| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on inner surface of sphere/bowl |
| Difficulty | Challenging +1.2 This is a standard M3 circular motion problem requiring energy conservation and Newton's second law in circular motion. Part (a) uses F = mv²/r + mg at the lowest point, part (b) applies energy conservation, and part (c) requires projectile motion analysis to show the particle clears the rim. While multi-step, these are well-practiced techniques for M3 students with no novel geometric insight required beyond recognizing the 120° angle geometry. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}mv^2 - \frac{1}{2}mU^2 = mgr(1+\cos 60)\) | M1A1A1 | Attempt energy equation from \(A\) to \(B\) with 2 KE terms and PE term including trig function. Correct KE terms (difference). Correct PE term all signs correct |
| \(R - mg = m\frac{v^2}{r}\) | M1A1A1 | Attempt NL2 along radius at \(B\). Correct difference of forces. Correct mass × accel term, completely correct equation |
| \(R - mg = 2mg \times \frac{3}{2} + m\frac{U^2}{r}\) | dM1 | Eliminate \(v\) between two equations. Dependent on both previous M marks |
| \(R = 4mg + m\frac{U^2}{r}\) | A1 | Correct expression for \(R\). Any equivalent form accepted, must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P\) leaves bowl with speed \(U\) at \(60°\) to horizontal | ||
| Vertical speed \(= U\sin 60 \left(= \frac{U\sqrt{3}}{2}\right)\) | B1 | Correct vertical speed on leaving bowl |
| \(0 = (U\sin 60)^2 - 2gs\) | M1 | Use \(v^2 = u^2 + 2as\) with initial vertical speed to find greatest height above rim |
| \(s = \frac{3U^2}{8g}\) | A1 | Solve to obtain correct greatest height above rim |
| Greatest height \(= \frac{3}{2}r + \frac{3U^2}{8g}\) | A1ft | Add \(\frac{3r}{2}\) to previous answer to obtain greatest height above floor |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0 = U\sin 60\, t - \frac{1}{2}gt^2\) | M1 | Use \(s = ut + \frac{1}{2}at^2\) with initial vertical speed and \(s=0\) to find time until \(P\) returns to level of rim |
| Time to level of rim \(= \frac{2U\sin 60}{g} \left(= \frac{U\sqrt{3}}{g}\right)\) | A1 | Correct time obtained |
| Horizontal speed \(= U\cos 60 = \left(\frac{1}{2}U\right)\) | ||
| Horizontal dist \(= \frac{2U^2\sin 60}{2g}\) | B1 | Correct horizontal distance. Allow with one, two trig functions or none |
| \(U^2 > 2gr \therefore\) horiz dist \(> \frac{4rg\sin 60}{2g} = 2r\sin 60\) | M1 | Compare horizontal distance with diameter of rim using inequality. Depends on previous M mark |
| \(\therefore P\) does not fall back into the bowl | A1cso | Correct conclusion from correct work |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0 = U\sin 60 - gt\), \(t = \frac{U\sqrt{3}}{2g}\) | M1, A1 | |
| Horizontal dist to highest point \(= \frac{U^2\sqrt{3}}{4g}\) | B1 | |
| \(U^2 > 2gr \therefore\) horiz dist \(> r\sin 60\), \(\therefore P\) does not fall back into bowl | M1, A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Time to \(A = \frac{2r\sqrt{3}}{U}\) | B1 | Use horizontal motion to find time to \(A\) |
| Vert dist \(= U\frac{\sqrt{3}}{2} \times \frac{2r\sqrt{3}}{U} - \frac{1}{2}g\frac{4r^2 \times 3}{U^2} \left(= 3r - \frac{6gr^2}{U^2}\right)\) | M1A1 | Find vertical height above \(A\) |
| \(U^2 > 2gr \therefore\) vert dist \(> 0\), \(\therefore P\) does not fall back into bowl | M1, A1cso | Use inequality to determine whether \(P\) is above rim |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 - \frac{1}{2}mU^2 = mgr(1+\cos 60)$ | M1A1A1 | Attempt energy equation from $A$ to $B$ with 2 KE terms and PE term including trig function. Correct KE terms (difference). Correct PE term all signs correct |
| $R - mg = m\frac{v^2}{r}$ | M1A1A1 | Attempt NL2 along radius at $B$. Correct difference of forces. Correct mass × accel term, completely correct equation |
| $R - mg = 2mg \times \frac{3}{2} + m\frac{U^2}{r}$ | dM1 | Eliminate $v$ between two equations. Dependent on both previous M marks |
| $R = 4mg + m\frac{U^2}{r}$ | A1 | Correct expression for $R$. Any equivalent form accepted, must be positive |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P$ leaves bowl with speed $U$ at $60°$ to horizontal | | |
| Vertical speed $= U\sin 60 \left(= \frac{U\sqrt{3}}{2}\right)$ | B1 | Correct vertical speed on leaving bowl |
| $0 = (U\sin 60)^2 - 2gs$ | M1 | Use $v^2 = u^2 + 2as$ with initial vertical speed to find greatest height above rim |
| $s = \frac{3U^2}{8g}$ | A1 | Solve to obtain correct greatest height above rim |
| Greatest height $= \frac{3}{2}r + \frac{3U^2}{8g}$ | A1ft | Add $\frac{3r}{2}$ to previous answer to obtain greatest height above floor |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0 = U\sin 60\, t - \frac{1}{2}gt^2$ | M1 | Use $s = ut + \frac{1}{2}at^2$ with initial vertical speed and $s=0$ to find time until $P$ returns to level of rim |
| Time to level of rim $= \frac{2U\sin 60}{g} \left(= \frac{U\sqrt{3}}{g}\right)$ | A1 | Correct time obtained |
| Horizontal speed $= U\cos 60 = \left(\frac{1}{2}U\right)$ | | |
| Horizontal dist $= \frac{2U^2\sin 60}{2g}$ | B1 | Correct horizontal distance. Allow with one, two trig functions or none |
| $U^2 > 2gr \therefore$ horiz dist $> \frac{4rg\sin 60}{2g} = 2r\sin 60$ | M1 | Compare horizontal distance with diameter of rim using inequality. Depends on previous M mark |
| $\therefore P$ does not fall back into the bowl | A1cso | Correct conclusion from correct work |
**ALT 1 (c) — Work to highest point:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0 = U\sin 60 - gt$, $t = \frac{U\sqrt{3}}{2g}$ | M1, A1 | |
| Horizontal dist to highest point $= \frac{U^2\sqrt{3}}{4g}$ | B1 | |
| $U^2 > 2gr \therefore$ horiz dist $> r\sin 60$, $\therefore P$ does not fall back into bowl | M1, A1cso | |
**ALT 2 (c) — Find vertical height when above $A$:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Time to $A = \frac{2r\sqrt{3}}{U}$ | B1 | Use horizontal motion to find time to $A$ |
| Vert dist $= U\frac{\sqrt{3}}{2} \times \frac{2r\sqrt{3}}{U} - \frac{1}{2}g\frac{4r^2 \times 3}{U^2} \left(= 3r - \frac{6gr^2}{U^2}\right)$ | M1A1 | Find vertical height above $A$ |
| $U^2 > 2gr \therefore$ vert dist $> 0$, $\therefore P$ does not fall back into bowl | M1, A1cso | Use inequality to determine whether $P$ is above rim |
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Could you please share the correct page(s) that contain the actual mark scheme content? I'd be happy to format it once I can see the questions, answers, mark allocations, and guidance notes.7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{85d8fc7d-8863-419e-8eef-8751a6fb6315-11_412_1054_260_447}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A hollow sphere has internal radius $r$ and centre $O$. A bowl with a plane circular rim is formed by removing part of the sphere. The bowl is fixed to a horizontal floor with the rim uppermost and horizontal. The point $B$ is the lowest point of the inner surface of the bowl. The point $A$, where angle $A O B = 120 ^ { \circ }$, lies on the rim of the bowl, as shown in Figure 4. A particle $P$ of mass $m$ is projected from $A$, with speed $U$ at $90 ^ { \circ }$ to $O A$, and moves on the smooth inner surface of the bowl. The motion of $P$ takes place in the vertical plane $O A B$.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m , g , U$ and $r$, the magnitude of the force exerted on $P$ by the bowl at the instant when $P$ passes through $B$.
\item Find, in terms of $g , U$ and $r$, the greatest height above the floor reached by $P$.
Given that $U > \sqrt { 2 g r }$
\item show that, after leaving the surface of the bowl, $P$ does not fall back into the bowl.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2017 Q7 [17]}}