| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Particle on inclined plane |
| Difficulty | Standard +0.3 Question 4 is a standard 2D force resolution and resultant calculation requiring Pythagoras and basic trigonometry—routine M1 material. Question 5 involves straightforward SUVAT application on an inclined plane to find acceleration, then using Newton's second law to find the angle—standard mechanics with no novel insight required, slightly easier than average overall. |
| Spec | 3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([Y_1^2 = 68^2 - (-60)^2,\ Y_3^2 = 100^2 - 96^2;\) \(Y_1 = 68\sin 28.1°,\ Y_3 = 100\sin 16.3°]\) | M1 | For using \(Y^2 = F^2 - X^2\), or finding angles \(\alpha\) and \(\beta\) between forces of magnitudes 68 and 100 and the \(x\)-axis, then finding two relevant magnitudes from \(68\sin\alpha\) and \(100\sin\beta\) |
| Correct magnitudes (32, 75, 28) | A1 | Can be scored by implication if final A1 is scored for correct answer to part (i) |
| Components are \(-32\), 75 and \(-28\) | A1ft | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([R^2 = (-60 + 0 + 96)^2 + (-32 + 75 - 28)^2]\) | M1 | For using \(R^2 = X^2 + Y^2\) |
| Magnitude is 39 N | A1 | |
| \([\theta = \tan^{-1}\{(-32 + 75 - 28) \div (-60 + 0 + 96)\}]\) | M1 | For using \(\theta = \tan^{-1}(Y/X)\) |
| Direction is \(22.6°\) (or \(0.395\ \text{rad}^c\)) anticlockwise from \(+ve\) \(x\)-axis | A1 | [4] Accept just '22.6 from \(x\)-axis' or just '\(\theta = 22.6\)' |
# Question 4:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[Y_1^2 = 68^2 - (-60)^2,\ Y_3^2 = 100^2 - 96^2;$ $Y_1 = 68\sin 28.1°,\ Y_3 = 100\sin 16.3°]$ | M1 | For using $Y^2 = F^2 - X^2$, or finding angles $\alpha$ and $\beta$ between forces of magnitudes 68 and 100 and the $x$-axis, then finding two relevant magnitudes from $68\sin\alpha$ and $100\sin\beta$ |
| Correct magnitudes (32, 75, 28) | A1 | Can be scored by implication if final A1 is scored for correct answer to part (i) |
| Components are $-32$, 75 and $-28$ | A1ft | **[3]** |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[R^2 = (-60 + 0 + 96)^2 + (-32 + 75 - 28)^2]$ | M1 | For using $R^2 = X^2 + Y^2$ |
| Magnitude is 39 N | A1 | |
| $[\theta = \tan^{-1}\{(-32 + 75 - 28) \div (-60 + 0 + 96)\}]$ | M1 | For using $\theta = \tan^{-1}(Y/X)$ |
| Direction is $22.6°$ (or $0.395\ \text{rad}^c$) anticlockwise from $+ve$ $x$-axis | A1 | **[4]** Accept just '22.6 from $x$-axis' or just '$\theta = 22.6$' |
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\includegraphics[max width=\textwidth, alt={}, center]{9fbb63e3-4017-461e-9110-500be2c20778-2_583_862_1343_644}
Three coplanar forces of magnitudes $68 \mathrm {~N} , 75 \mathrm {~N}$ and 100 N act at an origin $O$, as shown in the diagram. The components of the three forces in the positive $x$-direction are $- 60 \mathrm {~N} , 0 \mathrm {~N}$ and 96 N , respectively. Find\\
(i) the components of the three forces in the positive $y$-direction,\\
(ii) the magnitude and direction of the resultant of the three forces.\\
$5 A , B$ and $C$ are three points on a line of greatest slope of a plane which is inclined at $\theta ^ { \circ }$ to the horizontal, with $A$ higher than $B$ and $B$ higher than $C$. Between $A$ and $B$ the plane is smooth, and between $B$ and $C$ the plane is rough. A particle $P$ is released from rest on the plane at $A$ and slides down the line $A B C$. At time 0.8 s after leaving $A$, the particle passes through $B$ with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
\hfill \mbox{\textit{CAIE M1 2012 Q4 [7]}}