| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: time to ground |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT question requiring standard application of kinematic equations with constant acceleration. All parts follow routine procedures: (a) solve quadratic equation for time, (b) use v² = u² + 2as, (c) find max height then sum distances, (d) sketch linear v-t graph. No problem-solving insight needed, just careful arithmetic and sign conventions. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Complete method to find total time, e.g. \(-19.6 = 14.7t + \frac{1}{2}(-9.8)t^2\) using one equation | M1 | Complete method required |
| OR: \(0 = 14.7 - 9.8t_1 \Rightarrow t_1 = 1.5\) | M1 | Using four equations method |
| \(s_1 = 14.7 \times 1.5 - \frac{1}{2} \times 9.8 \times 1.5^2 = 11.025\) | ||
| \(30.625 = \frac{1}{2} \times 9.8 \times t_2^2 \Rightarrow t_2 = 2.5\) | ||
| \(t = t_1 + t_2 = 4\) s | ||
| There are two A marks for all equations used, \(-1\) each error | A1, M(A)1 | Many other methods allowed |
| \(t = 4\) s only | A1 | If using quadratic, ignore other solution even if incorrect. If they combine 2 solutions, A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v^2 = 14.7^2 + 2(-9.8)(-19.6)\) OR \(v = 14.7 + (-9.8) \times 4\) | M1 A1 | Complete method to find speed |
| Speed \(= 24.5\) or \(25\) \((\text{m s}^{-1})\) | A1 | Answer must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| e.g. \(0^2 = 14.7^2 + 2(-9.8)s\) or \(24.5^2 = 2 \times 9.8s\) | M1 | Method to find a relevant distance |
| \(s = 11.025\) (11 or better) \(\quad s = 30.625\) | A1 | A correct relevant distance |
| Total distance \(= 2 \times 11.025 + 19.6\) \(\quad\) Total distance \(= 2 \times 30.625 - 19.6\) | M1 | Method to find total distance |
| \(= 41.7\) (3 sf) or \(42\) (2 sf) m | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Straight line crossing \(t\)-axis | B1 | Line may be reflected in \(t\)-axis |
| Start point \((0, 14.7)\) or on axes | B1 | Correct appropriate coordinates for start point |
| End point \((4, -24.5)\) or on axes | B1ft | ft on answers to (a) and (b) |
## Question 2:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Complete method to find total time, e.g. $-19.6 = 14.7t + \frac{1}{2}(-9.8)t^2$ using one equation | M1 | Complete method required |
| OR: $0 = 14.7 - 9.8t_1 \Rightarrow t_1 = 1.5$ | M1 | Using four equations method |
| $s_1 = 14.7 \times 1.5 - \frac{1}{2} \times 9.8 \times 1.5^2 = 11.025$ | | |
| $30.625 = \frac{1}{2} \times 9.8 \times t_2^2 \Rightarrow t_2 = 2.5$ | | |
| $t = t_1 + t_2 = 4$ s | | |
| There are two A marks for all equations used, $-1$ each error | A1, M(A)1 | Many other methods allowed |
| $t = 4$ s only | A1 | If using quadratic, ignore other solution even if incorrect. If they combine 2 solutions, A0 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v^2 = 14.7^2 + 2(-9.8)(-19.6)$ **OR** $v = 14.7 + (-9.8) \times 4$ | M1 A1 | Complete method to find speed |
| Speed $= 24.5$ or $25$ $(\text{m s}^{-1})$ | A1 | Answer must be positive |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. $0^2 = 14.7^2 + 2(-9.8)s$ or $24.5^2 = 2 \times 9.8s$ | M1 | Method to find a relevant distance |
| $s = 11.025$ (11 or better) $\quad s = 30.625$ | A1 | A correct relevant distance |
| Total distance $= 2 \times 11.025 + 19.6$ $\quad$ Total distance $= 2 \times 30.625 - 19.6$ | M1 | Method to find total distance |
| $= 41.7$ (3 sf) or $42$ (2 sf) m | A1 | Correct answer |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Straight line crossing $t$-axis | B1 | Line may be reflected in $t$-axis |
| Start point $(0, 14.7)$ or on axes | B1 | Correct appropriate coordinates for start point |
| End point $(4, -24.5)$ or on axes | B1ft | ft on answers to (a) and (b) |
---2. A small ball is thrown vertically upwards with speed $14.7 \mathrm {~ms} ^ { - 1 }$ from a point that is 19.6 m above horizontal ground. The ball is modelled as a particle moving freely under gravity.
Find
\begin{enumerate}[label=(\alph*)]
\item the total time from when the ball is thrown to when it first hits the ground,
\item the speed of the ball immediately before it first hits the ground,
\item the total distance travelled by the ball from when it is thrown to when it first hits the ground.
\item Sketch a velocity-time graph for the motion of the ball from when it is thrown to when it first hits the ground.
State the coordinates of the start point and the coordinates of the end point of your graph.\\
DO NOT WRITEIN THIS AREA
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2020 Q2 [14]}}