| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Projectile energy - finding speed or height |
| Difficulty | Standard +0.3 This is a straightforward M2 question combining energy conservation with projectile motion. Part (a) uses basic energy conservation (a standard technique), part (b) requires resolving vertical motion with SUVAT, and part (c) asks for minimum speed (horizontal component). All steps are routine applications of well-practiced methods with no novel insight required, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | 2 | 4 |
Question 6:
6 | 2 | 46.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cf960066-46b8-42a3-8a8b-d8deb76e7c70-11_694_1004_264_529}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A ball is projected from a point $A$ which is 8 m above horizontal ground as shown in Figure 4. The ball is projected with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\theta ^ { \circ }$ above the horizontal. The ball moves freely under gravity and hits the ground at the point $B$. The speed of the ball immediately before it hits the ground is $2 u \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item By considering energy, find the value of $u$.
The time taken for the ball to move from $A$ to $B$ is 2 seconds. Find
\item the value of $\theta$,
\item the minimum speed of the ball on its path from $A$ to $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q6 [11]}}