| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on outer surface of sphere |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring energy conservation and Newton's second law in the radial direction. Part (a) is a straightforward 'show that' using energy methods, part (b) requires setting normal reaction to zero (standard technique), and part (c) needs another energy calculation. While it involves multiple steps, all techniques are textbook-standard for M3 with no novel insight required, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}mv^2 = mg(a\cos\alpha - a\cos\theta)\) | M1 A1 A1 | |
| \(v^2 = 2ga(\cos\alpha - \cos\theta)\) ★ | A1 | cso (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mg\cos\theta(-R) = \frac{mv^2}{a}\) with \(R=0\) | M1 A1=A1 | |
| \(g\cos\theta = 2g\left(\frac{3}{4} - \cos\theta\right)\) | M1 | |
| \(\cos\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}\) (accept 60°) | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| From \(A\) to \(B\): \(\frac{1}{2}mw^2 = mg(a + a\cos\alpha)\) | M1 A1 A1 | |
| \(w^2 = 2ga\left(1 + \frac{3}{4}\right) \Rightarrow w = \left(\frac{7ga}{2}\right)^{\frac{1}{2}}\) | A1 | (4 marks) [13 total] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| From \(P\) to \(C\): \(v_P^2 = 2ga\left(\frac{3}{4}-\frac{1}{2}\right) = \frac{ga}{2}\) | ||
| \(\frac{1}{2}mw^2 - \frac{1}{2}m\left(\frac{ga}{2}\right) = mg(a + a\cos\theta)\) | M1 A1 A1 | |
| \(w^2 - \frac{ga}{2} = 2mga\left(1+\frac{1}{2}\right) \Rightarrow w = \left(\frac{7ga}{2}\right)^{\frac{1}{2}}\) | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v_P = \left(\frac{ga}{2}\right)^{\frac{1}{2}}\) | ||
| \(u_y = \left(\frac{ga}{2}\right)^{\frac{1}{2}}\sin 60° = \left(\frac{3ga}{8}\right)^{\frac{1}{2}}\) | M1 A1 | |
| \(v_y^2 = u_y^2 + 2g \times \frac{3a}{2} = \frac{27ga}{8}\) | A1 | |
| \(u_x = \left(\frac{ga}{2}\right)^{\frac{1}{2}}\cos 60° = \left(\frac{ga}{8}\right)^{\frac{1}{2}}\) | A1 | |
| \(w^2 = u_x^2 + v_y^2 = \frac{ga}{8} + \frac{27ga}{8} = \frac{7ga}{2} \Rightarrow w = \left(\frac{7ga}{2}\right)^{\frac{1}{2}}\) | (4 marks) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = mg(a\cos\alpha - a\cos\theta)$ | M1 A1 A1 | |
| $v^2 = 2ga(\cos\alpha - \cos\theta)$ ★ | A1 | cso (4 marks) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mg\cos\theta(-R) = \frac{mv^2}{a}$ with $R=0$ | M1 A1=A1 | |
| $g\cos\theta = 2g\left(\frac{3}{4} - \cos\theta\right)$ | M1 | |
| $\cos\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}$ (accept 60°) | A1 | (5 marks) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| From $A$ to $B$: $\frac{1}{2}mw^2 = mg(a + a\cos\alpha)$ | M1 A1 A1 | |
| $w^2 = 2ga\left(1 + \frac{3}{4}\right) \Rightarrow w = \left(\frac{7ga}{2}\right)^{\frac{1}{2}}$ | A1 | (4 marks) [13 total] |
**Alternatives to 5(c):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| From $P$ to $C$: $v_P^2 = 2ga\left(\frac{3}{4}-\frac{1}{2}\right) = \frac{ga}{2}$ | | |
| $\frac{1}{2}mw^2 - \frac{1}{2}m\left(\frac{ga}{2}\right) = mg(a + a\cos\theta)$ | M1 A1 A1 | |
| $w^2 - \frac{ga}{2} = 2mga\left(1+\frac{1}{2}\right) \Rightarrow w = \left(\frac{7ga}{2}\right)^{\frac{1}{2}}$ | A1 | (4 marks) |
**Alternative using projectile motion from P:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v_P = \left(\frac{ga}{2}\right)^{\frac{1}{2}}$ | | |
| $u_y = \left(\frac{ga}{2}\right)^{\frac{1}{2}}\sin 60° = \left(\frac{3ga}{8}\right)^{\frac{1}{2}}$ | M1 A1 | |
| $v_y^2 = u_y^2 + 2g \times \frac{3a}{2} = \frac{27ga}{8}$ | A1 | |
| $u_x = \left(\frac{ga}{2}\right)^{\frac{1}{2}}\cos 60° = \left(\frac{ga}{8}\right)^{\frac{1}{2}}$ | A1 | |
| $w^2 = u_x^2 + v_y^2 = \frac{ga}{8} + \frac{27ga}{8} = \frac{7ga}{2} \Rightarrow w = \left(\frac{7ga}{2}\right)^{\frac{1}{2}}$ | | (4 marks) |
---5. A smooth solid sphere, with centre $O$ and radius $a$, is fixed to the upper surface of a horizontal table. A particle $P$ is placed on the surface of the sphere at a point $A$, where $O A$ makes an angle $\alpha$ with the upward vertical, and $0 < \alpha < \frac { \pi } { 2 }$. The particle is released from rest. When $O P$ makes an angle $\theta$ with the upward vertical, and $P$ is still on the surface of the sphere, the speed of $P$ is $v$.
\begin{enumerate}[label=(\alph*)]
\item Show that $v ^ { 2 } = 2 g a ( \cos \alpha - \cos \theta )$.
Given that $\cos \alpha = \frac { 3 } { 4 }$, find
\item the value of $\theta$ when $P$ loses contact with the sphere,
\item the speed of $P$ as it hits the table.\\
(Total 13 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2005 Q5 [13]}}