Investigation 1
[16 marks]This question asks you to find a power law from experimental data by plotting logarithms.
In a physics practical, Sofia releases a ball from rest and measures the time t seconds to roll a distance d metres down a fixed ramp.
\begin{array}{|l|r|r|r|r|r|r|}\hline d\ (\mathrm{m}) & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 & 3.0\\\hline t\ (\mathrm{s}) & 0.64 & 0.90 & 1.11 & 1.27 & 1.43 & 1.56\\\hline\end{array}
(a) Explain why a model t=kd^{n} gives a straight line when \ln t is plotted against \ln d.
[2]
(b) Find the least-squares line of \ln t against \ln d, and hence find k and n.
[4]
(c) Sofia's teacher says n should be \frac{1}{2}. Comment on her result.
[2]
(d) Predict the time to roll 5 m and comment on the reliability of the prediction.
[3]
(e) Using n=\frac{1}{2}, find the value of k that minimises the sum of squared residuals of t against \sqrt{d}.
[5]
Show solutionHide solution
(a) t=kd^{n}\Rightarrow\ln t=\ln k+n\ln d [M1]; this has the form Y=c+nX with Y=\ln t and X=\ln d, a straight line of gradient n and intercept \ln k [R1].
(b) regression of \ln t on \ln d for the six readings [M1]; \ln t=-0.102+0.498\ln d [A1]; n=0.498 [A1]; k=e^{-0.102}=0.903 [A1].
(c) 0.498 is very close to 0.5 [R1]; the difference is well within what measurement error in six readings can produce, so the result is consistent with the teacher's value [R1].
(d) t=0.903\times5^{0.498} [M1] =2.01 s [A1]; 5 m lies outside the measured range 0.5 m to 3 m, so this is extrapolation, and the ramp may not even be that long [R1].
(e) S(k)=\sum\left(t-k\sqrt{d}\right)^{2} [M1]; \dfrac{dS}{dk}=-2\sum\sqrt{d}\left(t-k\sqrt{d}\right)=0 [M1]; k=\dfrac{\sum t\sqrt{d}}{\sum d} [A1]; \sum t\sqrt{d}=9.47 and \sum d=10.5 [A1]; k=0.902 [A1].
Final answer
- (b) \ln t=-0.102+0.498\ln d, n=0.498, k=0.903
- (c) n is close to 0.5, consistent
- (d) 2.01 s, extrapolation beyond 3 m
- (e) k=\frac{\sum t\sqrt d}{\sum d}=0.902.
