| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – surface contact (sphere/track, leaving surface) |
| Difficulty | Standard +0.8 This M3 vertical circular motion question requires energy conservation, radial force analysis to find the leaving point, and projectile motion. Part (a) is a standard 'show that' using energy. Parts (b-c) require setting normal reaction to zero, solving a transcendental equation (5cos θ = 3), then tracking projectile motion back to ground level—multiple sophisticated steps beyond typical A-level questions. |
| Spec | 6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}mv^2 - \frac{1}{2}m\left(\frac{ag}{5}\right) = mga(1-\cos\theta)\) | M1A1A1 | Energy equation from start to general position - must have 2 KE terms and a loss of PE; A1 LHS correct; A1 RHS correct |
| \(v^2 = 2ag + \frac{ag}{5} - 2ag\cos\theta = \frac{ag}{5}(11-10\cos\theta)\) | A1 (4) | A1cso re-arrange to given result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mg\cos\alpha\ (-R) = m\frac{v^2}{a}\) | M1A1 | NL2 along radius, acceleration in either form, \(R\) need not be shown, weight must be resolved; A1 fully correct equation with or w/o \(R\), accel now \(\frac{v^2}{a}\) |
| \(g\cos\alpha = \frac{g}{5}(11-10\cos\alpha)\) or sub \(\cos\alpha = \frac{v^2}{ag}\) in energy equation | M1A1 | M1 elimination of \(v^2\) or \(\cos\alpha\); A1 correct equation after elimination |
| \(\cos\alpha = \frac{11}{15}\) | ||
| \(P\) leaves sphere with speed \(\sqrt{\frac{ag}{5}\left(11-\frac{22}{3}\right)} = \sqrt{\frac{11ag}{15}}\) | DM1A1 (6) | DM1 substitute their \(\cos\alpha\) to obtain expression for \(v^2\), dep on both previous M marks; A1 correct expression for \(v\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horiz comp \(= \sqrt{\frac{11ag}{15}} \times \cos\alpha = \sqrt{\frac{11ag}{15}} \times \frac{11}{15}\) | M1 | M1 obtaining expression for horiz comp of speed at \(P\) |
| By cons of energy from top: \(2mag = \frac{1}{2}mV^2 - \frac{1}{2}m\frac{ag}{5}\) | M1 | M1 use energy to obtain speed when particle hits the floor |
| \(V^2 = \frac{21ag}{5}\) | A1 | A1 correct speed at floor |
| \(\cos\theta = \sqrt{\frac{11ag}{15}} \times \frac{11}{15} \times \sqrt{\frac{5}{21ag}} = \sqrt{\frac{11}{63}} \times \frac{11}{15} = 0.30642...\) | M1 | M1 use horizontal speed and resultant speed to find angle |
| \(\theta = 72.155...\) Accept \(72°\) or better | A1 (5) | A1 correct angle 2 sf or more figures (\(g\) cancels) |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 - \frac{1}{2}m\left(\frac{ag}{5}\right) = mga(1-\cos\theta)$ | M1A1A1 | Energy equation from start to general position - must have 2 KE terms and a loss of PE; A1 LHS correct; A1 RHS correct |
| $v^2 = 2ag + \frac{ag}{5} - 2ag\cos\theta = \frac{ag}{5}(11-10\cos\theta)$ | A1 (4) | A1cso re-arrange to given result |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mg\cos\alpha\ (-R) = m\frac{v^2}{a}$ | M1A1 | NL2 along radius, acceleration in either form, $R$ need not be shown, weight must be resolved; A1 fully correct equation with or w/o $R$, accel now $\frac{v^2}{a}$ |
| $g\cos\alpha = \frac{g}{5}(11-10\cos\alpha)$ or sub $\cos\alpha = \frac{v^2}{ag}$ in energy equation | M1A1 | M1 elimination of $v^2$ or $\cos\alpha$; A1 correct equation after elimination |
| $\cos\alpha = \frac{11}{15}$ | | |
| $P$ leaves sphere with speed $\sqrt{\frac{ag}{5}\left(11-\frac{22}{3}\right)} = \sqrt{\frac{11ag}{15}}$ | DM1A1 (6) | DM1 substitute their $\cos\alpha$ to obtain expression for $v^2$, dep on both previous M marks; A1 correct expression for $v$ |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Horiz comp $= \sqrt{\frac{11ag}{15}} \times \cos\alpha = \sqrt{\frac{11ag}{15}} \times \frac{11}{15}$ | M1 | M1 obtaining expression for horiz comp of speed at $P$ |
| By cons of energy from top: $2mag = \frac{1}{2}mV^2 - \frac{1}{2}m\frac{ag}{5}$ | M1 | M1 use energy to obtain speed when particle hits the floor |
| $V^2 = \frac{21ag}{5}$ | A1 | A1 correct speed at floor |
| $\cos\theta = \sqrt{\frac{11ag}{15}} \times \frac{11}{15} \times \sqrt{\frac{5}{21ag}} = \sqrt{\frac{11}{63}} \times \frac{11}{15} = 0.30642...$ | M1 | M1 use horizontal speed and resultant speed to find angle |
| $\theta = 72.155...$ Accept $72°$ or better | A1 (5) | A1 correct angle 2 sf or more figures ($g$ cancels) |
---6. A smooth sphere, with centre $O$ and radius $a$, is fixed with its lowest point $A$ on a horizontal floor. A particle $P$ is placed on the surface of the sphere at the point $B$, where $B$ is vertically above $A$. The particle is projected horizontally from $B$ with speed $\sqrt { \frac { a g } { 5 } }$ and moves along the surface of the sphere. When $O P$ makes an angle $\theta$ with the upward vertical, and $P$ is still in contact with the sphere, the speed of $P$ is $v$.
\begin{enumerate}[label=(\alph*)]
\item Show that $v ^ { 2 } = \frac { a g } { 5 } ( 11 - 10 \cos \theta )$.
The particle leaves the surface of the sphere at the point $C$.\\
Find
\item the speed of $P$ at $C$ in terms of $a$ and $g$,
\item the size of the angle between the floor and the direction of motion of $P$ at the instant immediately before $P$ hits the floor.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2015 Q6 [15]}}