| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Speed at specific time or position |
| Difficulty | Standard +0.8 This is a multi-part M2 projectile question requiring systematic application of kinematic equations with given conditions at a specific time. Part (a) requires resolving velocities at point A and working backwards to find initial angle; part (b) involves energy calculations at maximum height; part (c) requires finding when speed equals 6 m/s both ascending and descending. While conceptually straightforward for M2 students, the multi-step nature, need to handle both velocity components carefully, and the less routine part (c) about the time interval elevate this above average difficulty. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| VIIV STHI NI JINM ION OC | VIAV SIHI NI JMAM/ION OC | VIAV SIHI NI JIIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Equate horizontal components of speeds: \(u\cos\theta° = 6\cos 45°\left(=3\sqrt{2}\right)\) \((4.24....)\) | M1 | |
| \(u\cos\theta° = 6\cos 45°\left(=3\sqrt{2}\right)\) \((4.24....)\) | A1 | Correct unsimplified |
| Use suvat for vertical speeds: \(u\sin\theta° - 2g = -6\sin 45°\) | M1 | Condone sign errors |
| \(\left(u\sin\theta = 2g - 3\sqrt{2}\right)\) | A1 | Correct unsimplified |
| Divide to find \(\tan\theta\): \(\tan\theta = \dfrac{2g - 6\sin 45}{6\cos 45}\) | DM1 | Dependent on previous 2 Ms. Follow their components. |
| \(\left(= \dfrac{2g-3\sqrt{2}}{3\sqrt{2}} = 3.61..\right) \Rightarrow \theta = 74.6\ (75)\) | A1 | \((u = 15.93....)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| At max height, speed \(= u\cos\theta\left(=3\sqrt{2}\ \text{m s}^{-1}\right)\) | B1 | |
| \(\text{KE} = \dfrac{1}{2} \times 0.7 \times \left(3\sqrt{2}\right)^2\ \text{(J)}\) | M1 | Correct for their \(v\) at the top, \(v \neq 0\) |
| \(= 6.3\ \text{(J)}\) | A1 | accept awrt 6.30. CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| When P is moving upwards at \(6\ \text{m s}^{-1}\) | M1 | Use suvat to find first time \(v=6\) |
| \(u\sin\theta - gt = 3\sqrt{2}\) | A1 | |
| \(2g - 3\sqrt{2} - gt = 3\sqrt{2}\) | M1 | Solve for \(t\) |
| \(t = \dfrac{2g - 6\sqrt{2}}{g} = 1.13....\) | A1 | Sensitive to premature approximation. Allow 1.14. |
| \(T = 2 - 1.13 = 0.87\) | A1 | CAO accept awrt 0.87 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(6\sin 45 = 0 + gt\) | M1A1 | find time from top to \(A\) |
| \(T = 2t = \dfrac{\dfrac{12\sqrt{2}}{2}}{g} = 0.87\) | M1, A1, A1 | Correct strategy; Correct unsimplified |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(u\sin\theta = gt\) (their \(u, \theta\)) | M1 | Time to top |
| \(t = 1.567....\) | A1 | |
| \(T = 2(2 - 1.567...)\) | M1A1 | |
| \(= 0.87\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Vertical speed at \(A = -\)(vertical speed at \(B\)) \(= \sqrt{36-\left(3\sqrt{2}\right)^2} = 3\sqrt{2}\) | M1, A1 | Or use the \(45°\) angle |
| Use \(v = u + at\) for \(A \rightarrow B\); \(-3\sqrt{2} = 3\sqrt{2} - gT\) | M1, A1 | Correct use for their values |
| \(T = 0.87\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(v^2 = \left(3\sqrt{2}\right)^2 + \left(u\sin\theta - gt\right)^2 \leq 36\) | M1 | Form expression for \(v^2\). Inequality not needed at this stage |
| A1 | Correct inequality for \(v^2\) | |
| \(-\sqrt{18} \leq u\sin\theta - gt \leq \sqrt{18}\) | M1 | |
| \(\dfrac{u\sin\theta - \sqrt{18}}{g} \leq t \leq \dfrac{u\sin\theta + \sqrt{18}}{g}\) | A1 | |
| \(T = \dfrac{u\sin\theta + \sqrt{18}}{g} - \dfrac{u\sin\theta - \sqrt{18}}{g} = \dfrac{2\sqrt{18}}{g} = 0.866\) | A1 |
## Question 7(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Equate horizontal components of speeds: $u\cos\theta° = 6\cos 45°\left(=3\sqrt{2}\right)$ $(4.24....)$ | M1 | |
| $u\cos\theta° = 6\cos 45°\left(=3\sqrt{2}\right)$ $(4.24....)$ | A1 | Correct unsimplified |
| Use suvat for vertical speeds: $u\sin\theta° - 2g = -6\sin 45°$ | M1 | Condone sign errors |
| $\left(u\sin\theta = 2g - 3\sqrt{2}\right)$ | A1 | Correct unsimplified |
| Divide to find $\tan\theta$: $\tan\theta = \dfrac{2g - 6\sin 45}{6\cos 45}$ | DM1 | Dependent on previous 2 Ms. Follow their components. |
| $\left(= \dfrac{2g-3\sqrt{2}}{3\sqrt{2}} = 3.61..\right) \Rightarrow \theta = 74.6\ (75)$ | A1 | $(u = 15.93....)$ |
**Total: (6)**
---
## Question 7(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| At max height, speed $= u\cos\theta\left(=3\sqrt{2}\ \text{m s}^{-1}\right)$ | B1 | |
| $\text{KE} = \dfrac{1}{2} \times 0.7 \times \left(3\sqrt{2}\right)^2\ \text{(J)}$ | M1 | Correct for their $v$ at the top, $v \neq 0$ |
| $= 6.3\ \text{(J)}$ | A1 | accept awrt 6.30. CSO |
**Total: (3)**
---
## Question 7(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| When P is moving upwards at $6\ \text{m s}^{-1}$ | M1 | Use suvat to find first time $v=6$ |
| $u\sin\theta - gt = 3\sqrt{2}$ | A1 | |
| $2g - 3\sqrt{2} - gt = 3\sqrt{2}$ | M1 | Solve for $t$ |
| $t = \dfrac{2g - 6\sqrt{2}}{g} = 1.13....$ | A1 | Sensitive to premature approximation. Allow 1.14. |
| $T = 2 - 1.13 = 0.87$ | A1 | CAO accept awrt 0.87 |
**Total: (5)**
**Alternative:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $6\sin 45 = 0 + gt$ | M1A1 | find time from top to $A$ |
| $T = 2t = \dfrac{\dfrac{12\sqrt{2}}{2}}{g} = 0.87$ | M1, A1, A1 | Correct strategy; Correct unsimplified |
**Total: (5)**
**Alternative (continued):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u\sin\theta = gt$ (their $u, \theta$) | M1 | Time to top |
| $t = 1.567....$ | A1 | |
| $T = 2(2 - 1.567...)$ | M1A1 | |
| $= 0.87$ | A1 | |
**Alternative (vertical speed at $A$):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Vertical speed at $A = -$(vertical speed at $B$) $= \sqrt{36-\left(3\sqrt{2}\right)^2} = 3\sqrt{2}$ | M1, A1 | Or use the $45°$ angle |
| Use $v = u + at$ for $A \rightarrow B$; $-3\sqrt{2} = 3\sqrt{2} - gT$ | M1, A1 | Correct use for their values |
| $T = 0.87$ | A1 | |
**Alternative 7d:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $v^2 = \left(3\sqrt{2}\right)^2 + \left(u\sin\theta - gt\right)^2 \leq 36$ | M1 | Form expression for $v^2$. Inequality not needed at this stage |
| | A1 | Correct inequality for $v^2$ |
| $-\sqrt{18} \leq u\sin\theta - gt \leq \sqrt{18}$ | M1 | |
| $\dfrac{u\sin\theta - \sqrt{18}}{g} \leq t \leq \dfrac{u\sin\theta + \sqrt{18}}{g}$ | A1 | |
| $T = \dfrac{u\sin\theta + \sqrt{18}}{g} - \dfrac{u\sin\theta - \sqrt{18}}{g} = \dfrac{2\sqrt{18}}{g} = 0.866$ | A1 | |
**Total: (5)**
**Overall Total: (14 marks)**7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f30ed5b8-880e-42de-860e-d1538fa68f11-24_549_1284_258_322}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
At time $t = 0$, a particle $P$ of mass 0.7 kg is projected with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a fixed point $O$ at an angle $\theta ^ { \circ }$ to the horizontal. The particle moves freely under gravity. At time $t = 2$ seconds, $P$ passes through the point $A$ with speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is moving downwards at $45 ^ { \circ }$ to the horizontal, as shown in Figure 4.
Find
\begin{enumerate}[label=(\alph*)]
\item the value of $\theta$,
\item the kinetic energy of $P$ as it reaches the highest point of its path.
For an interval of $T$ seconds, the speed, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, of $P$ is such that $v \leqslant 6$
\item Find the value of $T$.
\begin{center}
\begin{tabular}{|l|l|l|}
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VIIV STHI NI JINM ION OC & VIAV SIHI NI JMAM/ION OC & VIAV SIHI NI JIIYM ION OO \\
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\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q7 [14]}}