| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Topic | Motion on a slope |
| Type | Motion up rough slope |
| Difficulty | Standard +0.8 This is a two-part mechanics problem requiring resolution of forces on an inclined plane and kinematics. Part (i) involves setting up equations of motion with gravity component and solving for θ, which is moderately challenging. Part (ii) adds friction, requiring careful force analysis and integration of motion equations. The problem demands solid understanding of inclined plane mechanics and multi-step algebraic manipulation, placing it above average difficulty but not requiring exceptional insight. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| The net force opposing motion up the slope is \(mg\sin\phi\), so the deceleration is \(g\sin\theta\) | M1 M1 | |
| Using the formula \(v = u + at\) (\(v = 0, u = V, a = -g\sin\theta, t = 2V/g\cos\theta\)) | A1 | |
| \(V = V 2\sin\theta\cos\theta \Rightarrow \sin 2\theta = 1\) | A1 | |
| \(\theta = \pi/4\) | A1 AG | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Resolving forces perpendicular to the slope: \(R - mg\cos\theta = 0\) | M1 | |
| Applying Newton's law to motion parallel to the slope: \(ma = -mg\sin\theta - F\) where \(F = 0.5R\) | M1 | |
| Obtain: \(a = -g(\sin\theta + 0.5\cos\theta)\) | B1 | |
| Using \(v^2 = u^2 + 2as\), obtain: \(s = \frac{V^2}{2g(\sin\theta + 0.5\cos\theta)} = \frac{V^2\sqrt{2}}{3g}\) | M1 A1 | [5] |
### (i)
The net force opposing motion up the slope is $mg\sin\phi$, so the deceleration is $g\sin\theta$ | M1 M1 |
Using the formula $v = u + at$ ($v = 0, u = V, a = -g\sin\theta, t = 2V/g\cos\theta$) | A1 |
$V = V 2\sin\theta\cos\theta \Rightarrow \sin 2\theta = 1$ | A1 |
$\theta = \pi/4$ | A1 AG | [4]
### (ii)
Resolving forces perpendicular to the slope: $R - mg\cos\theta = 0$ | M1 |
Applying Newton's law to motion parallel to the slope: $ma = -mg\sin\theta - F$ where $F = 0.5R$ | M1 |
Obtain: $a = -g(\sin\theta + 0.5\cos\theta)$ | B1 |
Using $v^2 = u^2 + 2as$, obtain: $s = \frac{V^2}{2g(\sin\theta + 0.5\cos\theta)} = \frac{V^2\sqrt{2}}{3g}$ | M1 A1 | [5]A particle is projected from a point $P$ on an inclined plane, up the line of greatest slope through $P$, with initial speed $V$. The angle of the plane to the horizontal is $\theta$.
\begin{enumerate}[label=(\roman*)]
\item If the plane is smooth, and the particle travels for a time $\frac{2V}{g}\cos\theta$ before coming instantaneously to rest, show that $\theta = \frac{1}{4}\pi$. [4]
\item If the same plane is given a roughened surface, with a coefficient of friction 0.5, find the distance travelled before the particle comes instantaneously to rest. [5]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2010 Q10 [9]}}