June 2022 Paper 1 Q8
8
| \(t\) | 0 | 10 | 50 | |
| \(m\) | 1250 | 750 | 450 |
| Scheme | Marks | AO |
|---|---|---|
| 20 (minutes) | B1 | 3.3 |
| 97.2 (grams) | B1 | 3.4 |
| [2] |
Notes
B1: Obtain \(t = 20\)
Allow [19.9, 20.1] from setting up and using exponential model
B1: Obtain \(m = 97.2\)
Allow [97.1, 97.3] from setting up and using exponential model
| Scheme | Marks | AO |
|---|---|---|
| (i) \(160\mathrm{e}^{-0.055t} = 80\) | B1 | 3.4 |
| \(\mathrm{e}^{-0.055t} = 0.5\) \(-0.055t = \ln 0.5\) | M1 | 3.4 |
| \(t = 12.6\) (minutes) | A1 | 1.1 |
| [3] | ||
| (ii) \(\dfrac{\mathrm{d}m}{\mathrm{d}t} = -8.8\mathrm{e}^{-0.055t}\) | B1 | 3.4 |
| \(-8.8\mathrm{e}^{-0.055 \times 15}\) | M1 | 3.4 |
| \(= -3.86\), hence rate of decay is 3.86 grams/minute | A1 | 1.1 |
| [3] |
Notes
(b)(i)
B1: Equate given model to 80
soi, so could be \(\mathrm{e}^{-0.055t} = 0.5\)
M1: Attempt correct process to find value of \(t\), as far as dealing with exponential term
Rearrange to \(\mathrm{e}^{-0.055t} = k\), and hence obtain \(-0.055t = \ln k\)
If introducing logs straight away then need to get as far as \(\ln 160 - 0.055t = \ln(\text{their } 80)\)
A1: Obtain \(t = 12.6\), or better
If more sig fig given, then allow answers which round to 12.60 (the more accurate answer is 12.602676..)
(b)(ii)
B1: Correct derivative soi
Allow unsimplified
No need to see \(\frac{\mathrm{d}m}{\mathrm{d}t} =\)
M1: Substitute \(t = 15\) into their derivative
Must be of the form \(k\mathrm{e}^{-0.055t}\), with \(k \ne 160\)
Possibly still with \(k\) unsimplified
Substitution sufficient, no need to evaluate for M1
A1: Units required, and positive answer
Must follow correct derivative ie negative coefficient
No need to see \(-3.86\) first, but A0 if clear error
Accept 3.9 grams/minute
Accept g/m for grams/minute
| Scheme | Marks | AO |
|---|---|---|
| For \(A\), \(\dfrac{\mathrm{d}m}{\mathrm{d}t} = -63.9\mathrm{e}^{-0.0511t}\) Rate of decrease at \(t = 15\) is 29.7 g/min hence \(A\) decaying at a faster rate | B1 | 3.4 |
| [1] |
Notes
B1: State \(A\), with clear comparison
Insufficient to just say that \(A\) has a greater initial mass – needs to consider decay factor as well
Allow solutions that identify that \(B\) is decaying faster, with supporting evidence
eg after 10 minutes, \(B\)’s mass is 92.3g which is 58% of initial mass whereas \(A\) is 60% of initial mass so \(B\) decaying faster
eg \(A\)’s half-life is 13.6 so \(B\) is decaying faster
eg change initial mass in model \(B\) to 1250 then when \(t = 10\) \(B\)’s mass would be 721g which is less than 750 hence decaying faster
eg compare coefficients of \(t\) (for \(A\), coeff is \(-0.0511\)); \(B\)’s is of a greater magnitude hence decaying faster
For either solution, the conclusion and the supporting evidence must be consistent
Numerical supporting evidence must be correct, allowing for slight inaccuracies from using different numbers of sig fig (see appendix below)
Appendix: supporting evidence for Q8(c)
When comparing % remaining or percentage lost in \(t\) minutes, substance \(B\) is shown to be decreasing at a faster rate. (Choose \(t = 15\).)
Substance A
| time | 0 | 10 | 20 | 50 | 15 |
|---|---|---|---|---|---|
| mass (exact) | 1250 | 750 | 450 | 97.200 | 580.948 |
| mass (\(k = -0.0511\)) | 1250 | 750 | 450 | 97.115 | 580.796 |
| mass (\(k = -0.051\)) | 1250 | 750 | 450 | 97.602 | 581.667 |
| Percentage decreased at \(t =\) | 0 | 10 | 20 | 50 | 15 |
|---|---|---|---|---|---|
| Exact \(k\) value | 0% | 40% | 64% | 92.22% | 53.52% |
| \(k = -0.0511\) | 0% | 40% | 64% | 92.23% | 53.54% |
| \(k = -0.051\) | 0% | 40% | 64% | 92.19% | 53.47% |
| Percentage remaining at \(t =\) | 0 | 10 | 20 | 50 | 15 |
|---|---|---|---|---|---|
| Exact \(k\) value | 100% | 60% | 36% | 7.78% | 46.48% |
| \(k = -0.0511\) | 100% | 60% | 36% | 7.77% | 46.46% |
| \(k = -0.051\) | 100% | 60% | 36% | 7.81% | 46.53% |
Substance B
| time | 0 | 10 | 20 | 50 | 15 |
|---|---|---|---|---|---|
| mass (exact) | 160.00 | 92.312 | 53.259 | 10.228 | 70.118 |
| Percentage decreased at \(t =\) | 0 | 10 | 20 | 50 | 15 |
|---|---|---|---|---|---|
| Exact \(k\) value | 0% | 42.31% | 66.71% | 93.61% | 56.18% |
| Percentage remaining at \(t =\) | 0 | 10 | 20 | 50 | 15 |
|---|---|---|---|---|---|
| Exact \(k\) value | 100% | 57.69% | 33.29% | 6.39% | 43.82% |
When comparing RATE of decrease at \(t\) minutes, substance \(A\) is shown to be decreasing at a faster rate.
Substance A
| time | 0 | 10 | 20 | 50 | 15 |
|---|---|---|---|---|---|
| \(\mathrm{d}m/\mathrm{d}t\) (exact) | −63.853 | −38.312 | −22.987 | −4.965 | −29.676 |
| \(\mathrm{d}m/\mathrm{d}t\) (\(k = -0.0511\)) | −63.875 | −38.318 | −22.987 | −4.963 | −29.679 |
| \(\mathrm{d}m/\mathrm{d}t\) (\(k = -0.051\)) | −63.750 | −38.282 | −22.988 | −4.978 | −29.665 |
Substance B
| time | 0 | 10 | 20 | 50 | 15 |
|---|---|---|---|---|---|
| \(\mathrm{d}m/\mathrm{d}t\) (exact) | −8.800 | −5.077 | −2.929 | −0.563 | −3.856 |