| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Topic | Motion on a slope |
| Type | Motion down smooth slope |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring resolution of forces to find acceleration down the slope (g sin θ), then applying SUVAT equations twice: once to find position at t=2s, then solving for when the particle returns to that position. The calculations are routine with the given sin θ = 1/25 making arithmetic manageable. Slightly easier than average as it follows a standard template with no conceptual surprises. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
Total distance \(= 2 \times 7.8125 - 4.2 = 11.425\ \text{m}\)
**Question 10** [10 marks]
$a = -g\sin\theta = -0.4$ — B1
When $t = 2$:
$s = 2.5 \times 2 - \frac{1}{2} \times 0.4 \times 2^2 = 4.2\ \text{m}$
— M1: Use an appropriate 'suvat' equation (or could find $v\ (= 1.7\ \text{ms}^{-1})$). A1: Correct outcome.
When $s = 4.2$:
$4.2 = 2.5t - 0.2t^2$
$\therefore t^2 - 12.5t + 21 = 0$
$\therefore (t-2)(t-10.5) = 0$
$\therefore t = 10.5\ \text{s}$
— M1: Use another appropriate 'suvat' equation (or could use $v = -1.7\ \text{ms}^{-1}$, e.g. quadratic equation for $t$). A1: Solved. A1: Correct value of $t$ chosen. c.a.o.
At top of motion:
$t = \frac{1}{2}(2 + 10.5) = 6.25\ \text{s}$
$s = 2.5 \times 6.25 - \frac{1}{2} \times 0.4 \times 6.25^2 = 7.8125\ \text{m}$
— M1: Ft *their* 10.5. (Or $0^2 = 2.5^2 - 2 \times 0.4 \times s$, or find distance from the mark to the top $(= 3.6125)$.) A1.
Total distance $= 2 \times 7.8125 - 4.2 = 11.425\ \text{m}$
— M1: Or equivalent. A1: c.a.o.
**Total: [10]**10 A particle is projected up a long smooth slope at a speed of $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The slope is at an angle $\theta$ to the horizontal where $\sin \theta = \frac { 1 } { 25 }$. After 2 seconds it passes a mark on the slope.
Find the total time taken from the moment of projection until it passes the mark again and the total distance travelled in that time.
{www.cie.org.uk} after the live examination series.
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\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2015 Q10 [10]}}