| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion up then down slope |
| Difficulty | Standard +0.3 This is a standard M1 friction on an inclined plane question with straightforward application of Newton's second law and SUVAT equations. Part (i) is a 'show that' calculation requiring F = μR. Parts (ii) and (iii) involve routine resolution of forces and kinematics with no novel problem-solving required. The multi-part structure and 15 marks indicate moderate length, but all techniques are textbook-standard for M1, making it slightly easier than average overall. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03f Weight: W=mg3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(R = 0.5g\cos 40°\) \(F = 0.6 \times 0.5g\cos 40°\) Magnitude is 2.25N AG | B1 M1 A1 | \(R = 3.7536\) For using F = \(\mu\) R |
| (ii) \(-/+0.5g\sin 40° - F = 0.5a\) (a) Acceleration is \(10.8ms^{-2}\) (b) Acceleration is \(1.79ms^{-2}\) | M1 A1 A1 A1 | For applying Newton's second law (either case) //slope, two forces Either case Accept 10.8 from correct working (both forces have the same sign) Accept -1.79 from correct working (the forces have opposite sign) Accept ! 1.8(0) |
| (iii)a) \(0 = 4 + (-10.8)T_1\) \(T_1 = 0.370(3)\) | M1 A1 | Requires appropriate sign Accept 0.37 |
| b) \(0 = 4^2 + 2(-10.8)s\) or \(s = (0 + 4) \times 0.37/2or\) \(s = 4(0.370) + \frac{1}{2}(- 10.8)(0.370)^2\) \(0.7405 = \frac{1}{2}(1.79)T_2^2\) \(0.370 + 0.908\) \(= 1.28s\) | M1 A1 ft M1 A1ft M1 A1 | For complete method of finding distance from A to highest point using a(up) with appropriate sign ft a(up) and/or \(T_1\) (s = 0.7405) For method of finding time taken from highest point to A and not using a(up) ft a(down) and cv(0.7405) (\(T_2 = 0.908\) approx) Using T = \(T_1 + T_2\) with different values for \(T_1, T_2\) 3 significant figures cao |
(i) $R = 0.5g\cos 40°$ $F = 0.6 \times 0.5g\cos 40°$ Magnitude is 2.25N AG | B1 M1 A1 | $R = 3.7536$ For using F = $\mu$ R
(ii) $-/+0.5g\sin 40° - F = 0.5a$ (a) Acceleration is $10.8ms^{-2}$ (b) Acceleration is $1.79ms^{-2}$ | M1 A1 A1 A1 | For applying Newton's second law (either case) //slope, two forces Either case Accept 10.8 from correct working (both forces have the same sign) Accept -1.79 from correct working (the forces have opposite sign) Accept ! 1.8(0)
(iii)a) $0 = 4 + (-10.8)T_1$ $T_1 = 0.370(3)$ | M1 A1 | Requires appropriate sign Accept 0.37
b) $0 = 4^2 + 2(-10.8)s$ or $s = (0 + 4) \times 0.37/2or$ $s = 4(0.370) + \frac{1}{2}(- 10.8)(0.370)^2$ $0.7405 = \frac{1}{2}(1.79)T_2^2$ $0.370 + 0.908$ $= 1.28s$ | M1 A1 ft M1 A1ft M1 A1 | For complete method of finding distance from A to highest point using a(up) with appropriate sign ft a(up) and/or $T_1$ (s = 0.7405) For method of finding time taken from highest point to A and not using a(up) ft a(down) and cv(0.7405) ($T_2 = 0.908$ approx) Using T = $T_1 + T_2$ with different values for $T_1, T_2$ 3 significant figures caoA particle $P$ of mass 0.5 kg moves upwards along a line of greatest slope of a rough plane inclined at an angle of $40°$ to the horizontal. $P$ reaches its highest point and then moves back down the plane. The coefficient of friction between $P$ and the plane is 0.6.
\begin{enumerate}[label=(\roman*)]
\item Show that the magnitude of the frictional force acting on $P$ is 2.25 N, correct to 3 significant figures. [3]
\item Find the acceleration of $P$ when it is moving
\begin{enumerate}[label=(\alph*)]
\item up the plane,
\item down the plane.
\end{enumerate}
[4]
\item When $P$ is moving up the plane, it passes through a point $A$ with speed $4 \text{ m s}^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Find the length of time before $P$ reaches its highest point.
\item Find the total length of time for $P$ to travel from the point $A$ to its highest point and back to $A$.
\end{enumerate}
[8]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2007 Q7 [15]}}