| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Particle on sphere or circular surface |
| Difficulty | Challenging +1.2 This is a standard M3 circular motion problem combining energy conservation with projectile motion. Part (a) is routine application of energy conservation (given initial speed and height change from tan α = 3/4). Part (b) requires projectile motion from point B with given initial conditions. The problem is well-structured with clear steps and standard techniques, making it above average difficulty due to the multi-stage nature and need for careful geometry/trigonometry, but not requiring novel insight. |
| Spec | 3.02i Projectile motion: constant acceleration model6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use of Energy at A = energy at B | ||
| \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgh\), \(\frac{1}{2}mgr = \frac{1}{2}mv^2 + mg\times r\times(1-\cos\alpha) = \frac{1}{2}mv^2 + mg\times r\times\frac{1}{5}\) | M1 A1A1 | |
| \(v^2 = gr - \frac{2gr}{5} = \frac{3gr}{5}\) | ||
| \(v = \sqrt{\frac{3gr}{5}}\) \*AG\* | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horizontal component of speed at \(B\) and at \(C\) = their \(v\cos\theta\) | M1 | |
| Vertical component of speed at \(B\) = their \(v\sin\theta\) | M1 | |
| Conservation of energy gives speed at \(C = \sqrt{\frac{2g}{5}}\) | ||
| Vertical component of speed at \(C = \sqrt{\frac{2g}{5} - \frac{16\times 6g}{25^2}} \approx 1.5539\) | M1A1 | |
| \(v = u + at \Rightarrow t = \frac{1.5539...+0.92017...}{g} \approx 0.252\ \text{seconds}\) | M1A1 | |
| Horizontal distance \(= \frac{3}{5}\times 0.4 + 1.22689...\times 0.252... = 0.55\ \text{(m)}\) | M1A1 | (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horizontal component of speed at \(B\) and at \(C\) = their \(v\cos\theta\) | M1 | |
| Vertical component of speed at \(B\) = their \(v\sin\theta\) | M1 | |
| \(s = ut + \frac{1}{2}at^2:\ -\frac{1}{5}\times 0.4 = -\frac{2}{25} = \sqrt{\frac{6g}{25}}\times\frac{3}{5}t - \frac{1}{2}gt^2\) | M1A1 | |
| \(4.9t^2 - 0.92017...t - 0.08 = 0\) | ||
| \(t = \frac{0.920 + \sqrt{0.920^2 + 0.32\times 4.9}}{9.8} = 0.252......\) | M1A1 | |
| Horizontal distance \(= \frac{3}{5}\times 0.4 + 1.22689...\times 0.252... = 0.55\ \text{(m)}\) | M1A1 | (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horizontal component of speed at \(B\) and at \(C\) = their \(v\cos\theta\) | M1 | |
| Vertical component of speed at \(B\) = their \(v\sin\theta\) | M1 | |
| \(s = ut + \frac{1}{2}at^2:\ -\frac{1}{5}\times 0.4 = -\frac{2}{25} = \sqrt{\frac{6g}{25}}\times\frac{3}{5}t - \frac{1}{2}gt^2\) | M1A1 | |
| \(4.9t^2 - 0.92017t - 0.08 = 0\) | ||
| Horizontal distance from \(B = 1.22689...\times t = x\) | ||
| Form quadratic in \(x\) by substituting for \(t\) above | M1 | |
| \(3.255x^2 - 0.75x - 0.08 = 0\) | ||
| \(x = \frac{0.75 + \sqrt{0.75^2 + 4\times 3.255\times 0.08}}{2\times 3.255} = 0.3097...\) | M1A1 | |
| Horizontal distance \(= \frac{3}{5}\times 0.4 + 0.3097... = 0.55\ \text{(m)}\) | A1 | (8) [12] |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of Energy at A = energy at B | | |
| $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgh$, $\frac{1}{2}mgr = \frac{1}{2}mv^2 + mg\times r\times(1-\cos\alpha) = \frac{1}{2}mv^2 + mg\times r\times\frac{1}{5}$ | M1 A1A1 | |
| $v^2 = gr - \frac{2gr}{5} = \frac{3gr}{5}$ | | |
| $v = \sqrt{\frac{3gr}{5}}$ \*AG\* | A1 | **(4)** |
### Part (b) — v1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Horizontal component of speed at $B$ and at $C$ = their $v\cos\theta$ | M1 | |
| Vertical component of speed at $B$ = their $v\sin\theta$ | M1 | |
| Conservation of energy gives speed at $C = \sqrt{\frac{2g}{5}}$ | | |
| Vertical component of speed at $C = \sqrt{\frac{2g}{5} - \frac{16\times 6g}{25^2}} \approx 1.5539$ | M1A1 | |
| $v = u + at \Rightarrow t = \frac{1.5539...+0.92017...}{g} \approx 0.252\ \text{seconds}$ | M1A1 | |
| Horizontal distance $= \frac{3}{5}\times 0.4 + 1.22689...\times 0.252... = 0.55\ \text{(m)}$ | M1A1 | **(8)** |
### Part (b) — v2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Horizontal component of speed at $B$ and at $C$ = their $v\cos\theta$ | M1 | |
| Vertical component of speed at $B$ = their $v\sin\theta$ | M1 | |
| $s = ut + \frac{1}{2}at^2:\ -\frac{1}{5}\times 0.4 = -\frac{2}{25} = \sqrt{\frac{6g}{25}}\times\frac{3}{5}t - \frac{1}{2}gt^2$ | M1A1 | |
| $4.9t^2 - 0.92017...t - 0.08 = 0$ | | |
| $t = \frac{0.920 + \sqrt{0.920^2 + 0.32\times 4.9}}{9.8} = 0.252......$ | M1A1 | |
| Horizontal distance $= \frac{3}{5}\times 0.4 + 1.22689...\times 0.252... = 0.55\ \text{(m)}$ | M1A1 | **(8)** |
### Part (b) — v3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Horizontal component of speed at $B$ and at $C$ = their $v\cos\theta$ | M1 | |
| Vertical component of speed at $B$ = their $v\sin\theta$ | M1 | |
| $s = ut + \frac{1}{2}at^2:\ -\frac{1}{5}\times 0.4 = -\frac{2}{25} = \sqrt{\frac{6g}{25}}\times\frac{3}{5}t - \frac{1}{2}gt^2$ | M1A1 | |
| $4.9t^2 - 0.92017t - 0.08 = 0$ | | |
| Horizontal distance from $B = 1.22689...\times t = x$ | | |
| Form quadratic in $x$ by substituting for $t$ above | M1 | |
| $3.255x^2 - 0.75x - 0.08 = 0$ | | |
| $x = \frac{0.75 + \sqrt{0.75^2 + 4\times 3.255\times 0.08}}{2\times 3.255} = 0.3097...$ | M1A1 | |
| Horizontal distance $= \frac{3}{5}\times 0.4 + 0.3097... = 0.55\ \text{(m)}$ | A1 | **(8) [12]** |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{daa795f0-2c5e-4617-a295-fbe74c22be4a-08_504_1429_212_264}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Part of a hollow spherical shell, centre $O$ and radius $r$, forms a bowl with a plane circular rim. The bowl is fixed to a horizontal surface at $A$ with the rim uppermost and horizontal.
The point $A$ is the lowest point of the bowl. The point $B$, where $\angle A O B = \alpha$ and $\tan \alpha = \frac { 3 } { 4 }$, is on the rim of the bowl, as shown in Figure 2. A small smooth marble $M$ is placed inside the bowl at $A$, and given an initial horizontal speed $\sqrt { } ( g r )$. The motion of $M$ takes place in the vertical plane $O A B$.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $M$ as it reaches $B$ is $\sqrt { } \left( \frac { 3 } { 5 } g r \right)$.
After leaving the surface of the bowl at $B , M$ moves freely under gravity and first strikes the horizontal surface at the point $C$. Given that $r = 0.4 \mathrm {~m}$,
\item find the distance $A C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2013 Q5 [12]}}