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Changing Units — Question 3

Section 1-3 · Physics pages 56–57

Question 3 Density: the small volume unit is in the denominator A uniform polymer has density ρ = 1.20 g/cm3. A small piece has volume V = 250 mm3. Use the model m = ρV and the exact relations 1 g = 10−3 kg, 1 cm = 10−2 m = 10 mm. Treat the supplied density and volume as exact for the calculation. 1 cm One edge: 1 cm = 10 mm Three edges determine a volume. Sample: V = 250 mm3 A density denominator is a cubic unit: all three length factors change. (a) Convert the density to kg/m3 with an explicit chain of unit factors. Explain why the numerical density increases even though grams are converted to kilograms. (b) Calculate the piece’s mass in kilograms using only metre–kilogram units. Independently check by converting its volume to cm3 and calculating in grams. (c) A program converts 1.20 g/cm3 to 0.00120 kg/m3. It then uses the correctly converted volume in m3. By what factor is its predicted mass wrong? Identify the missing operation. (d) A second material has density 950 kg/m3. Which has greater mass for equal volume? Give the density ratio without comparing bare numbers in different units.
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Question 3 - Solution Density: the small volume unit is in the denominator 1 cm One edge: 1 cm = 10 mm Three edges determine a volume. Sample: V = 250 mm3 (a) Convert both parts of the quotient. 1.20 g cm3 10−3 kg 1 g (100 cm 1 m )3 = 1200 kg/m3 . The mass-unit change supplies 10−3, but the denominator conversion supplies 106. Their net factor is 103. One cubic metre contains one million cubic centimetres; the density describes the mass per that much larger volume. (b) Use two complete routes. Since 1 mm3 = 10−9 m3, V = 2.50 × 10−7 m3, m = (1200)(2.50 × 10−7) kg = 3.00 × 10−4 kg . Alternatively 1 cm3 = 1000 mm3, so V = 0.250 cm3. Then m = (1.20 g/cm3)(0.250 cm3) = 0.300 g = 3.00 × 10−4 kg. (c) Trace the missing denominator conversion. ρwrong ρcorrect = 0.00120 1200 = 10−6, mwrong = 3.00 × 10−10 kg. The erroneous density is 10−6 times the correct one, so the resulting mass is also one million times too small: 3.00 × 10−10 kg. Converting grams alone gives 0.00120 kg/cm3, not kg/m3. Relabeling the denominator omits its factor 106. (d) Compare like with like. The polymer has the greater density and hence the greater mass at equal volume: ρpolymer ρsecond = 1200 950 = 24 19 ≈1.263 . Comparing 1.20 directly with 950 would compare incompatible numerical representations.

Question 3

Density: the small volume unit is in the denominator

A uniform polymer has density ρ=1.20g/cm3\rho=1.20\ \mathrm{g/cm^3}. A small piece has volume V=250mm3V=250\ \mathrm{mm^3}. Use the model m=ρVm=\rho V and the exact relations 1g=10−3kg,1cm=10−2m=10mm.1\ \mathrm{g}=10^{-3}\ \mathrm{kg},\qquad 1\ \mathrm{cm}=10^{-2}\ \mathrm{m}=10\ \mathrm{mm}. Treat the supplied density and volume as exact for the calculation.

Diagram shown in the original SVG worksheet.

  1. Convert the density to kg/m3\mathrm{kg/m^3} with an explicit chain of unit factors. Explain why the numerical density increases even though grams are converted to kilograms.

  2. Calculate the piece’s mass in kilograms using only metre–kilogram units. Independently check by converting its volume to cm3\mathrm{cm^3} and calculating in grams.

  3. A program converts 1.20g/cm31.20\ \mathrm{g/cm^3} to 0.00120kg/m30.00120\ \mathrm{kg/m^3}. It then uses the correctly converted volume in m3\mathrm{m^3}. By what factor is its predicted mass wrong? Identify the missing operation.

  4. A second material has density 950kg/m3950\ \mathrm{kg/m^3}. Which has greater mass for equal volume? Give the density ratio without comparing bare numbers in different units.

Question 3 - Solution

Density: the small volume unit is in the denominator

Diagram shown in the original SVG worksheet.

(a) Convert both parts of the quotient.

1.20gcm310−3kg1g(100cm1m)3=1200kg/m3.1.20\ \frac{\mathrm{g}}{\mathrm{cm^3}} \frac{10^{-3}\ \mathrm{kg}}{1\ \mathrm{g}} \left(\frac{100\ \mathrm{cm}}{1\ \mathrm{m}}\right)^3 =\boxed{1200\ \mathrm{kg/m^3}}. The mass-unit change supplies 10−310^{-3}, but the denominator conversion supplies 10610^6. Their net factor is 10310^3. One cubic metre contains one million cubic centimetres; the density describes the mass per that much larger volume.

(b) Use two complete routes.

Since 1mm3=10−9m31\ \mathrm{mm^3}=10^{-9}\ \mathrm{m^3}, V=2.50×10−7m3,m=(1200)(2.50×10−7)kg=3.00×10−4kg.V=2.50\times10^{-7}\ \mathrm{m^3},\qquad m=(1200)(2.50\times10^{-7})\ \mathrm{kg} =\boxed{3.00\times10^{-4}\ \mathrm{kg}}. Alternatively 1cm3=1000mm31\ \mathrm{cm^3}=1000\ \mathrm{mm^3}, so V=0.250cm3V=0.250\ \mathrm{cm^3}. Then m=(1.20g/cm3)(0.250cm3)=0.300g=3.00×10−4kg.m=(1.20\ \mathrm{g/cm^3})(0.250\ \mathrm{cm^3}) =0.300\ \mathrm{g}=3.00\times10^{-4}\ \mathrm{kg}.

(c) Trace the missing denominator conversion.

ρwrongρcorrect=0.001201200=10−6,mwrong=3.00×10−10kg.\frac{\rho_{\mathrm{wrong}}}{\rho_{\mathrm{correct}}}=\frac{0.00120}{1200}=10^{-6},\qquad m_{\mathrm{wrong}}=3.00\times10^{-10}\ \mathrm{kg}.

The erroneous density is 10−610^{-6} times the correct one, so the resulting mass is also one million times too small: 3.00×10−10kg3.00\times10^{-10}\ \mathrm{kg}. Converting grams alone gives 0.00120kg/cm30.00120\ \mathrm{kg/cm^3}, not kg/m3\mathrm{kg/m^3}. Relabeling the denominator omits its factor 10610^6.

(d) Compare like with like.

The polymer has the greater density and hence the greater mass at equal volume: ρpolymerρsecond=1200950=2419≈1.263.\frac{\rho_{\mathrm{polymer}}}{\rho_{\mathrm{second}}}=\frac{1200}{950} =\frac{24}{19}\approx\boxed{1.263}. Comparing 1.20 directly with 950 would compare incompatible numerical representations.

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