| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Standard +0.3 This is a standard mechanics question on resolving forces and finding resultants. Part (a) requires setting up a component equation and solving for an angle, then calculating the resultant—straightforward application of resolution. Part (b) is routine calculation of resultant magnitude and direction using Pythagoras and inverse tan. Both parts follow standard textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
| 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) | 20cos30=25cos60+10cosα |
| Answer | Marks | Guidance |
|---|---|---|
| 17.32=12.5+10cosα, → cosα=0.4821 | M1 | For resolving forces horizontally, all relevant terms included |
| α = 61.2 | A1 | From α = 61.18 |
| Answer | Marks | Guidance |
|---|---|---|
| =10+8.761−21.651 | M1 | For resolving forces vertically, all relevant terms included |
| Magnitude of resultant force = 2.89 N | A1 | A0 for –2.89 N or for ±2.89 N. |
| Answer | Marks |
|---|---|
| 6(b) | X =25cos60+10cos45−20cos30 |
| Answer | Marks | Guidance |
|---|---|---|
| =10+7.07107−21.65064=−4.57957 | M1 | For either horizontal or vertical component, correct number of |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | For both correct, allow unsimplified | |
| R= X2 +Y2 | M1 | OE. Using a method to find the resultant force, using expressions |
| Answer | Marks | Guidance |
|---|---|---|
| X | M1 | OE. A method to find the direction, using expressions for X and Y |
| Answer | Marks | Guidance |
|---|---|---|
| Direction = 63.8° below positive x-axis | A1 | For both correct, angle clearly explained. |
| 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 A Level – Mark Scheme May/June 2021
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 2021 Page 5 of 14
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | 20cos30=25cos60+10cosα
[ ]
17.32=12.5+10cosα, → cosα=0.4821 | M1 | For resolving forces horizontally, all relevant terms included
α = 61.2 | A1 | From α = 61.18
Resultant =20sin30+10sin61.2−25sin60
[ ]
=10+8.761−21.651 | M1 | For resolving forces vertically, all relevant terms included
Magnitude of resultant force = 2.89 N | A1 | A0 for –2.89 N or for ±2.89 N.
Allow 2.89 N downwards
4
--- 6(b) ---
6(b) | X =25cos60+10cos45−20cos30
=12.5+7.07107−17.32051=2.25056
Y =20sin30+10sin45−25sin60
=10+7.07107−21.65064=−4.57957 | M1 | For either horizontal or vertical component, correct number of
relevant terms. Allow ±X and/or ±Y
A1 | For both correct, allow unsimplified
R= X2 +Y2 | M1 | OE. Using a method to find the resultant force, using expressions
for X and Y with at least 5 relevant terms.
Y
α=tan −1
X | M1 | OE. A method to find the direction, using expressions for X and Y
with at least 5 relevant terms.
Resultant = 5.10 N,
Direction = 63.8° below positive x-axis | A1 | For both correct, angle clearly explained.
May use a diagram with a correct arrow and arc for angle.
Allow angle 296° (measured anticlockwise from +ve x-axis)
5
Question | Answer | Marks | Guidance\includegraphics{figure_6}
Three coplanar forces of magnitudes 10 N, 25 N and 20 N act at a point $O$ in the directions shown in the diagram.
\begin{enumerate}[label=(\alph*)]
\item Given that the component of the resultant force in the $x$-direction is zero, find $\alpha$, and hence find the magnitude of the resultant force. [4]
\item Given instead that $\alpha = 45$, find the magnitude and direction of the resultant of the three forces. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q6 [9]}}