| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 19 |
| 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 question on motion up a rough slope requiring resolution of forces, application of F=ma, and use of SUVAT equations. Part (a) is a 'show that' requiring routine force resolution (weight components and friction). Parts (b) and (c) are straightforward applications of kinematic equations once the acceleration is known. The question involves multiple steps but uses well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03r Friction: concept and vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(sin\alpha = \frac{3}{5}\) (3,4,5 Pythag. triple) so \(cos\alpha = \frac{4}{5}\) resolve perp. to plane: \(R - mgcos\alpha = 0\) so \(R = \frac{3}{5}mg\) \(F = \mu R = \frac{1}{3}mg\) Resolve up the plane: \(F - mg\sin\alpha = ma\) \(\frac{1}{3}mg - \frac{3}{5}mg = ma\) so \(a = \frac{4}{15}g\) i.e. \(a = \frac{4}{15}g\) and is directed down the slope | B1, M1 A1, M1 A1, M1, A1, A1 | |
| (b) \(u = 20\), \(v = 0\), \(a = -\frac{4}{3}g\) use \(v^2 = u^2 + 2as\) \(0 = 400 - \frac{8}{3}gs \Rightarrow s = 25.51 \text{ m}\), i.e. \(4.49 \text{ m}\) (nearest cm) from top | M1, M2 A1 | |
| (c) friction now acts up slope but \(R\) (and hence \(F\)) as in part (a) \(mg\sin\alpha - F = ma \Rightarrow \frac{2}{5}mg - \frac{1}{3}mg = ma\) \(a = \frac{2}{5}g\) \(u = 0\), \(s = 25.51\), \(a = \frac{2}{5}g\), \(s = ut + \frac{1}{2}at^2\) \(25.51 = \frac{1}{5}gt^2\) i.e. \(t = 3.61 \text{ seconds}\) | B1, M2, A1, M1, M1 A1 | (19) |
| (a) $sin\alpha = \frac{3}{5}$ (3,4,5 Pythag. triple) so $cos\alpha = \frac{4}{5}$ resolve perp. to plane: $R - mgcos\alpha = 0$ so $R = \frac{3}{5}mg$ $F = \mu R = \frac{1}{3}mg$ Resolve up the plane: $F - mg\sin\alpha = ma$ $\frac{1}{3}mg - \frac{3}{5}mg = ma$ so $a = \frac{4}{15}g$ i.e. $a = \frac{4}{15}g$ and is directed down the slope | B1, M1 A1, M1 A1, M1, A1, A1 | |
| (b) $u = 20$, $v = 0$, $a = -\frac{4}{3}g$ use $v^2 = u^2 + 2as$ $0 = 400 - \frac{8}{3}gs \Rightarrow s = 25.51 \text{ m}$, i.e. $4.49 \text{ m}$ (nearest cm) from top | M1, M2 A1 | |
| (c) friction now acts up slope but $R$ (and hence $F$) as in part (a) $mg\sin\alpha - F = ma \Rightarrow \frac{2}{5}mg - \frac{1}{3}mg = ma$ $a = \frac{2}{5}g$ $u = 0$, $s = 25.51$, $a = \frac{2}{5}g$, $s = ut + \frac{1}{2}at^2$ $25.51 = \frac{1}{5}gt^2$ i.e. $t = 3.61 \text{ seconds}$ | B1, M2, A1, M1, M1 A1 | (19) |7. A machine fires ball-bearings up the line of greatest slope of a rough plane inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 3 } { 5 }$.
The coefficient of friction between the ball-bearings and the plane is $\frac { 1 } { 4 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the acceleration of the ball-bearings is $\frac { 4 } { 5 } g$ and state its direction.
Given that the machine is placed at a point $A , 30 \mathrm {~m}$ from the top edge of the plane, and the ball-bearings are projected with an initial speed of $20 \mathrm {~ms} ^ { - 1 }$,
\item find, giving your answer to the nearest cm , how close the ball-bearings get to the top edge of the plane.
\item How long does it take for a ball-bearing to travel from the highest point it reaches back down to the point $A$ again?
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [19]}}