| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question involving vector resolution and friction. Part (i) requires resolving forces in perpendicular directions and using Pythagoras (routine technique), while part (ii) applies F=ma with friction calculation. Both parts follow textbook methods with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 3.03d Newton's second law: 2D vectors3.03p Resultant forces: using vectors3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks |
|---|---|
| Marks | Guidance |
| B1 | |
| M1 | Uses appropriate Pythagoras |
| M1 | Attempt to solve 3 term QE "=0" |
| A1 | |
| M1 | Targets any relevant angle appropriately |
| A1√ | ft cv(P) |
| A1 | Must be angle with vertical |
| [7] |
| Answer | Marks |
|---|---|
| Marks | Guidance |
| M1* | Bracketed terms must have opposite signs |
| A1√ | ft 29.4-2xcv(P(i)) OR 29.4 -25cos(cv(\(\theta\)(i))) |
| D*M1 | N2L, cv(12) cv(5.4) should be acceptable |
| A1 | |
| B1 | Allow bearing (0)90° |
| [5] |
## Part i
**Answer/Working:**
Perpendicular components of (2P) and +/-(5-P)
$(P-5)^2 + (2P)^2 = 25^2$
$5P^2 - 10P - 600 = 0$
$P=12$
$\cos\theta = (2x12)/25$, $\tan\theta = (12-5)/ 2x12$ etc.
Angle with vertical = 16.3°
| **Marks** | **Guidance** |
|-----------|-------------|
| B1 | |
| M1 | Uses appropriate Pythagoras |
| M1 | Attempt to solve 3 term QE "=0" |
| A1 | |
| M1 | Targets any relevant angle appropriately |
| A1√ | ft cv(P) |
| A1 | Must be angle with vertical |
| [7] | |
## Part ii
**Answer/Working:**
$R = +/-(3x9.8 - 2x12)$ OR $R = +/-(3x9.8 - 25\cos(\text{Ans}(i)))$
$R = 5.4$ N (may be implied)
$12 - 5 - 0.15x5.4 = 3a$ OR $25\sin(\text{cv}(\theta(i))) - 0.15x5.4 = 3a$
$a = 2.06$ m s$^{-2}$
Direction East
| **Marks** | **Guidance** |
|-----------|-------------|
| M1* | Bracketed terms must have opposite signs |
| A1√ | ft 29.4-2xcv(P(i)) OR 29.4 -25cos(cv($\theta$(i))) |
| D*M1 | N2L, cv(12) cv(5.4) should be acceptable |
| A1 | |
| B1 | Allow bearing (0)90° |
| [5] | |
---Three forces act on a particle. The first force has magnitude $P\text{ N}$ and acts horizontally due east. The second force has magnitude $5\text{ N}$ and acts horizontally due west. The third force has magnitude $2P\text{ N}$ and acts vertically upwards. The resultant of these three forces has magnitude $25\text{ N}$.
\begin{enumerate}[label=(\roman*)]
\item Calculate $P$ and the angle between the resultant and the vertical. [7]
\end{enumerate}
The particle has mass $3\text{ kg}$ and rests on a rough horizontal table. The coefficient of friction between the particle and the table is $0.15$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the acceleration of the particle, and state the direction in which it moves. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2016 Q5 [12]}}