| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2001 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Kinematics: displacement-velocity-acceleration |
| Difficulty | Standard +0.3 This is a standard M2 mechanics problem applying Newton's second law and kinematics to motion on an inclined plane. Part (a) requires setting up forces (weight component, resistance), using v²=u²+2as to find acceleration, then solving for R—straightforward but multi-step. Part (b) simply reapplies the same method with a different angle. The question is slightly easier than average A-level because it follows a very standard template with clear guidance through parts (a) and (b), requiring no novel insight beyond routine mechanics procedures. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| (i) P.E. = \(\pm 0.5gh\), \(\pm g \sin 20°\); (ii) K.E. = \(\frac{1}{2} \times 0.5 \times 25\) | M1A1;B1 M1A1 \(\sqrt{M1A1}\) | (7) |
| Answer | Marks |
|---|---|
| Solving for R; R = 1.45 or 1.4 | |
| Note: \(2(R + 0.5 \times 9.8 \times \sin 20°) = \frac{1}{2}(0.5)25\) scores first 5 marks, mark as scheme |
| Answer | Marks |
|---|---|
| Speed equation for a: \(0 = 25 \pm 2a\) (2) \((a = \pm6.25)\) | M1A1 |
| Equation of motion: \((R + 0.5 \times 9.8 \times \sin 20°) = \pm0.5a\) | M1A1 |
| Totally correct equation: \(-(R + 0.5g \sin 40°) = 0.5a\) (\(a = -9.2\)) and \(0 = 25 + 2as\) | A1 M1A1 |
| Solving for R | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(s = 1.36\) m or 1.4\(^+\)m | M1A1√ A1 | (3) |
**(a)** Using work/energy equation:
(i) P.E. = $\pm 0.5gh$, $\pm g \sin 20°$; (ii) K.E. = $\frac{1}{2} \times 0.5 \times 25$ | M1A1;B1 M1A1 $\sqrt{M1A1}$ | (7)
$\frac{1}{2} \times 0.5 \times 25 = 0.5gh + 2R$
Solving for R; **R = 1.45** or **1.4** | |
**Note:** $2(R + 0.5 \times 9.8 \times \sin 20°) = \frac{1}{2}(0.5)25$ scores first 5 marks, mark as scheme | |
**Alternative method:**
Speed equation for a: $0 = 25 \pm 2a$ (2) $(a = \pm6.25)$ | M1A1 |
Equation of motion: $(R + 0.5 \times 9.8 \times \sin 20°) = \pm0.5a$ | M1A1 |
Totally correct equation: $-(R + 0.5g \sin 40°) = 0.5a$ ($a = -9.2$) and $0 = 25 + 2as$ | A1 M1A1 |
Solving for R | M1A1 |
**(b)** Complete method for $s$
[Work/energy equation: $\frac{1}{2} \times 0.5 \times 25 = s \cdot R + 0.5 \times 9.8 \times \sin 40°$
or $-(R + 0.5g \sin 40°) = 0.5a$ ($a = -9.2$) and $0 = 25 + 2as$]
Answer: **$s = 1.36$ m** or **1.4$^+$m** | M1A1√ A1 | (3)
---A child is playing with a small model of a fire-engine of mass 0.5 kg and a straight, rigid plank. The plank is inclined at an angle $\alpha$ to the horizontal. The fire-engine is projected up the plank along a line of greatest slope. The non-gravitational resistance to the motion of the fire-engine is constant and has magnitude $R$ newtons.
When $\alpha = 20°$ the fire-engine is projected with an initial speed of 5 m s$^{-1}$ and first comes to rest after travelling 2 m.
\begin{enumerate}[label=(\alph*)]
\item Find, to 3 significant figures, the value of $R$.
[7]
\end{enumerate}
When $\alpha = 40°$ the fire-engine is again projected with an initial speed of 5 m s$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find how far the fire-engine travels before first coming to rest.
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2001 Q5 [10]}}