| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Standard +0.3 This is a standard M1 kinematics question requiring integration of velocity to find displacement, using given conditions to find a constant, and then finding minimum velocity using calculus. The steps are routine (integrate, apply boundary conditions, differentiate to find minimum) with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks |
|---|---|
| 6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 6(a) | ( ) |
| k t2 −10t+21 dt | M1 |
| Answer | Marks |
|---|---|
| 3 | A1 |
| Answer | Marks |
|---|---|
| 3 3 | M1 |
| Answer | Marks |
|---|---|
| (A1 for both) | A1 |
| Solving for k | M1 |
| k = 0.05 | A1 |
| Answer | Marks |
|---|---|
| 3 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(b) | Differentiating v or completing the square for v | M1 |
| a = 0.05(2t –10) | A1 | |
| Min value of v is at t = 5. | M1 | |
| Displacement at t = 5 is 2.58 m (2.5833...) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PPMMTT
9709/41 Cambridge International AS & A Level – Mark Scheme May/June 2020
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no “follow through” from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2020 Page 5 of 12
Question | Answer | Marks
--- 6(a) ---
6(a) | ( )
k t2 −10t+21 dt | M1
1
s=k t3 +5t2 +21t +C
3 | A1
1 1
2.85=k ×33 −5×32 +21×3 +C or 2.4=k ×63 −5×62 +21×6 +C
3 3 | M1
2.85 = 27k + C, 2.4 = 18k + C
(A1 for both) | A1
Solving for k | M1
k = 0.05 | A1
1
s=0.05 t3 −5t2 +21t +1.5
3 | A1
7
--- 6(b) ---
6(b) | Differentiating v or completing the square for v | M1
a = 0.05(2t –10) | A1
Min value of v is at t = 5. | M1
Displacement at t = 5 is 2.58 m (2.5833...) | A1
4
Question | Answer | MarksA particle moves in a straight line $AB$. The velocity $v\text{ m s}^{-1}$ of the particle $t\text{ s}$ after leaving $A$ is given by $v = t(5 - 2t)$ where $k$ is a constant. The displacement of the particle from $A$, in the direction towards $B$, is $2.5\text{ m}$ when $t = 3$ and is $2.4\text{ m}$ when $t = 6$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$. Hence find an expression, in terms of $t$, for the displacement of the particle from $A$. [7]
\item Find the displacement of the particle from $A$ when its velocity is a minimum. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q6 [11]}}