| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Session | Specimen |
| Marks | 5 |
| Topic | Friction |
| Type | Particle on inclined plane - force parallel to slope |
| Difficulty | Standard +0.3 Part (i) is a standard inclined plane equilibrium problem requiring routine resolution of forces parallel and perpendicular to the plane, with straightforward algebra yielding μ = tan(30°). Part (ii) adds vertical acceleration requiring a pseudo-force in the reference frame, but remains a methodical application of Newton's second law with clear resolution directions. Both parts follow standard mechanics procedures without requiring novel insight. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03v Motion on rough surface: including inclined planes |
**(i)** The box is in equilibrium on a slope of 30 degrees, so resolving along and at right angles to the slope: M1
$F - 20g\sin 30 = 0$
$R - 20g\cos 30 = 0$ A1
The box is on the point of sliding, so: M1
$\mu = \frac{F}{R} = \tan 30$ A1
$= \frac{1}{\sqrt{3}}$ **AG** **[4]**
**(ii)** Resolving vertically and horizontally and applying Newton's law of motion: M1
$R\cos 10 + F\sin 10 - 20g = 20 \times \frac{1}{2}g$ A1
$F\cos 10 - R\sin 10 = 0$ A1
(Note: During any vertical movement at $< 30°$, the ratio of friction to normal reaction is less than the coefficient of sliding friction, so there is no horizontal acceleration.)
Solving: M1
$F = \frac{300\sin 10}{\cos^2 10 + \sin^2 10} = 300\sin 10 = 52.1$ N
$R = 300\cos 10 = 295$ N A1 **[5]**12 (i) Whilst a helicopter is hovering, the floor of its cargo hold maintains an angle of $30 ^ { \circ }$ to the horizontal. There is a box of mass 20 kg on the floor. If the box is just on the point of sliding, show by resolving forces that the coefficient of friction between the box and the floor is $\frac { 1 } { \sqrt { 3 } }$.\\
(ii) The helicopter ascends at a constant acceleration 0.5 g . If the cargo hold is now maintained at $10 ^ { \circ }$ to the horizontal, determine the frictional force and the normal reaction between the box and the floor.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 Q12 [5]}}