Edexcel M2 2005 January — Question 7 15 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2005
SessionJanuary
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeProjection from elevated point - angle above horizontal
DifficultyStandard +0.3 This is a standard M2 projectiles question requiring systematic application of SUVAT equations with vertical projection from height. While it has multiple parts (4 marks worth), each part follows directly from standard techniques: resolving initial velocity, using s=ut+½at² for time of flight, horizontal distance calculation, and finding velocity components at a given height. The sin α = 3/5 setup makes calculations cleaner. Slightly above average difficulty due to the multi-part nature and need for careful bookkeeping, but no novel problem-solving required.
Spec3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model

7. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 4} \includegraphics[alt={},max width=\textwidth]{a9e00b5b-3804-4f8d-9cc8-7d1170027726-6_568_1582_360_239}
\end{figure} A particle \(P\) is projected from a point \(A\) with speed \(32 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of elevation \(\alpha\), where \(\sin \alpha = \frac { 3 } { 5 }\). The point \(O\) is on horizontal ground, with \(O\) vertically below \(A\) and \(O A = 20 \mathrm {~m}\). The particle \(P\) moves freely under gravity and passes through a point \(B\), which is 16 m above ground, before reaching the ground at the point \(C\), as shown in Figure 4. Calculate
  1. the time of the flight from \(A\) to \(C\),
  2. the distance \(O C\),
  3. the speed of \(P\) at \(B\),
  4. the angle that the velocity of \(P\) at \(B\) makes with the horizontal.

Question 7:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\uparrow\ u_y = 32 \times \frac{3}{5}\) \((= 19.2)\)B1
\(-20 = 19.2t - 4.9t^2\)M1 A2(1,0) \(-1\) each error
\(t \approx 4.8\) or \(4.77\) (s)A1
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\rightarrow\ u_x = 32 \times \frac{4}{5}\) \((= 25.6)\)B1
\(d = 25.6 \times 4.77...\)M1
\(\approx 120\) or \(122\) (m)A1
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\uparrow\ v_y^2 = 19.2^2 + 2 \times 9.8 \times 4\) \([v_y^2 = 447.04,\ v_y \approx 21.14]\)M1
\(V^2 = 447.04 + 25.6^2\)M1 A1
\(V = 33\) or \(33.2 \text{ ms}^{-1}\)A1
Alternative for (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{1}{2}m(V^2 - 32^2) = mg \times 4\)M1 A1
\(V^2 = 1102.4\)M1
\(V = 33\) or \(33.2 \text{ ms}^{-1}\)A1
Part (d):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\tan\theta = \frac{21.14}{25.6}\) \(\left(\text{or } \cos\theta = \frac{25.6}{33.2}, \ldots\right)\)M1 A1ft ft their components or resultant
\(\theta \approx 40°\) or \(39.6°\)A1
## Question 7:

### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\uparrow\ u_y = 32 \times \frac{3}{5}$ $(= 19.2)$ | B1 | |
| $-20 = 19.2t - 4.9t^2$ | M1 A2(1,0) | $-1$ each error |
| $t \approx 4.8$ or $4.77$ (s) | A1 | |

### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\rightarrow\ u_x = 32 \times \frac{4}{5}$ $(= 25.6)$ | B1 | |
| $d = 25.6 \times 4.77...$ | M1 | |
| $\approx 120$ or $122$ (m) | A1 | |

### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\uparrow\ v_y^2 = 19.2^2 + 2 \times 9.8 \times 4$ $[v_y^2 = 447.04,\ v_y \approx 21.14]$ | M1 | |
| $V^2 = 447.04 + 25.6^2$ | M1 A1 | |
| $V = 33$ or $33.2 \text{ ms}^{-1}$ | A1 | |

### Alternative for (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}m(V^2 - 32^2) = mg \times 4$ | M1 A1 | |
| $V^2 = 1102.4$ | M1 | |
| $V = 33$ or $33.2 \text{ ms}^{-1}$ | A1 | |

### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\theta = \frac{21.14}{25.6}$ $\left(\text{or } \cos\theta = \frac{25.6}{33.2}, \ldots\right)$ | M1 A1ft | ft their components or resultant |
| $\theta \approx 40°$ or $39.6°$ | A1 | |
7.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 4}
  \includegraphics[alt={},max width=\textwidth]{a9e00b5b-3804-4f8d-9cc8-7d1170027726-6_568_1582_360_239}
\end{center}
\end{figure}

A particle $P$ is projected from a point $A$ with speed $32 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of elevation $\alpha$, where $\sin \alpha = \frac { 3 } { 5 }$. The point $O$ is on horizontal ground, with $O$ vertically below $A$ and $O A = 20 \mathrm {~m}$. The particle $P$ moves freely under gravity and passes through a point $B$, which is 16 m above ground, before reaching the ground at the point $C$, as shown in Figure 4.

Calculate
\begin{enumerate}[label=(\alph*)]
\item the time of the flight from $A$ to $C$,
\item the distance $O C$,
\item the speed of $P$ at $B$,
\item the angle that the velocity of $P$ at $B$ makes with the horizontal.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2005 Q7 [15]}}