| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Ball bouncing on horizontal surface |
| Difficulty | Challenging +1.8 This is a challenging M3 mechanics problem requiring multiple stages: projectile motion with a constraint (string), collision with restitution, energy considerations to find the minimum coefficient, then a second projectile phase after string breaks. It demands careful geometric reasoning about the trajectory, application of conservation principles, and integration of several mechanics concepts. The multi-stage nature and the need to work backwards from the condition that P passes through A after rebounding makes this significantly harder than standard projectile or collision questions, though the individual techniques are within M3 scope. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\dfrac{1}{2}mv^2 - \dfrac{1}{2}m \times 10ag = mag\) | M1A1 | Energy equation from \(A\) to \(B\). Must have 2 KE terms and 1 PE term. Fully correct equation |
| \(v^2 = 12ag\) | ||
| After impact: \(V = e\sqrt{12ag}\) | B1 | Correct speed immediately after impact |
| Energy to top: \(\dfrac{1}{2}me^2(12ag) - \dfrac{1}{2}mW^2 = mag\) | M1 | Energy equation from \(B\) to \(A\), with speed after impact at \(B\) |
| At top: \(T + mg = m\dfrac{W^2}{a}\) | M1A1 | Equation of motion along radius at \(A\). Correct equation at \(A\) |
| \(T \geq 0 \Rightarrow W^2 \geq ag\) | dM1 | Use tension at \(A \geq 0\) to obtain inequality for \((\text{speed at } A)^2\) |
| \(\dfrac{1}{2}me^2(12ag) - mag\left(= \dfrac{1}{2}mW^2\right) \geq \dfrac{1}{2}mag\) | dM1 | Use energy equation to obtain \(e^2 \geq \ldots\) Depends on second and third M marks |
| \(e^2 \geq \dfrac{1}{4}\), \(\therefore e \geq \dfrac{1}{2}\) | A1cso (9) | Reach the given answer with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Energy to string breaking: \(\dfrac{1}{2}m\dfrac{3}{4}(12ag) - \dfrac{1}{2}mX^2 = mag\cos 30°\) | M1 | Energy equation from \(B\) to \(C\) or from \(A\) to \(C\) (\(\frac{3}{4}\) or \(e^2\) used) |
| \(X^2 = 9ag - 2ag\cos 30°\) | A1 | Correct expression for \((\text{speed at } C)^2\). Must have \(3/4\) now |
| Horiz speed \(= X\cos 30° = \cos 30°\sqrt{ag(9 - 2\cos 30°)}\) | M1 | Attempt horizontal component of speed at \(C\) with their speed at \(C\) |
| Speed at \(D\) = speed at \(B\) (after rebounding) \(= \dfrac{\sqrt{3}}{2}\sqrt{12ag}\ (= 3\sqrt{ag})\) | M1 | Obtain speed at \(D\) or vertical speed at \(D\) by finding vertical speed at \(C\) and using SUVAT. NB: This is an A mark on e-PEN |
| \(\cos\theta = \dfrac{\cos 30°\sqrt{ag(9-2\cos 30°)}}{3\sqrt{ag}}\) OR \(\tan\theta = \dfrac{\text{vert speed}}{\text{horiz speed}}\) | M1 | Use horizontal speed of \(P\) with (resultant) speed or vertical speed of \(P\) at \(D\) to form expression for cos or tan of required angle |
| \(\theta = \cos^{-1}\!\left(\dfrac{\cos 30°\sqrt{(9-2\cos 30°)}}{3}\right) = 38.89\ldots°\) | A1 (6) | Correct size of angle in degrees or radians. 2 sf minimum (\(g\) cancels) |
# Question 7:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\dfrac{1}{2}mv^2 - \dfrac{1}{2}m \times 10ag = mag$ | M1A1 | Energy equation from $A$ to $B$. Must have 2 KE terms and 1 PE term. Fully correct equation |
| $v^2 = 12ag$ | | |
| After impact: $V = e\sqrt{12ag}$ | B1 | Correct speed immediately after impact |
| Energy to top: $\dfrac{1}{2}me^2(12ag) - \dfrac{1}{2}mW^2 = mag$ | M1 | Energy equation from $B$ to $A$, with speed after impact at $B$ |
| At top: $T + mg = m\dfrac{W^2}{a}$ | M1A1 | Equation of motion along radius at $A$. Correct equation at $A$ |
| $T \geq 0 \Rightarrow W^2 \geq ag$ | dM1 | Use tension at $A \geq 0$ to obtain inequality for $(\text{speed at } A)^2$ |
| $\dfrac{1}{2}me^2(12ag) - mag\left(= \dfrac{1}{2}mW^2\right) \geq \dfrac{1}{2}mag$ | dM1 | Use energy equation to obtain $e^2 \geq \ldots$ Depends on second and third M marks |
| $e^2 \geq \dfrac{1}{4}$, $\therefore e \geq \dfrac{1}{2}$ | A1cso (9) | Reach the given answer with no errors seen |
**ALT for last 3 marks:** $m\dfrac{W^2}{a} - mg \geq 0$
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| Energy to string breaking: $\dfrac{1}{2}m\dfrac{3}{4}(12ag) - \dfrac{1}{2}mX^2 = mag\cos 30°$ | M1 | Energy equation from $B$ to $C$ or from $A$ to $C$ ($\frac{3}{4}$ or $e^2$ used) |
| $X^2 = 9ag - 2ag\cos 30°$ | A1 | Correct expression for $(\text{speed at } C)^2$. Must have $3/4$ now |
| Horiz speed $= X\cos 30° = \cos 30°\sqrt{ag(9 - 2\cos 30°)}$ | M1 | Attempt horizontal component of speed at $C$ with their speed at $C$ |
| Speed at $D$ = speed at $B$ (after rebounding) $= \dfrac{\sqrt{3}}{2}\sqrt{12ag}\ (= 3\sqrt{ag})$ | M1 | Obtain speed at $D$ or vertical speed at $D$ by finding vertical speed at $C$ and using SUVAT. **NB: This is an A mark on e-PEN** |
| $\cos\theta = \dfrac{\cos 30°\sqrt{ag(9-2\cos 30°)}}{3\sqrt{ag}}$ OR $\tan\theta = \dfrac{\text{vert speed}}{\text{horiz speed}}$ | M1 | Use horizontal speed of $P$ with (resultant) speed or vertical speed of $P$ at $D$ to form expression for cos or tan of required angle |
| $\theta = \cos^{-1}\!\left(\dfrac{\cos 30°\sqrt{(9-2\cos 30°)}}{3}\right) = 38.89\ldots°$ | A1 (6) | Correct size of angle in degrees or radians. 2 sf minimum ($g$ cancels) |
Accept $39°$, $38.9°$ or better; or $0.68$, $0.679$ radians or better.7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a67e3644-13fa-4196-a2ef-ea1e26f5726c-20_442_967_283_486}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A light inextensible string of length $a$ has one end attached to a fixed point $O$ on a horizontal plane. A particle $P$ is attached to the other end of the string. The particle is held at the point $A$, where $A$ is vertically above $O$ and $O A = a$. The particle is then projected horizontally with speed $\sqrt { 10 a g }$, as shown in Figure 2. The particle strikes the plane at the point $B$. After rebounding from the plane, $P$ passes through $A$. The coefficient of restitution between the plane and $P$ is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that $e \geqslant \frac { 1 } { 2 }$
The point $C$ is above the horizontal plane such that $O C = a$ and angle $C O B = 120 ^ { \circ }$
As the particle reaches $C$, the string breaks. The particle now moves freely under gravity and strikes the plane at the point $D$.\\
Given that $e = \frac { \sqrt { 3 } } { 2 }$
\item find the size of the angle between the horizontal and the direction of motion of $P$ at $D$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2017 Q7 [15]}}