| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring resolution of forces in two perpendicular directions and application of F=ma. Part (i) involves setting up equations from the condition that the resultant is horizontal (vertical components cancel), then finding the resultant and acceleration. Part (ii) is even simpler - just combining perpendicular forces. All techniques are standard M1 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = P\cos20° - 0.04g\) | B1 | |
| For setting \(V = 0\) | M1 | |
| \(P = 0.417\) | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = P\sin20°\) | M1 | For using \(R\) = horizontal component of \(P\) |
| Magnitude is \(0.143\) N | A1ft | ft value of \(P\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.143 = 0.04a\) | M1 | For using Newton's second law |
| Acceleration is \(3.57\) ms\(^{-2}\) | A1ft | ft magnitude of the resultant |
| Answer | Marks | Guidance |
|---|---|---|
| \(R^2 = 0.08^2 + (0.04g)^2\) | M1 | For using \(R^2 = P^2 + W^2\) |
| Magnitude is \(0.400\) N (or \(0.40\) or \(0.4\)) | A1 | |
| \(\tan\theta = \pm0.04g/0.08\) or \(\tan(90°-\theta) = \pm0.08/0.04g\) | M1 | For using \(\tan\theta = Y/X\) or \(\tan(90°-\theta) = X/Y\) |
| Angle made with horizontal is \(78.5°\) or \(1.37\) radians, or angle made with vertical is \(11.5°\) or \(0.201\) radians | A1 | |
| Downwards or below horizontal | B1 | Direction may alternatively be shown clearly on a diagram or given as a bearing |
# Question 6:
## Part (i)(a)
| $V = P\cos20° - 0.04g$ | B1 | |
| For setting $V = 0$ | M1 | |
| $P = 0.417$ | A1 | **Total: 3** |
## Part (i)(b)
| $R = P\sin20°$ | M1 | For using $R$ = horizontal component of $P$ |
| Magnitude is $0.143$ N | A1ft | ft value of $P$ | **Total: 2** |
## Part (i)(c)
| $0.143 = 0.04a$ | M1 | For using Newton's second law |
| Acceleration is $3.57$ ms$^{-2}$ | A1ft | ft magnitude of the resultant | **Total: 2** |
## Part (ii)
| $R^2 = 0.08^2 + (0.04g)^2$ | M1 | For using $R^2 = P^2 + W^2$ |
| Magnitude is $0.400$ N (or $0.40$ or $0.4$) | A1 | |
| $\tan\theta = \pm0.04g/0.08$ or $\tan(90°-\theta) = \pm0.08/0.04g$ | M1 | For using $\tan\theta = Y/X$ or $\tan(90°-\theta) = X/Y$ |
| Angle made with horizontal is $78.5°$ or $1.37$ radians, or angle made with vertical is $11.5°$ or $0.201$ radians | A1 | |
| Downwards or below horizontal | B1 | Direction may alternatively be shown clearly on a diagram or given as a bearing | **Total: 5** |
---6 A particle of mass 0.04 kg is acted on by a force of magnitude $P \mathrm {~N}$ in a direction at an angle $\alpha$ to the upward vertical.\\
(i) The resultant of the weight of the particle and the force applied to the particle acts horizontally. Given that $\alpha = 20 ^ { \circ }$ find
\begin{enumerate}[label=(\alph*)]
\item the value of $P$,
\item the magnitude of the resultant,
\item the magnitude of the acceleration of the particle.\\
(ii) It is given instead that $P = 0.08$ and $\alpha = 90 ^ { \circ }$. Find the magnitude and direction of the resultant force on the particle.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2005 Q6 [12]}}