OCR M1 2013 January — Question 1 5 marks

Exam BoardOCR
ModuleM1 (Mechanics 1)
Year2013
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeBearing and compass direction problems
DifficultyModerate -0.8 This is a straightforward vector addition problem using bearings. Students resolve forces into components (north/east directions are perpendicular, making calculations simple), sum them, then find magnitude using Pythagoras and bearing using inverse tan. It's more routine than average A-level questions since the bearings are along cardinal directions (no awkward angles) and requires only standard mechanics techniques with no problem-solving insight.
Spec3.03p Resultant forces: using vectors

1 Three horizontal forces, acting at a single point, have magnitudes \(12 \mathrm {~N} , 14 \mathrm {~N}\) and 5 N and act along bearings \(000 ^ { \circ } , 090 ^ { \circ }\) and \(270 ^ { \circ }\) respectively. Find the magnitude and bearing of their resultant.

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(X = 14 - 5\)B1 Or \(5 - 14\)
\(R^2 = (14-5)^2 + 12^2\)M1 Pythagoras, \(R\) as hypotenuse, 3 squared terms
\(R = 15\) NA1
\(\tan\theta = (14-5)/12\)M1 Any correct trig, angle between 12 and \(R\) targetted
\(\theta = 36.9°\)A1 Accept 37, 037
[5]
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $X = 14 - 5$ | B1 | Or $5 - 14$ |
| $R^2 = (14-5)^2 + 12^2$ | M1 | Pythagoras, $R$ as hypotenuse, 3 squared terms |
| $R = 15$ N | A1 | |
| $\tan\theta = (14-5)/12$ | M1 | Any correct trig, angle between 12 and $R$ targetted |
| $\theta = 36.9°$ | A1 | Accept 37, 037 |
| **[5]** | | |
1 Three horizontal forces, acting at a single point, have magnitudes $12 \mathrm {~N} , 14 \mathrm {~N}$ and 5 N and act along bearings $000 ^ { \circ } , 090 ^ { \circ }$ and $270 ^ { \circ }$ respectively. Find the magnitude and bearing of their resultant.

\hfill \mbox{\textit{OCR M1 2013 Q1 [5]}}