Maximizing Space: Estimating the Number of Ping-Pong Balls in a Room

[albinoboy]

Homework Statement

Estimate the number of Ping - Pong balls that would fit into a typical-size 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)

Homework Equations

75% of 1000m^3 = 750m^3
750m^3(1 000 000 cm^3/1m^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!

 

[albinoboy]

Anyone?

 

[willem2]

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.

 

[Sirsh]

volume of pingpong ball = 4/3$$\pi$$r³

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($$\pi$$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.

 

[albinoboy]

volume of pingpong ball = 4/3$$\pi$$r³

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($$\pi$$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.

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?

 

[Sirsh]

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.

 

[DaveC426913]

volume of pingpong ball = 4/3$$\pi$$r³

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($$\pi$$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.

Sirsh, this is in violation of https://www.physicsforums.com/showthread.php?t=5374".

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

Please avoid this in the future.

 

[Sirsh]

Sirsh, this is in violation of PF rules

I'll be sure to not let it happen again.

 

[collinsmark]

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?

I haven't gone through the math myself. But one thing I've noticed that might help is that the program stipulated in the question (c) "order of magnitude only". However you have mentioned that you were entering things like "1.57e6". That's far more precise than just the order of magnitude.

 

[ideasrule]

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%.

That's an epic programming fail. The question asks for an order of magnitude. 10-100% is clearly within an order of magnitude.

 

[ginsengsniper]

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 base-10. Still seems like a sensitive answer to write in online submission though. Note that x #/ping pong balls is clearly separate from volume.