| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2021 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on inner surface of sphere/bowl |
| Difficulty | Challenging +1.8 This is a substantial Further Mechanics question requiring circular motion analysis with impulse, energy methods, and constraint conditions for leaving the surface. Parts (a)-(c) involve standard techniques (energy conservation, normal force = 0 condition), but part (d) requires careful work with resistance over a curved path, and the multi-part nature with symbolic manipulation throughout elevates difficulty. However, the methods are well-established for Further Mechanics students, making it challenging but not exceptional. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(I = mu \Rightarrow u = I/m\) | B1 | Use of Impulse \(=\) change of momentum |
| Init PE \(= mgr - mgr\cos\frac{\pi}{3}\) | M1 | \(\left(= \frac{1}{2}mgr\right)\). Attempt to use '\(mgh\)' to find initial PE. Could use edge as zero PE level (so init PE \(= -\frac{1}{2}mgr\)) but must be clear and signs consistent |
| \(\frac{1}{2}mu^2 + \frac{1}{2}mgr = \frac{1}{2}mv^2 + mgr\) | M1 | Conservation of energy; KE & PE considered on both sides |
| \(v^2 = u^2 - gr \Rightarrow v = \sqrt{\frac{I^2}{m^2} - gr}\) | A1 | oe e.g. \(v = \frac{\sqrt{I^2 - m^2 gr}}{m}\) |
| Alternative method: | ||
| \(u = I/m\) | B1 | |
| \(\Delta PE = mgr\cos\frac{\pi}{3}\ \left(= \frac{1}{2}mgr\right)\) | M1 | |
| \(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - \frac{1}{2}mgr\) | M1 | Subtract gain in PE |
| \(v^2 = u^2 - gr \Rightarrow v = \sqrt{\frac{I^2}{m^2} - gr}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2}mv^2 = mgh \Rightarrow h = \frac{1}{2g}\left(\frac{I^2}{m^2} - gr\right) = \frac{I^2}{2gm^2} - \frac{r}{2}\) | B1 | oe e.g. \(h = \frac{I^2 - m^2 gr}{2m^2 g}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Consider the case where \(h \to 0\) | M1 | e.g. \(\frac{I^2}{m^2} = gr\) |
| Maximum possible value of \(m\) is \(\frac{I}{\sqrt{gr}}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Work done against \(R = r\left(\frac{\pi}{2} + \frac{\pi}{3}\right)R\) | M1 | |
| \(\frac{1}{2}mu^2 + \frac{1}{2}mgr = \frac{1}{2}mv^2 + mgr + r\left(\frac{\pi}{2} + \frac{\pi}{3}\right)R\) or \(\frac{I^2}{2m} + \frac{1}{2}mgr = \frac{1}{2}mv^2 + mgr + r\left(\frac{\pi}{2} + \frac{\pi}{3}\right)R\) | M1 | Revising the energy equation (condone incorrect initial energy from (a)) to include an energy loss term (work done against \(R\)). Could already be in terms of \(I\) rather than \(u\). Could be expressed as an inequality |
| Need \(v > 0\) so \(I > \sqrt{m^2 gr + \frac{5\pi mrR}{3}}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([I] = MLT^{-1}\) and \([RHS] = (M^2LT^{-2}L + MLMLT^{-2})^{1/2}\) | M1 | Attempt dimensional analysis on both sides |
| Hence \([RHS] = MLT^{-1} = [I]\) so the inequality is dimensionally consistent | A1 |
# Question 6:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $I = mu \Rightarrow u = I/m$ | B1 | Use of Impulse $=$ change of momentum |
| Init PE $= mgr - mgr\cos\frac{\pi}{3}$ | M1 | $\left(= \frac{1}{2}mgr\right)$. Attempt to use '$mgh$' to find initial PE. Could use edge as zero PE level (so init PE $= -\frac{1}{2}mgr$) but must be clear and signs consistent |
| $\frac{1}{2}mu^2 + \frac{1}{2}mgr = \frac{1}{2}mv^2 + mgr$ | M1 | Conservation of energy; KE & PE considered on both sides |
| $v^2 = u^2 - gr \Rightarrow v = \sqrt{\frac{I^2}{m^2} - gr}$ | A1 | oe e.g. $v = \frac{\sqrt{I^2 - m^2 gr}}{m}$ |
| **Alternative method:** | | |
| $u = I/m$ | B1 | |
| $\Delta PE = mgr\cos\frac{\pi}{3}\ \left(= \frac{1}{2}mgr\right)$ | M1 | |
| $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - \frac{1}{2}mgr$ | M1 | Subtract gain in PE |
| $v^2 = u^2 - gr \Rightarrow v = \sqrt{\frac{I^2}{m^2} - gr}$ | A1 | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}mv^2 = mgh \Rightarrow h = \frac{1}{2g}\left(\frac{I^2}{m^2} - gr\right) = \frac{I^2}{2gm^2} - \frac{r}{2}$ | B1 | oe e.g. $h = \frac{I^2 - m^2 gr}{2m^2 g}$ |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Consider the case where $h \to 0$ | M1 | e.g. $\frac{I^2}{m^2} = gr$ |
| Maximum possible value of $m$ is $\frac{I}{\sqrt{gr}}$ | A1 | |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Work done against $R = r\left(\frac{\pi}{2} + \frac{\pi}{3}\right)R$ | M1 | |
| $\frac{1}{2}mu^2 + \frac{1}{2}mgr = \frac{1}{2}mv^2 + mgr + r\left(\frac{\pi}{2} + \frac{\pi}{3}\right)R$ or $\frac{I^2}{2m} + \frac{1}{2}mgr = \frac{1}{2}mv^2 + mgr + r\left(\frac{\pi}{2} + \frac{\pi}{3}\right)R$ | M1 | Revising the energy equation (condone incorrect initial energy from (a)) to include an energy loss term (work done against $R$). Could already be in terms of $I$ rather than $u$. Could be expressed as an inequality |
| Need $v > 0$ so $I > \sqrt{m^2 gr + \frac{5\pi mrR}{3}}$ | A1 | AG |
## Part (e):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[I] = MLT^{-1}$ and $[RHS] = (M^2LT^{-2}L + MLMLT^{-2})^{1/2}$ | M1 | Attempt dimensional analysis on both sides |
| Hence $[RHS] = MLT^{-1} = [I]$ so the inequality is dimensionally consistent | A1 | |6 A smooth hemispherical shell of radius $r \mathrm {~m}$ is held with its circular rim horizontal and uppermost. The centre of the rim is at the point $O$ and the point on the inner surface directly below $O$ is $A$.
A small object $P$ of mass $m \mathrm {~kg}$ is held at rest on the inner surface of the shell so that $\angle \mathrm { POA } = \frac { 1 } { 3 } \pi$ radians. At the instant that $P$ is released, an impulse is applied to $P$ in the direction of the tangent to the surface at $P$ in the vertical plane containing $O , A$ and $P$. The magnitude of the impulse is denoted by $I$ Ns.\\
$P$ immediately starts to move along the surface towards $A$ (see diagram).\\
$X$ is a point on the circular rim. $P$ leaves the shell at $X$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a65c4b75-b8b4-4a51-8abb-f857dc278271-5_512_860_829_242}
In an initial model of the motion of $P$ it is assumed that $P$ experiences no resistance to its motion.
\begin{enumerate}[label=(\alph*)]
\item Find in terms of $r , g , m$ and $I$ an expression for the speed of $P$ at the instant that it leaves the shell at $X$.
\item Find in terms of $r , g , m$ and $I$ an expression for the maximum height attained by $P$ above $X$ after it has left the shell.
\item Find an expression for the maximum mass of $P$ for which $P$ still leaves the shell.
In a revised model it is assumed that $P$ experiences a resistive force of constant magnitude $R$ while it is moving.
\item Show that, in order for $P$ to still leave the shell at $X$ under the revised model,
$$I > \sqrt { m ^ { 2 } g r + \frac { 5 \pi m r R } { 3 } } .$$
\item Show that the inequality from part (d) is dimensionally consistent.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2021 Q6 [12]}}