Mass — Question 5

Section 5-4 · Physics pages 680–681

Question 5 An unknown resistance can bias a one-point mass estimate In two positive-motion tests, a cart’s applied force P and acceleration a are (5 N, 1 m/s2) and (9 N, 3 m/s2). An unknown constant resistance R opposes motion, so P −R = ma. a (m/s2) P (N) (1, 5) (3, 9) Two tests with the same unknown resistance; dashed line joins observations. (a) Find m and R. (b) What mass would the naive ratio P/a give for each test? (c) Explain the advantage and limitation of the two-test method.
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Question 5 - Solution An unknown resistance can bias a one-point mass estimate a (m/s2) P (N) (1, 5) (3, 9) (a) Subtract the equations. 9 −5 = m(3 −1) ⇒ m = 2 kg , R = 5 −2(1) = 3 N 9 −5 = m(3 −1) gives m = 2 kg. Then R = 5 −2(1) = 3 N. (b) Expose the bias. mnaive,1 = 5/1 = 5 kg, mnaive,2 = 9/3 = 3 kg The naive masses are 5 kg and 3 kg, neither the correct 2 kg. The applied force is not the net force. (c) State the shared-offset assumption. ∆P = m∆a + ∆R, ∆P ∆a = m only if ∆R = 0 Subtraction removes a resistance that is the same in both tests. If resistance changes with speed or conditions, two observations do not justify this fixed-R fit without more evidence.

Question 5

An unknown resistance can bias a one-point mass estimate

In two positive-motion tests, a cart’s applied force P and acceleration a are (5N,1m/s2)(5\ \mathrm N,1\ \mathrm{m/s^2}) and (9N,3m/s2)(9\ \mathrm N,3\ \mathrm{m/s^2}). An unknown constant resistance R opposes motion, so P−R=maP-R=ma.

Diagram shown in the original SVG worksheet.

  1. Find m and R.

  2. What mass would the naive ratio P/aP/a give for each test?

  3. Explain the advantage and limitation of the two-test method.

Question 5 - Solution

An unknown resistance can bias a one-point mass estimate

Diagram shown in the original SVG worksheet.

(a) Subtract the equations.

9−5=m(3−1)⇒m=2kg,R=5−2(1)=3N9-5=m(3-1)\ \Rightarrow\ \boxed{m=2\,\mathrm{kg}},\quad R=5-2(1)=3\,\mathrm N 9−5=m(3−1)9-5=m(3-1) gives m=2m=2 kg. Then R=5−2(1)=3R=5-2(1)=3 N.

(b) Expose the bias.

mnaive,1=5/1=5kg,mnaive,2=9/3=3kgm_{\mathrm{naive,1}}=5/1=5\,\mathrm{kg},\quad m_{\mathrm{naive,2}}=9/3=3\,\mathrm{kg} The naive masses are 5 kg and 3 kg, neither the correct 2 kg. The applied force is not the net force.

(c) State the shared-offset assumption.

ΔP=mΔa+ΔR,ΔPΔa=monly if ΔR=0\Delta P=m\Delta a+\Delta R,\quad\frac{\Delta P}{\Delta a}=m\quad\text{only if }\Delta R=0 Subtraction removes a resistance that is the same in both tests. If resistance changes with speed or conditions, two observations do not justify this fixed-R fit without more evidence.