
 Quote  Modify
|
Here's a link to an ingenious physics puzzle: The Falling Ball Puzzle [Broken link removed]. Remarkably, the answer is independent of both the initial velocity of the ball, and the acceleration due to gravity.
Not much physics knowledge is required. The fact that the collision is perfectly elastic means that the ball's speed immediately before and immediately after impact is the same. No energy is lost. Also, angle of incidence equals angle of reflection, as measured from the normal.
The Falling Ball Puzzle
A small ball of mass m is launched horizontally with initial velocity V from the lip of the semi-cylindrical depression of radius R. The ball makes a perfectly elastic collision with the depression and is observed to rise straight up.
How high will it rise above the lip of the depression?
(Consider the ball to be a point mass.)
Puzzle due to Dr. Akaske, formerly at Hamline University
//Broken link removed and puzzle statement added by Icarus
|
| « Last Edit: Sep 3rd, 2003, 6:56pm by Icarus » |
 IP Logged |
Nick's Mathematical Puzzles
|
|
|
SWF
Uberpuzzler
Posts: 879
|
|
Re: Falling Ball (some physics required)
« Reply #1 on: Feb 10th, 2003, 7:46pm » |
Quote Modify
|
This one looks easy at first, but the algebra gets tricky if you are not clever. The height above the lip I come up with is about 0.44R or exactly:
|
|
IP Logged |
|
|
|
jabhiji
Newbie
Gender:
Posts: 24
|
|
Re: Falling Ball (some physics required)
« Reply #2 on: Feb 11th, 2003, 11:21am » |
Quote Modify
|
Nice to know that an analytical solution can be found, SWF.
But I was lazy and did this numerically. The ball hits the circle about 17 degrees from the bottommost point and rises 0.437 R, is what my computer claims.
The expression for rise above the lip was
(1 + Sin(x))^2
----------------- R
Cos(x)
where the angle x is the solution of
1 + Sin(x)
Tan(2x) = ---------------
2 Cos(x)
which turned out to be x = 0.29725 (17 deg.)
I think that the problem statement is a bit misleading when it says the answer is independent of both V and g. There is only a specific V and g that makes the ball hit and bounce up vertically.
|
|
IP Logged |
|
|
|
jabhiji
Newbie
Gender:
Posts: 24
|
|
Re: Falling Ball (some physics required)
« Reply #3 on: Feb 11th, 2003, 2:43pm » |
Quote Modify
|
Here is a slight extension to the problem:
What shape should the pit be so that no matter what the initial horizontal launch velocity V, the ball always bounces in the vertical direction after hitting the pit ?
An extreme case: for V ---> infinity, the trajectory is almost horizontal, and thus the curved pit makes an angle of 45 deg. when it meets this horizontal.
|
|
IP Logged |
|
|
|
redPEPPER
Full Member
Posts: 160
|
|
Re: Falling Ball (some physics required)
« Reply #4 on: Feb 12th, 2003, 6:10am » |
Quote Modify
|
on Feb 11th, 2003, 11:21am, jabhiji wrote:| I think that the problem statement is a bit misleading when it says the answer is independent of both V and g. There is only a specific V and g that makes the ball hit and bounce up vertically. |
|
More precisely, for a given g, there's only a specific V that makes the ball bounce vertically. And vice-versa.
|
|
IP Logged |
|
|
|
aero_guy
Senior Riddler
Gender:
Posts: 513
|
|
Re: Falling Ball (some physics required)
« Reply #5 on: Feb 12th, 2003, 12:32pm » |
Quote Modify
|
OK, I tried to solve this one when it first came out, but had a bit of a problem. I solved fo the height it reached at the end using a basic energy equation. This is dependent upon something like V/g^2, or vice versa, I don't recall. Then I use two triangles, one describing the depth from impact point to lip/center to impact point (directly)/center to impact point (horizontally), and another describing the velocity components at impact. These equations along with a relation of the height fallen to distance travelled should be enough to solve it. However, when I solve for the horizontal distance travelled (an intermediate step to the answer), I always get that it either lands in the near half of the circle (in which case it cannot bounce up) or that it flies outside the circle (a clear impossibility). The reasoning is solid, so I assume there is a minor logic error somewhere. Generally I was wondering, though, if this was the long way around to the answer and if there was a simple, more elegant way of accomplishing it.
|
|
IP Logged |
|
|
|
SWF
Uberpuzzler
Posts: 879
|
|
Re: Falling Ball (some physics required)
« Reply #6 on: Feb 12th, 2003, 5:26pm » |
Quote Modify
|
For Aero guy, here is more detail. Use a coordinate system with origin at center of the circle, and let t be the time it takes for the particle to travel between starting point and the impact point. Also, angle phi equals the angle of incidence and reflection from the impact point.
At impact the x and y components of velocity are V and -g*t, squaring these and adding gives magnitude of velocity. Magnitude of velocity also equals V/sin(2*phi) because before impact, x component of velocity is V and direction is angle 2*phi from vertical. This allows t to be expressed as t=V/g/tan(2*phi). Also, conservation of energy immediately gives maximum height after bouncing off the circle, h=V2/2/g.
The x coordinate of the point of impact is V*t-R=R*cos(phi). The y coordinate of impact point is -gt2/2 = R*sin(phi). Substituting the expressions found for t and h gives two equations:
Dividing one by the other eliminates h and R to give an equation that involves only phi. The expression can be put in terms of just sin(phi) but simplifying with various trig identities and factoring out (1+sin(phi)). This leaves a quadradic in terms of sin(phi), whose only physically reasonable solution is:
After using this to give cos(phi) and tan(2*phi), either of the two equations above to gives h in terms of R.
|
|
IP Logged |
|
|
|
aero_guy
Senior Riddler
Gender:
Posts: 513
|
|
Re: Falling Ball (some physics required)
« Reply #7 on: Feb 13th, 2003, 12:08pm » |
Quote Modify
|
Thanks SWF, the process was a little different than I used, but I found my error. It is interesting. In most riddles you are not supposed to give the answer, but rather describe the process in the forum. For these math type puzzles, though, the 'answer' IS the process, and the number you get at the end is just a consequence of it.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print