| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question on resultant forces and equilibrium. Part (i) requires routine application of the cosine rule and sine rule with given angle and magnitudes. Parts (ii) and (iii) involve straightforward equilibrium conditions on a smooth surface (normal reaction perpendicular to surface, resolving forces). While multi-part, each step follows standard textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03a Force: vector nature and diagrams3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = +/-(10 - 6\cos70)\), \(y = 6\sin70\) | B1, B1 | \(10\cos55 + 6\cos55\ (= 9.177)\); Uses cosine rule M1 |
| *OR* \(+/-(10\cos70 - 6)\), \(10\sin70\) | \(+/-(10\sin55 - 6\sin55)\ (= +/-3.2766)\); \(R^2 = 6^2 + 10^2 - 2\times6\times10\cos\ \) B1; Uses angle of \(70\) B1 | |
| *OR* correct resolving in 2 perpendicular directions | ||
| \(R^2 = \{+/-(10-6\cos70)\}^2 + (6\sin70)^2\) | M1 | \(R^2 = (10\cos55 + 6\cos55)^2 + (10\sin55 - 6\sin55)^2\) |
| \(R = 9.74\) N | A1 | \(R = 9.74\) N A1 |
| \(\tan\alpha = (6\cos20)/(10 - 6\sin20)\) | M1 | \(\sin\alpha/6 = \sin70/9.744\) M1 |
| \(\alpha = 35.4°\) | A1 | \(\alpha = 35.4°\) A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Force \(= (20 - 9.74) = 10.3\) N | B1ft | Difference of weight and Resultant; ft \(20 -\) cv(9.74) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tan\theta = +/-(10 - 6\cos70)/6\sin70\) | M1 | Uses resultant is vertical; Angle \(= 90 -\) cv(35.4) M1 |
| *OR* \(\tan\varphi = +/-(6\sin70)/(10 - 6\cos70)\) | Angle \(= 54.6°\) A1 | |
| Angle \(= 54.6°\) | A1 |
## Question 4:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = +/-(10 - 6\cos70)$, $y = 6\sin70$ | B1, B1 | $10\cos55 + 6\cos55\ (= 9.177)$; Uses cosine rule M1 |
| *OR* $+/-(10\cos70 - 6)$, $10\sin70$ | | $+/-(10\sin55 - 6\sin55)\ (= +/-3.2766)$; $R^2 = 6^2 + 10^2 - 2\times6\times10\cos\ $ B1; Uses angle of $70$ B1 |
| *OR* correct resolving in 2 perpendicular directions | | |
| $R^2 = \{+/-(10-6\cos70)\}^2 + (6\sin70)^2$ | M1 | $R^2 = (10\cos55 + 6\cos55)^2 + (10\sin55 - 6\sin55)^2$ |
| $R = 9.74$ N | A1 | $R = 9.74$ N A1 |
| $\tan\alpha = (6\cos20)/(10 - 6\sin20)$ | M1 | $\sin\alpha/6 = \sin70/9.744$ M1 |
| $\alpha = 35.4°$ | A1 | $\alpha = 35.4°$ A1 |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Force $= (20 - 9.74) = 10.3$ N | B1ft | Difference of weight and Resultant; ft $20 -$ cv(9.74) |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan\theta = +/-(10 - 6\cos70)/6\sin70$ | M1 | Uses resultant is vertical; Angle $= 90 -$ cv(35.4) M1 |
| *OR* $\tan\varphi = +/-(6\sin70)/(10 - 6\cos70)$ | | Angle $= 54.6°$ A1 |
| Angle $= 54.6°$ | A1 | |
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\includegraphics[max width=\textwidth, alt={}, center]{8b79facc-e37f-45c3-95c0-9f2a30ca8fe4-3_394_963_276_552}
Two forces of magnitudes 6 N and 10 N separated by an angle of $110 ^ { \circ }$ act on a particle $P$, which rests on a horizontal surface (see diagram).\\
(i) Find the magnitude of the resultant of the 6 N and 10 N forces, and the angle between the resultant and the 10 N force.
The two forces act in the same vertical plane. The particle $P$ has weight 20 N and rests in equilibrium on the surface. Given that the surface is smooth, find\\
(ii) the magnitude of the force exerted on $P$ by the surface,\\
(iii) the angle between the surface and the 10 N force.
\hfill \mbox{\textit{OCR M1 2015 Q4 [9]}}