| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical motion: velocity-time graph |
| Difficulty | Standard +0.3 This is a straightforward SUVAT problem requiring standard kinematic equations applied in two phases. Parts (a)-(c) involve routine calculations using v²=u²+2as and s=vt. Part (d) requires basic physical reasoning about air resistance and instantaneous velocity changes. Slightly easier than average due to clear setup and standard methods. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(u = 0\), \(s = 2200 - 240 = 1960\), \(a = 9.8\) use \(v^2 = u^2 + 2as\) \(v^2 = 0 + 2(9.8)(1960)\) so \(v = 196 \text{ ms}^{-1}\) | M2, M1 A1 | |
| (b) \(s = ut + \frac{1}{2}at^2\) \(1960 = 0 + 4.9t^2 \Rightarrow t = 20 \text{ seconds}\) | M1, M1 A1 | |
| (c) \(140 - 20 = 120\) seconds to travel \(240 \text{ m}\) speed \(= 2 \text{ ms}^{-1}\) | M1, A1 | |
| (d) e.g. no air resistance; velocity on opening parachute will not immediately reduce e.g. if air resistance included, value in (a) would be much lower and consequently value in (b) much higher | B2, B2 | (13) |
| (a) $u = 0$, $s = 2200 - 240 = 1960$, $a = 9.8$ use $v^2 = u^2 + 2as$ $v^2 = 0 + 2(9.8)(1960)$ so $v = 196 \text{ ms}^{-1}$ | M2, M1 A1 | |
| (b) $s = ut + \frac{1}{2}at^2$ $1960 = 0 + 4.9t^2 \Rightarrow t = 20 \text{ seconds}$ | M1, M1 A1 | |
| (c) $140 - 20 = 120$ seconds to travel $240 \text{ m}$ speed $= 2 \text{ ms}^{-1}$ | M1, A1 | |
| (d) e.g. no air resistance; velocity on opening parachute will not immediately reduce e.g. if air resistance included, value in (a) would be much lower and consequently value in (b) much higher | B2, B2 | (13) |6. A student attempts to sketch the acceleration-time graph of a parachutist who jumps from a plane at a height of 2200 m above the ground.
The student assumes that the parachutist falls freely from rest under gravity until she is 240 m from the ground at which point she opens her parachute. The student makes the assumption that, at this point, the velocity of the parachutist is immediately reduced to a value which remains constant until she reaches the ground 140 seconds after she left the plane.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e0de1908-cf67-460f-9473-b2dfded95b33-4_314_1013_598_383}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
The student decides to ignore air resistance and his sketch is shown in Figure 3. The value $t _ { 1 }$ is used by the student to denote the time at which the parachute is opened.
Using the model proposed by the student, calculate
\begin{enumerate}[label=(\alph*)]
\item the speed of the parachutist immediately before she opens her parachute,
\item the value of $t _ { 1 }$,
\item the speed of the parachutist after the parachute is opened.
\item Comment on two features of the student's model which are unrealistic and say what effect taking account of these would have had on the values which you calculated in parts (a) and (b).\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q6 [13]}}