| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2005 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Difficulty | Challenging +1.8 This is a challenging M3 question requiring energy conservation for circular motion, momentum conservation during collision, and circular motion dynamics. It involves multiple stages (swing down, catch ball, swing up) with geometric setup requiring careful angle work. The multi-step nature, combination of topics, and need to track energy/momentum through three distinct phases makes this significantly harder than average, though the individual techniques are standard M3 content. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| \(\frac{1}{2}(v^2 - 15) = m5\cos(1-\cos 60°)\) | M1 | |
| \(v = 8 \text{ m s}^{-1}\) | A1 A1 | (4+) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| \(\frac{1}{2}mw^2 = mg5(1-\cos 60°)\) | M1 | |
| \(w = 7 \text{ m s}^{-1}\) | A1 | |
| Check: \(60x - 3m = (60+m)7\) | M1 | |
| \(480 - 3m = 420 + 7m\) | M1 A1 | |
| \(60 = 10m\) | M1 | |
| \(C = m\) | A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| \(T - 66_5 = \frac{66x \cdot 7^2}{5}\) | M1 | |
| \(T = 1323 = 1290 (1294) \text{ N}\) | A1 | (3), (14) |
## Part (a)
| Content | Marks | Notes |
|---------|-------|-------|
| $\frac{1}{2}(v^2 - 15) = m5\cos(1-\cos 60°)$ | M1 | |
| $v = 8 \text{ m s}^{-1}$ | A1 A1 | (4+) |
## Part (b)
| Content | Marks | Notes |
|---------|-------|-------|
| $\frac{1}{2}mw^2 = mg5(1-\cos 60°)$ | M1 | |
| $w = 7 \text{ m s}^{-1}$ | A1 | |
| Check: $60x - 3m = (60+m)7$ | M1 | |
| $480 - 3m = 420 + 7m$ | M1 A1 | |
| $60 = 10m$ | M1 | |
| $C = m$ | A1 | (7) |
## Part (c)
| Content | Marks | Notes |
|---------|-------|-------|
| $T - 66_5 = \frac{66x \cdot 7^2}{5}$ | M1 | |
| $T = 1323 = 1290 (1294) \text{ N}$ | A1 | (3), (14) |\includegraphics{figure_6}
A trapeze artiste of mass 60 kg is attached to the end $A$ of a light inextensible rope $OA$ of length 5 m. The artiste must swing in an arc of a vertical circle, centre $O$, from a platform $P$ to another platform $Q$, where $PQ$ is horizontal. The other end of the rope is attached to the fixed point $O$ which lies in the vertical plane containing $PQ$, with $\angle POQ = 120^{\circ}$ and $OP = OQ = 5$ m, as shown in Figure 6.
As part of her act, the artiste projects herself from $P$ with speed $\sqrt{15}$ m s$^{-1}$ in a direction perpendicular to the rope $OA$ and in the plane $POQ$. She moves in a circular arc towards $Q$. At the lowest point of her path she catches a ball of mass $m$ kg which is travelling towards her with speed 3 m s$^{-1}$ and parallel to $QP$. After catching the ball, she comes to rest at the point $Q$.
By modelling the artiste and the ball as particles and ignoring her air resistance, find
\begin{enumerate}[label=(\alph*)]
\item the speed of the artiste immediately before she catches the ball, [4]
\item the value of $m$, [7]
\item the tension in the rope immediately after she catches the ball. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2005 Q7 [14]}}