# M&M Lab

Topics: Measurement, Systems of measurement, Sphere Pages: 3 (533 words) Published: January 22, 2013
Table 1 - Indirect Measurement|
Trial | Volume (CM3) | Diameter (CM) | Radius (CM) | M&M Thickness (CM) | 1 | 75| 11.34| 5.67| 0.722|
2 | 83| 12.68| 6.34| 0.658|

Table 2 - Direct Measurement|
Trial | M&M Thickness (CM) |
1 | 0.642|
2 | 0.741|
3 | 0.683|

Table 3 - Calculated Averages|
Method | Calculated Average Thickness (cm) |
Indirect (from Table 1)| 0.700|
Direct (from Table 2)| 0.689|

Questions:
1. When you performed Step 2 of the procedure, you actually made a cylinder of M&Ms. The cylinder was rather "smushed," and the height of the cylinder was the thickness of an M&M. Recall that the equation for the volume of a cylinder is V = (3.14)r2h. A. Rearrange the equation for "h." Show your work.

V= (3.14) r2 h
(3.14)r2
(3.14)r2
___V___ = _ (3.14)r2 h_

(3.14)r2
__V___ = h

(3.14)r2
h = __V__

B. Using the data from Table 1 and your equation, calculate the average thickness (height) of an M&M for each trial. Record your calculated values in Table 1. Hint: Students often forget that they must use the radius, and not the diameter, in the equation. Copy Table 1 into the assignment. Trail 1: h = 75 / (3.14 x 5.67)2

h = 75 / (3.14 x 32.1489)
h = 75 / 100.947546
h = 0.743 cm
Trail 2 h = 83 / (3.14 x 6.34)2
h = 83 / (3.14 x 40.1956)
h = 83 / 126.214184
h = 0.658 cm

C. You now have two values for the thickness of an M&M in Table 1. Determine the average M&M thickness using these values and record your value in Table 3. (0.743 + 0.658) / 2
= 1.401 / 2
= 0.700 cm

D. You have just determined a value for the thickness of an M&M using the indirect method. What makes this method "indirect"? I measured the thickness through other means than directly measuring the thickness of a single M&M. 2. When Step 4 of the procedure was performed, a vernier...