| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2016 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Collision/impulse during circular motion |
| Difficulty | Challenging +1.2 This is a standard M4 rigid body dynamics question covering moment of inertia by integration, conservation of angular momentum, energy methods, and forces in rotation. While it requires multiple techniques and careful bookkeeping across five parts, each individual step follows well-established procedures taught in M4. The integration in part (i) is routine, the collision uses standard angular impulse-momentum, and the energy equation follows a predictable pattern. This is moderately above average difficulty due to its length and the need to track expressions through multiple parts, but it doesn't require novel insight beyond applying learned methods systematically. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.03f Impulse-momentum: relation6.04b Find centre of mass: using symmetry6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Mass per unit length is \(\frac{2m}{4a}\) | B1 | |
| \(I = \sum\frac{m}{2a}x^2\delta x = \frac{m}{2a}\int x^2 dx\) | M1 | M1 for \(\int x^2 dx\); Limits not required for M mark |
| \(= \frac{m}{2a}\int_0^{4a} x^2 dx = \frac{m}{2a}\left[\frac{x^3}{3}\right]_0^{4a}\) | A1 | A1 for correct integration with limits |
| \(= \frac{32}{3}ma^2\) | A1 | AG Correctly shown |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Angular momentum of particle before impact = \(m(4av)\) | B1 | SC: If M0 then B1 for correct M of I |
| Angular momentum after impact = \(\left(\frac{32}{3}ma^2 + m(4a)^2\right)\omega\) | B1 | |
| By conservation of angular momentum: | ||
| \(4mav = \frac{80}{3}ma^2\omega\) | M1 | |
| \(\omega = \frac{3v}{20a}\) | A1 | 0.15\(va^{-1}\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| By conservation of energy | M1 | Correct number of terms; SC: If M0 then B1 for each part |
| \(\frac{1}{2}\left(\frac{80}{3}ma^2\right)\left[\dot{\theta}^2 - \left(\frac{3v}{20a}\right)^2\right] = ...\) | A1ft | Kinetic energy terms using their \(I\) and \(\omega\); \(\frac{40}{3}ma^2\dot{\theta}^2 - \frac{3}{10}mv^2 = ...\) |
| \(... = 8mga(\cos\theta - 1)\) | A1 | Potential energy terms |
| \(\dot{\theta}^2 = \frac{3g}{5a}(\cos\theta - 1) + \frac{9v^2}{400a^2}\) | A1 | \(k = \frac{3}{5}\); \(\dot{\theta}^2 = \omega^2 + \frac{16mga(\cos\theta - 1)}{I}\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3g}{5a}(-2) + \frac{9v^2}{400a^2} > 0\) | M1 | Setting \(\left(\frac{d\theta}{dt}\right)^2 > 0\) when \(\theta = \pi\); Condone for the M mark = or \(\geq\) |
| \(v^2 > \frac{160}{3}ga\) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\ddot{\theta} = \frac{3g}{5a}(-\sin\theta)\dot{\theta}\) | M1 | Differentiating \(\dot{\theta}\) with respect to \(t\); \(-2mg(2a) - mg(4a) = \frac{80}{3}ma^2\ddot{\theta}\) |
| \(\ddot{\theta} = -\frac{3g}{10a}\) | A1 | allow \(\frac{3g}{10a}\) |
| \(R - 3mg = 3m\left(\frac{8}{3}a\ddot{\theta}\right)\) | M1 | For transverse acceleration; \(r\sigma\) - mass must be \(3m\); Allow \(r = a\) for the M mark |
| \(R = \frac{3}{5}mg\) | A1 | |
| [4] |
**(i)**
Mass per unit length is $\frac{2m}{4a}$ | B1 |
$I = \sum\frac{m}{2a}x^2\delta x = \frac{m}{2a}\int x^2 dx$ | M1 | M1 for $\int x^2 dx$; Limits not required for M mark
$= \frac{m}{2a}\int_0^{4a} x^2 dx = \frac{m}{2a}\left[\frac{x^3}{3}\right]_0^{4a}$ | A1 | A1 for correct integration with limits
$= \frac{32}{3}ma^2$ | A1 | AG Correctly shown
| [4] |
**(ii)**
Angular momentum of particle before impact = $m(4av)$ | B1 | SC: If M0 then B1 for correct M of I
Angular momentum after impact = $\left(\frac{32}{3}ma^2 + m(4a)^2\right)\omega$ | B1 |
By conservation of angular momentum: |
$4mav = \frac{80}{3}ma^2\omega$ | M1 |
$\omega = \frac{3v}{20a}$ | A1 | 0.15$va^{-1}$
| [4] |
**(iii)**
By conservation of energy | M1 | Correct number of terms; SC: If M0 then B1 for each part
$\frac{1}{2}\left(\frac{80}{3}ma^2\right)\left[\dot{\theta}^2 - \left(\frac{3v}{20a}\right)^2\right] = ...$ | A1ft | Kinetic energy terms using their $I$ and $\omega$; $\frac{40}{3}ma^2\dot{\theta}^2 - \frac{3}{10}mv^2 = ...$
$... = 8mga(\cos\theta - 1)$ | A1 | Potential energy terms
$\dot{\theta}^2 = \frac{3g}{5a}(\cos\theta - 1) + \frac{9v^2}{400a^2}$ | A1 | $k = \frac{3}{5}$; $\dot{\theta}^2 = \omega^2 + \frac{16mga(\cos\theta - 1)}{I}$
| [4] |
**(iv)**
$\frac{3g}{5a}(-2) + \frac{9v^2}{400a^2} > 0$ | M1 | Setting $\left(\frac{d\theta}{dt}\right)^2 > 0$ when $\theta = \pi$; Condone for the M mark = or $\geq$
$v^2 > \frac{160}{3}ga$ | A1 |
| [2] |
**(v)**
$2\ddot{\theta} = \frac{3g}{5a}(-\sin\theta)\dot{\theta}$ | M1 | Differentiating $\dot{\theta}$ with respect to $t$; $-2mg(2a) - mg(4a) = \frac{80}{3}ma^2\ddot{\theta}$
$\ddot{\theta} = -\frac{3g}{10a}$ | A1 | allow $\frac{3g}{10a}$
$R - 3mg = 3m\left(\frac{8}{3}a\ddot{\theta}\right)$ | M1 | For transverse acceleration; $r\sigma$ - mass must be $3m$; Allow $r = a$ for the M mark
$R = \frac{3}{5}mg$ | A1 |
| [4] |A uniform rod $AB$ has mass $2m$ and length $4a$.
\begin{enumerate}[label=(\roman*)]
\item Show by integration that the moment of inertia of the rod about an axis perpendicular to the rod through $A$ is $\frac{32}{3}ma^2$. [4]
\end{enumerate}
The rod is initially at rest with $B$ vertically below $A$ and it is free to rotate in a vertical plane about a smooth fixed horizontal axis through $A$. A particle of mass $m$ is moving horizontally in the plane in which the rod is free to rotate. The particle has speed $v$, and strikes the rod at $B$. In the subsequent motion the particle adheres to the rod and the combined rigid body $Q$, consisting of the rod and the particle, starts to rotate.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find, in terms of $v$ and $a$, the initial angular speed of $Q$. [4]
\end{enumerate}
At time $t$ seconds the angle between $Q$ and the downward vertical is $\theta$ radians.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Show that $\dot{\theta}^2 = k\frac{g}{a}(\cos \theta - 1) + \frac{9v^2}{400a^2}$, stating the value of the constant $k$. [4]
\item Find, in terms of $a$ and $g$, the set of values of $v^2$ for which $Q$ makes complete revolutions. [2]
\end{enumerate}
When $Q$ is horizontal, the force exerted by the axis on $Q$ has vertically upwards component $R$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{4}
\item Find $R$ in terms of $m$ and $g$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR M4 2016 Q5 [18]}}