| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical motion: velocity-time graph |
| Difficulty | Moderate -0.3 This is a standard M1 SUVAT question with multiple parts requiring systematic application of kinematic equations across three distinct phases of motion. While it involves several steps and careful bookkeeping of the different motion phases, each individual calculation is routine (free fall, constant deceleration, constant velocity). The question guides students through parts (a)-(d) sequentially, and the techniques required are core M1 content with no novel problem-solving insight needed. Slightly easier than average due to its structured, methodical nature. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks |
|---|---|
| \(v = 2.5\times9.8 = 24.5\) (m s\(^{-1}\)) — Allow \(2.5g\) | B1 (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Velocity-time graph with correct shape (second line less steep than first); peak at 24.5; time values at 2.5, \(2.5+T\), and 20; horizontal velocity \(V\) shown | B1 shape, B1 figures (2) | B0 if solid vertical line at \(t=20\) |
| Answer | Marks | Guidance |
|---|---|---|
| OR \(V^2 = 24.5^2+2\times(-3.9)\times73.75\) and then \(5=24.5-3.9T\) | M1, A1A1M1 | Complete method to obtain equation in \(T\) only; A1A1M1 for correct equations, \(-1\) each error |
| \(T=5\) | A1 (5) | Must be single answer; other root (7.56) must be clearly rejected |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \left(\dfrac{1}{2}\times24.5\times2.5\right)+73.75+(20-2.5-5)\times(24.5-3.9\times5)\) | M1A2 | Complete method using triangle + trapezium + rectangle; \(-1\) each error |
| \(= 167\) (m) nearest metre | A1 (4) |
## Question 7:
### Part 7(a):
$v = 2.5\times9.8 = 24.5$ (m s$^{-1}$) — Allow $2.5g$ | B1 (1) |
### Part 7(b):
Velocity-time graph with correct shape (second line less steep than first); peak at 24.5; time values at 2.5, $2.5+T$, and 20; horizontal velocity $V$ shown | B1 shape, B1 figures (2) | B0 if solid vertical line at $t=20$
### Part 7(c):
$73.75 = \dfrac{(24.5+(24.5-3.9T))T}{2}$
**OR** $73.75 = 24.5T - \dfrac{1}{2}\times3.9T^2$
**OR** $73.75 = (24.5-3.9T)T + \dfrac{1}{2}\times3.9T\times T$
**OR** $V^2 = 24.5^2+2\times(-3.9)\times73.75$ and then $5=24.5-3.9T$ | M1, A1A1M1 | Complete method to obtain equation in $T$ only; A1A1M1 for correct equations, $-1$ each error
$T=5$ | A1 (5) | Must be single answer; other root (7.56) must be clearly rejected
### Part 7(d):
Height = Total area under graph
$= \left(\dfrac{1}{2}\times24.5\times2.5\right)+73.75+(20-2.5-5)\times(24.5-3.9\times5)$ | M1A2 | Complete method using triangle + trapezium + rectangle; $-1$ each error
$= 167$ (m) nearest metre | A1 (4) |
---7. A helicopter is hovering at rest above horizontal ground at the point $H$. A parachutist steps out of the helicopter and immediately falls vertically and freely under gravity from rest for 2.5 s . His parachute then opens and causes him to immediately decelerate at a constant rate of $3.9 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ for $T$ seconds ( $T < 6$ ), until his speed is reduced to $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. He then moves with this constant speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ until he hits the ground. While he is decelerating, he falls a distance of 73.75 m . The total time between the instant when he leaves $H$ and the instant when he hits the ground is 20 s .
The parachutist is modelled as a particle.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of the parachutist at the instant when his parachute opens.
\item Sketch a speed-time graph for the motion of the parachutist from the instant when he leaves $H$ to the instant when he hits the ground.
\item Find the value of $T$.
\item Find, to the nearest metre, the height of the point $H$ above the ground.\\
7. A helicopter is hovering at rest above horizontal ground at the point $H$. A parachutist steps
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2021 Q7 [12]}}