| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | June |
| 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 classic M3 circular motion problem requiring energy conservation, Newton's second law in polar form, and differentiation. Parts (i)-(ii) are standard bookwork. Parts (iii)-(iv) require careful application of concepts at the point of leaving contact, but follow directly from the setup with no novel insight needed. Slightly above average due to the multi-step nature and the less routine final parts. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}mv^2 = mga(1 - \cos\theta)\) | M1 | For using KE gain = PE loss |
| \(aw^2 = 2g(1 - \cos\theta)\) | A1 | |
| B1 | 3 | AG From \(v = wr\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mv^2/a = mg\cos\theta - R\) or \(maw^2 = mg\cos\theta - R\) | M1 | For using Newton's second law radially (3 terms required) with accel \(= v^2/r\) or \(w^2r\) |
| A1 | ||
| \([2mg(1-\cos\theta) = mg\cos\theta - R]\) | DM1 | For eliminating \(v^2\) or \(w^2\); depends on at least one previous M1 |
| \(R = mg(3\cos\theta - 2)\) | A1ft | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([\text{mg}\sin\theta = m(\text{accel.})\) or \(2a(\dot{\theta})\ddot{\theta} = 2g\sin\theta(\dot{\theta})]\) | M1 | For using Newton's second law tangentially or differentiating \(aw^2 = 2g(1-\cos\theta)\) w.r.t. \(t\) |
| Accel. \(= a\ddot{\theta} = g\sin\theta\) | A1 | |
| \([\theta = \cos^{-1}(2/3)]\) | M1 | For using \(R = 0\); ft from incorrect \(R\) of the form \(mg(A\cos\theta + B)\), \(A\neq 0\), \(B\neq 0\) |
| Acceleration is \(7.30\text{ ms}^{-2}\) | A1ft | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(dR/dt = (-3mg\sin\theta)\sqrt{2g(1-\cos\theta)/a}\) | M1 | For using rate of change \(= (dR/d\theta)(d\theta/dt)\); ft from incorrect \(R\) of the form \(mg(A\cos\theta + B)\), \(A\neq 0\) |
| A1ft | ||
| M1 | For using \(\cos\theta = 2/3\) | |
| Rate of change is \(-mg\sqrt{\dfrac{10g}{3a}}\text{ Ns}^{-1}\) | A1ft | 4 |
# Question 7:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = mga(1 - \cos\theta)$ | M1 | For using KE gain = PE loss |
| $aw^2 = 2g(1 - \cos\theta)$ | A1 | |
| | B1 | 3 | AG From $v = wr$ |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mv^2/a = mg\cos\theta - R$ or $maw^2 = mg\cos\theta - R$ | M1 | For using Newton's second law radially (3 terms required) with accel $= v^2/r$ or $w^2r$ |
| | A1 | |
| $[2mg(1-\cos\theta) = mg\cos\theta - R]$ | DM1 | For eliminating $v^2$ or $w^2$; depends on at least one previous M1 |
| $R = mg(3\cos\theta - 2)$ | A1ft | 4 | ft sign error in N2 equation |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[\text{mg}\sin\theta = m(\text{accel.})$ or $2a(\dot{\theta})\ddot{\theta} = 2g\sin\theta(\dot{\theta})]$ | M1 | For using Newton's second law tangentially or differentiating $aw^2 = 2g(1-\cos\theta)$ w.r.t. $t$ |
| Accel. $= a\ddot{\theta} = g\sin\theta$ | A1 | |
| $[\theta = \cos^{-1}(2/3)]$ | M1 | For using $R = 0$; ft from incorrect $R$ of the form $mg(A\cos\theta + B)$, $A\neq 0$, $B\neq 0$ |
| Acceleration is $7.30\text{ ms}^{-2}$ | A1ft | 4 | accept $g\sqrt{5}/3$ |
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $dR/dt = (-3mg\sin\theta)\sqrt{2g(1-\cos\theta)/a}$ | M1 | For using rate of change $= (dR/d\theta)(d\theta/dt)$; ft from incorrect $R$ of the form $mg(A\cos\theta + B)$, $A\neq 0$ |
| | A1ft | |
| | M1 | For using $\cos\theta = 2/3$ |
| Rate of change is $-mg\sqrt{\dfrac{10g}{3a}}\text{ Ns}^{-1}$ | A1ft | 4 | Any correct form of $\dot{R}$ with $\cos\theta = 2/3$ used; ft with $\square$ from incorrect $R$ of the form $mg(A\cos\theta + B)$, $A\neq 0$, $B\neq 0$ |7\\
\includegraphics[max width=\textwidth, alt={}, center]{85402f4a-8d55-47d8-ba48-5b837609b0f4-4_517_677_267_733}
A particle $P$ of mass $m \mathrm {~kg}$ is slightly disturbed from rest at the highest point on the surface of a smooth fixed sphere of radius $a$ m and centre $O$. The particle starts to move downwards on the surface. While $P$ remains on the surface $O P$ makes an angle of $\theta$ radians with the upward vertical and has angular speed $\omega$ rad s $^ { - 1 }$ (see diagram). The sphere exerts a force of magnitude $R \mathrm {~N}$ on $P$.\\
(i) Show that $a \omega ^ { 2 } = 2 g ( 1 - \cos \theta )$.\\
(ii) Find an expression for $R$ in terms of $m , g$ and $\theta$.
At the instant that $P$ loses contact with the surface of the sphere, find\\
(iii) the transverse component of the acceleration of $P$,\\
(iv) the rate of change of $R$ with respect to time $t$, in terms of $m , g$ and $a$.
\hfill \mbox{\textit{OCR M3 2008 Q7 [15]}}