| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done against air resistance - vertical motion |
| Difficulty | Standard +0.8 This is a multi-step mechanics problem requiring energy methods and kinematics across two scenarios. Students must find time T from free fall, then apply energy conservation with resistance forces in liquid, and finally calculate terminal speed. The connection between the two scenarios and handling the constant resistance force elevates this above routine M2 questions, though the individual techniques are standard. |
| Spec | 3.03a Force: vector nature and diagrams3.03b Newton's first law: equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(s = \frac{1}{2}at^2\); if time is doubled, acceleration is divided by 4 | M1 | |
| so net acc. \(= \frac{1}{4}g\) | ||
| \(mg - R = ma\) | A1 M1 A1 | |
| \(R = \frac{3}{4}g(0.6) = 4.41 \text{ N}\) | ||
| (b) \(v^2 = 2as = \frac{1}{2}g(2) = 9.8\) | M1 A1 | |
| \(v = 3.13 \text{ ms}^{-1}\) | A1 | 7 marks |
(a) $s = \frac{1}{2}at^2$; if time is doubled, acceleration is divided by 4 | M1 |
so net acc. $= \frac{1}{4}g$ | |
$mg - R = ma$ | A1 M1 A1 |
$R = \frac{3}{4}g(0.6) = 4.41 \text{ N}$ | |
(b) $v^2 = 2as = \frac{1}{2}g(2) = 9.8$ | M1 A1 |
$v = 3.13 \text{ ms}^{-1}$ | A1 | 7 marks4. A small stone, of mass 600 grams, is released from rest a height of 2 metres above ground level and falls under gravity. The time it takes to reach the ground is $T$ seconds. The stone is then again released from rest at the surface of a tank containing a 2 metre depth of liquid and reaches the bottom after $2 T$ seconds. It may be assumed that the resisting force acting on the stone is constant.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the resisting force exerted on the stone by the liquid.
\item Find the speed with which the stone hits the bottom of the tank.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q4 [7]}}