| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion up rough slope |
| Difficulty | Standard +0.3 This is a standard M1 mechanics problem requiring resolution of forces on an inclined plane with friction. While it involves multiple parts and both dynamic and static cases, the techniques are routine: resolving perpendicular to find normal reaction, resolving parallel using F=ma to find friction, then calculating coefficient and contact force. The conceptual demand is modest—students follow a well-practiced procedure with no novel insight required. Slightly above average difficulty due to the two-part nature (motion vs rest) and multiple calculations. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| i Wt cmpts: // plane \(\quad 0.6g\sin30°\) Perp plane \(0.6g\cos30°\) \(0.6g\sin30 +/- X = 0.6 \times 10\) X = +/- 3.06 \(\mu = 3.06 / 5.09(22..)\) \(\mu = 0.601\) OR \(3.06 = \mu \times 5.09(22..)\) \(\mu = 0.601\) | B1 B1 M1 A1 A1 M1 A1 [7] | +/-2.94 +/-5.09(22..) = R N2L // plane, 2 force terms and ma (allow no g) Both weight cmpt and accen signs same May be implied (Fr =0.6x10-0.6g\sin30 used) Uses \(\mu = Fr/R\) both terms same sign 0.6 |
| ii a) \(C^2 = 3.06^2 + 5.09^*\) \(C = 5.94 \text{ N}\) \(\tan\theta = 3.06/5.09(22..)\) Angle = (31) + 90 Angle = 121° OR \(\tan\theta = 5.09(22..)/3.06\) Angle = 180 – (59) Angle = 121° | M1 A1 M1* D*M1 A1 [5] | Pythagoras with Fr and R, to find hypotenuse Accept 5.9, 5.95 but not 6(,0) Or \(\tan\theta = \mu\) Not 120 \(\tan\theta = 1 / \mu\) Not 120 |
| b) \(C = (0.6 \times 9.8) = 5.88 \text{ N}\) Angle = 60° | B1 B1 [2] | 5.9 |
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| **i** Wt cmpts: // plane $\quad 0.6g\sin30°$ Perp plane $0.6g\cos30°$ $0.6g\sin30 +/- X = 0.6 \times 10$ X = +/- 3.06 $\mu = 3.06 / 5.09(22..)$ $\mu = 0.601$ OR $3.06 = \mu \times 5.09(22..)$ $\mu = 0.601$ | B1 B1 M1 A1 A1 M1 A1 [7] | +/-2.94 +/-5.09(22..) = R N2L // plane, 2 force terms and ma (allow no g) Both weight cmpt and accen signs same May be implied (Fr =0.6x10-0.6g\sin30 used) Uses $\mu = Fr/R$ both terms same sign 0.6 | **Accept Fr for X** **Accept Fr = [X]** **Accept Fr = [X]** |
| **ii** **a)** $C^2 = 3.06^2 + 5.09^*$ $C = 5.94 \text{ N}$ $\tan\theta = 3.06/5.09(22..)$ Angle = (31) + 90 Angle = 121° OR $\tan\theta = 5.09(22..)/3.06$ Angle = 180 – (59) Angle = 121° | M1 A1 M1* D*M1 A1 [5] | Pythagoras with Fr and R, to find hypotenuse Accept 5.9, 5.95 but not 6(,0) Or $\tan\theta = \mu$ Not 120 $\tan\theta = 1 / \mu$ Not 120 |
| **b)** $C = (0.6 \times 9.8) = 5.88 \text{ N}$ Angle = 60° | B1 B1 [2] | 5.9 | **No working needed as C is vertical** **No working needed as C is vertical** |7 A particle $P$ of mass 0.6 kg is projected up a line of greatest slope of a plane inclined at $30 ^ { \circ }$ to the horizontal. $P$ moves with deceleration $10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and comes to rest before reaching the top of the plane.\\
(i) Calculate the frictional force acting on $P$, and the coefficient of friction between $P$ and the plane.\\
(ii) Find the magnitude of the contact force exerted on $P$ by the plane and the angle between the contact force and the upward direction of the line of greatest slope,
\begin{enumerate}[label=(\alph*)]
\item when $P$ is in motion,
\item when $P$ is at rest.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2011 Q7 [14]}}