| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring resolution of forces into components and finding a resultant. The steps are clearly signposted (resolve parallel, resolve perpendicular, combine), requiring only standard trigonometry (cos/sin of 60°) and Pythagoras. This is easier than average A-level content as it's highly structured with no problem-solving insight needed. |
| Spec | 3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
**(i)** Attempt to find component — M1
$15\cos60 = 7.5$ N — A1 [2]
**(ii)** $15\sin60 = \frac{15}{2}\sqrt{3} = 13.0$ N to 3 sf — B1 [1]
**(iii)** Use Pythagoras, or cosine rule — M1
Obtain magnitude $\sqrt{475} = 5\sqrt{19} = 21.8$ N to 3 sf — A1
Use inverse tan, or sine rule — M1
Obtain angle $36.6°$ to 3 sf — A1 [4]
**Total: [7]**
*Allow sin/cos error. Accept any correct (unsimplified) form. c.a.o.*7\\
\includegraphics[max width=\textwidth, alt={}, center]{f0c32e07-f3a0-4d58-bd00-c266177ceaac-3_343_401_1439_872}
The diagram shows two forces of magnitudes 10 N and 15 N acting in a horizontal plane on a particle $P$.\\
(i) Find the component of the 15 N force which is parallel to the 10 N force.\\
(ii) Write down the component of the 15 N force which is perpendicular to the 10 N force.\\
(iii) Hence, or otherwise, calculate the magnitude and direction of the resultant force on $P$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2012 Q7 [7]}}