| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Speed at specific time or position |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question requiring resolution of velocity components and use of kinematic equations. Part (a) uses the condition at point A to find the launch angle, part (b) applies kinetic energy at maximum height (straightforward since vertical component is zero), and part (c) requires finding when speed equals 6 m/s using the velocity-time relationship. All techniques are routine for M2 students with no novel problem-solving 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 concepts6.02e Calculate KE and PE: using formulae |
7.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{e6d100ff-dd4a-4591-a0a3-81761773045e-13_552_1296_255_317}
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\caption{Figure 4}
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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 $T$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2016 Q7 [14]}}