| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: time to ground |
| Difficulty | Moderate -0.5 This is a straightforward SUVAT kinematics problem using a velocity-time graph. Students need to recognize that the initial velocity is read from the graph at t=0, and the height is found from the area under the graph (displacement). Both parts require only direct application of standard mechanics principles with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0 = u - 9.8 \times 2.5\) oe using gradient of graph. Allow \(g\) or 9.81 instead of 9.8 | M1 | Complete method using suvat or graph to produce an equation in \(u\) only, correct no. of terms, condone sign errors |
| \(u = 24.5\) or \(25\ (\text{m s}^{-1})\). Allow \(2.5g\) | A1 | cao (must be positive) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s = 24.5 \times 2 + \dfrac{1}{2} \times 9.8 \times 2^2\) OR \(s = 24.5 \times 7 - \dfrac{1}{2} \times 9.8 \times 7^2\) OR \(s = \dfrac{1}{2} \times 9.8 \times 4.5^2 - \left(24.5 \times 2.5 + \dfrac{1}{2} \times (-9.8) \times 2.5^2\right)\) OR \(s = \dfrac{1}{2} \times 9.8 \times 4.5^2 - \dfrac{1}{2} \times 9.8 \times 2.5^2\) | M1A1ft | Complete method to give final displacement, condone sign errors within a suvat equation. A1ft correct equation ft on their \(u\) |
| \(68.6\) or \(69\ \text{(m)}\) | A1 | cao |
# Question 6:
## Part 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0 = u - 9.8 \times 2.5$ oe using gradient of graph. Allow $g$ or 9.81 instead of 9.8 | M1 | Complete method using suvat or graph to produce an equation in $u$ only, correct no. of terms, condone sign errors |
| $u = 24.5$ or $25\ (\text{m s}^{-1})$. Allow $2.5g$ | A1 | cao (must be positive) |
## Part 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s = 24.5 \times 2 + \dfrac{1}{2} \times 9.8 \times 2^2$ OR $s = 24.5 \times 7 - \dfrac{1}{2} \times 9.8 \times 7^2$ OR $s = \dfrac{1}{2} \times 9.8 \times 4.5^2 - \left(24.5 \times 2.5 + \dfrac{1}{2} \times (-9.8) \times 2.5^2\right)$ OR $s = \dfrac{1}{2} \times 9.8 \times 4.5^2 - \dfrac{1}{2} \times 9.8 \times 2.5^2$ | M1A1ft | Complete method to give final displacement, condone sign errors within a suvat equation. A1ft correct equation ft on their $u$ |
| $68.6$ or $69\ \text{(m)}$ | A1 | cao |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{151d9232-5a78-4bc1-a57e-6c9cae80e473-22_428_993_251_479}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A small ball is thrown vertically upwards at time $t = 0$ from a point $A$ which is above horizontal ground. The ball hits the ground 7 s later.
The ball is modelled as a particle moving freely under gravity.\\
The velocity-time graph shown in Figure 3 represents the motion of the ball for $0 \leqslant t \leqslant 7$
\begin{enumerate}[label=(\alph*)]
\item Find the speed with which the ball is thrown.
\item Find the height of $A$ above the ground.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2021 Q6 [5]}}