Register to reply 
PingPong balls in a room? 
Share this thread: 
#1
Feb1710, 10:45 PM

P: 13

1. The problem statement, all variables and given/known data
Estimate the number of Ping  Pong balls that would fit into a typicalsize room (without being crushed). In your solution state the quantities you measure or estimate and the values you take for them. (Assume 25% of the space cannot be filled due to spherical packing.) (a) Find the volume of the room (in m^3) (b) Find the volume of a ball (in cm^3) (c) Find the number of balls (order of magnitude only) (balls) 2. Relevant equations 75% of 1000m^3 = 750m^3 750m^3(1 000 000 cm^3/1m^3) 3. The attempt at a solution (a)I said that an average room had 1000m^3 It said "Your response differs from the correct answer by orders of magnitude." (b) I said the volume of a ball had 100cm^3 It said "Your response differs from the correct answer by more than 100%." (c) I said the number of balls equaled 7.5e6 It said "Your response differs from the correct answer by more than 100%." Any help would be greatly appreciated! 


#2
Feb1710, 11:43 PM

P: 13

Anyone?



#3
Feb1810, 01:03 AM

P: 1,396

a) Do you really think a typical room is 1000m^3? that would be 20 x 20 x 2.5 m^3
b) google for "size of pingpong ball". Use formula for volume of sphere. 


#4
Feb1810, 03:46 AM

P: 139

PingPong balls in a room?
volume of pingpong ball = 4/3[tex]\pi[/tex]r^{3}
Volume of room = Length*Width*Height let's say that a standard room has a length of 7m width of 4m and a height of 2.5m. and the diameter of a pingpong ball is 4cm or 0.04m. Volume of room = 7*4*2.5 = 70m^2 Volume of room @ 75% = 70*0.75 = 52.5m^3 volume of a pingpong ball = 4/3([tex]\pi[/tex]0.02^3) = 0.0000335m^3 Fitting pingpong balls into room = volume of room / volume of pingpong ball. = 52.5/0.0000335 = 1,567,164 ping pong balls. 


#5
Feb1810, 11:57 AM

P: 13

All of this was very helpful. I did the math on my own and my answer matched yours. I went to enter it in and this is what I got. Attempt 1: (c) Find the number of balls (order of magnitude only) 1.56e6 Balls Your response differs from the correct answer by 10% to 100%. Attempt 2: (I tried rounding) (c) Find the number of balls (order of magnitude only) 1.57e6 Balls Your response differs from the correct answer by 10% to 100%. I only have one more shot at the answer and I know I am correct with the math. What am I doing wrong? 


#6
Feb1810, 05:21 PM

P: 139

Do they not state the size of the room? and specific size of the ball? that's probably why the answers are wrong, because the values we're using are just random.



#7
Feb1810, 06:34 PM

P: 15,319



#8
Feb1810, 07:30 PM

P: 139




#9
Feb1810, 07:39 PM

HW Helper
PF Gold
P: 1,949




#10
Feb1810, 07:46 PM

HW Helper
P: 2,322




#11
Aug2812, 05:22 PM

P: 1

I think the issue here is answer format. A correct answer would be written as " ~ 10^{x} #/balls in a room of x*y*z size ", where x clearly denotes the order of magnitude on base10. Still seems like a sensitive answer to write in online submission though. Note that x #/ping pong balls is clearly separate from volume.



Register to reply 
Related Discussions  
Reliable Method for Launching Ping Pong Balls at Varying Speeds?  Classical Physics  1  
How to hallucinate with pingpong balls and a radio  General Discussion  3  
New uses for pingpong balls  General Discussion  4  
Video object tracking: pingpong balls  Computing & Technology  1  
Quantum Ping Pong Balls  Quantum Physics  0 