| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.3 This is a standard circular motion problem requiring resolution of forces and application of F=ma radially. Part (a) involves finding the normal reaction using vertical equilibrium, part (b) uses horizontal circular motion equation. The geometry is straightforward (given d and a) and the method is textbook-standard for M3, making it slightly easier than average overall. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(R\cos\theta = mg\) or \(R\sin\theta = mg\) | M1A1 | Resolve vertically. Dimensionally correct, correct number of terms, condone sign errors and sin/cos confusion |
| \(R = \frac{mga}{d}\) | A1 | (3) Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(R\sin\theta = \frac{mv^2}{r}\) or \(R\cos\theta = \frac{mv^2}{r}\) | M1A1A1 | Form horizontal equation of motion. Accept \(\frac{v^2}{r}\) or \(r\omega^2\) for acceleration. Condone use of \(a\) for radius but M0 if \(a\) used for acceleration. A1: at most one error (incorrect form of acceleration or radius). A1: correct form of acceleration and correct radius |
| \(\frac{mga}{d} \times \frac{\sqrt{a^2-d^2}}{a} = \frac{mv^2}{\sqrt{a^2-d^2}}\) | DM1 | Dependent on previous M. Eliminate \(R\) and trig to form equation in \(v\), \(g\), \(a\) and \(d\) |
| \(v = \sqrt{\frac{g(a^2-d^2)}{d}}\) | A1 | (5) Correct answer ISW |
## Question 2:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $R\cos\theta = mg$ or $R\sin\theta = mg$ | M1A1 | Resolve vertically. Dimensionally correct, correct number of terms, condone sign errors and sin/cos confusion |
| $R = \frac{mga}{d}$ | A1 | (3) Correct answer |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $R\sin\theta = \frac{mv^2}{r}$ or $R\cos\theta = \frac{mv^2}{r}$ | M1A1A1 | Form horizontal equation of motion. Accept $\frac{v^2}{r}$ or $r\omega^2$ for acceleration. Condone use of $a$ for radius but M0 if $a$ used for acceleration. A1: at most one error (incorrect form of acceleration or radius). A1: correct form of acceleration and correct radius |
| $\frac{mga}{d} \times \frac{\sqrt{a^2-d^2}}{a} = \frac{mv^2}{\sqrt{a^2-d^2}}$ | DM1 | Dependent on previous M. Eliminate $R$ and trig to form equation in $v$, $g$, $a$ and $d$ |
| $v = \sqrt{\frac{g(a^2-d^2)}{d}}$ | A1 | (5) Correct answer ISW |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{176ae8f8-7de9-4993-825a-6067614526ae-04_351_563_296_751}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A thin hemispherical shell, with centre $O$ and radius $a$, is fixed with its open end uppermost and horizontal.
A particle $P$ of mass $m$ moves in a horizontal circle on the smooth inner surface of the shell. The vertical distance of $P$ below the level of $O$ is $d$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m , g , d$ and $a$, the magnitude of the force exerted on $P$ by the inner surface of the hemisphere.
The particle moves with constant speed $v$.
\item Find $v$ in terms of $g , a$ and $d$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2024 Q2 [8]}}