| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on outer surface of sphere |
| Difficulty | Challenging +1.2 This is a standard M3 circular motion problem requiring energy conservation and Newton's second law to find the leaving angle, followed by projectile motion. The 'show that' part guides students through the hardest step, and the projectile motion calculation is routine. More challenging than basic C1/C2 questions but follows a well-established template for this topic. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let speed at \(C\) be \(u\) | ||
| CE: \(\frac{1}{2}mu^2 - \frac{1}{2}m\left(\frac{ag}{4}\right) = mga(1-\cos\theta)\) | M1 A1 | |
| \(u^2 = \frac{9ga}{4} - 2ga\cos\theta\) | ||
| \(mg\cos\theta\ (+R) = \frac{mu^2}{a}\) | M1 A1 | |
| \(mg\cos\theta = \frac{9mg}{4} - 2mg\cos\theta\) | M1 | eliminating \(u\) |
| Leading to \(\cos\theta = \frac{3}{4}\) * | M1 A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At \(C\): \(u^2 = \frac{9ga}{4} - 2ga \times \frac{3}{4} = \frac{3}{4}ga\) | B1 | |
| \((\rightarrow)\ u_x = u\cos\theta = \sqrt{\left(\frac{3ga}{4}\right)} \times \frac{3}{4} = \sqrt{\left(\frac{27ga}{64}\right)} = 2.033\sqrt{a}\) | M1 A1ft | |
| \((\downarrow)\ u_y = u\sin\theta = \sqrt{\left(\frac{3ga}{4}\right)} \times \frac{\sqrt{7}}{4} = \sqrt{\left(\frac{21ga}{64}\right)} = 1.792\sqrt{a}\) | M1 | |
| \(v_y^2 = u_y^2 + 2gh \Rightarrow v_y^2 = \frac{21}{64}ga + 2g \times \frac{7}{4}a = \frac{245}{64}ga\) | M1 A1 | |
| \(\tan\psi = \frac{v_y}{u_x} = \sqrt{\left(\frac{245}{27}\right)} \approx 3.012\ldots\) | M1 | |
| \(\psi \approx 72°\) | A1 | awrt \(72°\) |
| Or \(1.3^c\ (1.2502^c)\) | awrt \(1.3^c\) (8) [15] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let speed at \(P\) be \(v\) | ||
| CE: \(\frac{1}{2}mv^2 - \frac{1}{2}m\left(\frac{ag}{4}\right) = mg \times 2a\) | M1 | or equivalent |
| \(v^2 = \frac{17mga}{4}\) | M1 A1 | |
| \(\cos\psi = \frac{u_x}{v} = \sqrt{\left(\frac{27}{64} \times \frac{4}{17}\right)} = \sqrt{\left(\frac{27}{272}\right)} \approx 0.315\) | M1 | |
| \(\psi \approx 72°\) | A1 | awrt \(72°\) |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let speed at $C$ be $u$ | | |
| CE: $\frac{1}{2}mu^2 - \frac{1}{2}m\left(\frac{ag}{4}\right) = mga(1-\cos\theta)$ | M1 A1 | |
| $u^2 = \frac{9ga}{4} - 2ga\cos\theta$ | | |
| $mg\cos\theta\ (+R) = \frac{mu^2}{a}$ | M1 A1 | |
| $mg\cos\theta = \frac{9mg}{4} - 2mg\cos\theta$ | M1 | eliminating $u$ |
| Leading to $\cos\theta = \frac{3}{4}$ * | M1 A1 | (7) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| At $C$: $u^2 = \frac{9ga}{4} - 2ga \times \frac{3}{4} = \frac{3}{4}ga$ | B1 | |
| $(\rightarrow)\ u_x = u\cos\theta = \sqrt{\left(\frac{3ga}{4}\right)} \times \frac{3}{4} = \sqrt{\left(\frac{27ga}{64}\right)} = 2.033\sqrt{a}$ | M1 A1ft | |
| $(\downarrow)\ u_y = u\sin\theta = \sqrt{\left(\frac{3ga}{4}\right)} \times \frac{\sqrt{7}}{4} = \sqrt{\left(\frac{21ga}{64}\right)} = 1.792\sqrt{a}$ | M1 | |
| $v_y^2 = u_y^2 + 2gh \Rightarrow v_y^2 = \frac{21}{64}ga + 2g \times \frac{7}{4}a = \frac{245}{64}ga$ | M1 A1 | |
| $\tan\psi = \frac{v_y}{u_x} = \sqrt{\left(\frac{245}{27}\right)} \approx 3.012\ldots$ | M1 | |
| $\psi \approx 72°$ | A1 | awrt $72°$ |
| Or $1.3^c\ (1.2502^c)$ | | awrt $1.3^c$ **(8) [15]** |
## Part (b) Alternative (last five marks):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let speed at $P$ be $v$ | | |
| CE: $\frac{1}{2}mv^2 - \frac{1}{2}m\left(\frac{ag}{4}\right) = mg \times 2a$ | M1 | or equivalent |
| $v^2 = \frac{17mga}{4}$ | M1 A1 | |
| $\cos\psi = \frac{u_x}{v} = \sqrt{\left(\frac{27}{64} \times \frac{4}{17}\right)} = \sqrt{\left(\frac{27}{272}\right)} \approx 0.315$ | M1 | |
| $\psi \approx 72°$ | A1 | awrt $72°$ |
> **Note:** The time of flight from $C$ to $P$ is $\dfrac{\sqrt{235}-\sqrt{21}}{8}\sqrt{\left(\dfrac{a}{g}\right)} \approx 1.38373\sqrt{\left(\dfrac{a}{g}\right)}$7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8374fa0f-cb28-497f-8696-877d7d0762f1-11_671_1077_276_429}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
A particle is projected from the highest point $A$ on the outer surface of a fixed smooth sphere of radius $a$ and centre $O$. The lowest point $B$ of the sphere is fixed to a horizontal plane. The particle is projected horizontally from $A$ with speed $\frac { 1 } { 2 } \sqrt { } ( g a )$. The particle leaves the surface of the sphere at the point $C$, where $\angle A O C = \theta$, and strikes the plane at the point $P$, as shown in Figure 5.
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos \theta = \frac { 3 } { 4 }$.
\item Find the angle that the velocity of the particle makes with the horizontal as it reaches $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2009 Q7 [15]}}