| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical motion with resistance force |
| Difficulty | Standard +0.3 This is a standard M1 two-stage motion problem requiring SUVAT equations for free fall, then Newton's second law with resistance force. All steps are routine: (a) uses s=ut+½at², (b) requires finding acceleration with resistance then using v²=u²+2as, (c) adds times from both phases, (d) is a standard sketch. Slightly easier than average due to straightforward application of well-practiced techniques with no conceptual surprises. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| VJYV SIHI NI JIIYM ION OC | Vayv sthin NI JLIYM ION OA | VJYV SIHI NI JAIVM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s = \frac{1}{2} \times 9.8 \times 5^2\) | M1 | Complete method to find the distance |
| \(= 123\) or \(120\) (m) | A1 (2) | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v = 9.8 \times 5 = 49\) OR \(v = \sqrt{2 \times 9.8 \times 122.5} = 49\) OR \(122.5 = 5v - \frac{1}{2} \times 9.8 \times 5^2 \Rightarrow v = 49\) | B1 | B1: 49 or \(-49\); allow \(5g\) or \(-5g\) or \(49^2\) (2401) seen |
| \(250g - 3200 = \pm 250a\) | M1A1 | M1: Equation of motion, correct terms, condone sign errors |
| Correct value for \(a\) (3 or \(-3\)) | A1 | Correct equation (allow \(+\) or \(-\)) |
| \(v^2 = 49^2 - 2 \times 3 \times (520 - 122.5)\) | M1 A1ft | M1: Complete method to find speed at ground (must have found new \(a\)); M0 if \(u=0\) used |
| \(v = 4\) (m s\(^{-1}\)) | A1 (7) | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4 = 49 - 3t\) OR \((520-122.5) = \frac{(49+4)}{2}t\) OR \((520-122.5) = 49t - \frac{1}{2} \times 3t^2\) OR \((520-122.5) = 4t + \frac{1}{2} \times 3t^2\) | M1 | |
| \(t = 15\) (other root \(\frac{53}{3}\) leads to \(v < 0\)) | A1 | |
| Total time \(= 5 + 15 = 20\) (s) | A1ft (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Graph: straight line from 49 decreasing to 4, then horizontal line at 4 to \(t = 20\) | B1 shape | Correct shape |
| Values 49, 4, 20 marked correctly | B1ft figs (2) | ft on their figures |
# Question 5:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s = \frac{1}{2} \times 9.8 \times 5^2$ | M1 | Complete method to find the distance |
| $= 123$ or $120$ (m) | A1 (2) | cao |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v = 9.8 \times 5 = 49$ **OR** $v = \sqrt{2 \times 9.8 \times 122.5} = 49$ **OR** $122.5 = 5v - \frac{1}{2} \times 9.8 \times 5^2 \Rightarrow v = 49$ | B1 | B1: 49 or $-49$; allow $5g$ or $-5g$ or $49^2$ (2401) seen |
| $250g - 3200 = \pm 250a$ | M1A1 | M1: Equation of motion, correct terms, condone sign errors |
| Correct value for $a$ (3 or $-3$) | A1 | Correct equation (allow $+$ or $-$) |
| $v^2 = 49^2 - 2 \times 3 \times (520 - 122.5)$ | M1 A1ft | M1: Complete method to find speed at ground (must have found new $a$); M0 if $u=0$ used |
| $v = 4$ (m s$^{-1}$) | A1 (7) | cao |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4 = 49 - 3t$ **OR** $(520-122.5) = \frac{(49+4)}{2}t$ **OR** $(520-122.5) = 49t - \frac{1}{2} \times 3t^2$ **OR** $(520-122.5) = 4t + \frac{1}{2} \times 3t^2$ | M1 | |
| $t = 15$ (other root $\frac{53}{3}$ leads to $v < 0$) | A1 | |
| Total time $= 5 + 15 = 20$ (s) | A1ft (3) | |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Graph: straight line from 49 decreasing to 4, then horizontal line at 4 to $t = 20$ | B1 shape | Correct shape |
| Values 49, 4, 20 marked correctly | B1ft figs (2) | ft on their figures |\begin{enumerate}
\item A parachute is used to deliver a box of supplies. The parachute is attached to the box.
\end{enumerate}
\begin{itemize}
\item the parachute and box are dropped from rest from a helicopter that is hovering at a height of 520 m above the ground
\item the parachute and box fall vertically and freely under gravity for 5 seconds, then the parachute opens
\item from the instant the parachute opens, it provides a resistance to motion of magnitude 3200 N
\item the parachute and box continue to fall vertically downwards after the parachute opens
\item the parachute and box are modelled throughout the motion as a particle $P$ of mass 250 kg\\
(a) Find the distance fallen by $P$ in the first 5 seconds.\\
(b) Find the speed with which $P$ lands on the ground.\\
(c) Find the total time from the instant when $P$ is dropped from the helicopter to the instant when $P$ lands on the ground.\\
(d) Sketch a speed-time graph for the motion of $P$ from the instant when $P$ is dropped from the helicopter to the instant when $P$ lands on the ground.
\end{itemize}
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VJYV SIHI NI JIIYM ION OC & Vayv sthin NI JLIYM ION OA & VJYV SIHI NI JAIVM ION OC \\
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\hfill \mbox{\textit{Edexcel M1 2024 Q5 [14]}}