| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2020 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle above horizontal |
| Difficulty | Standard +0.3 This is a standard projectiles question with straightforward application of SUVAT equations. Part (a) requires setting up horizontal and vertical equations with given angle and solving simultaneously—routine for A-level mechanics. Part (b) finds maximum height using standard methods. Parts (c) and (d) test conceptual understanding of modeling assumptions, which is basic. The 45° angle simplifies calculations, and all steps follow textbook procedures with no novel problem-solving required. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Using horizontal motion | M1 | Complete method to give equation in \(U\) and \(t\) only, condone sin/cos confusion and sign errors |
| \(U\cos 45° \cdot t = 100\) | A1 | Correct equation |
| Using vertical motion | M1 | Complete method to give equation in \(U\) and \(t\) only, condone sin/cos confusion and sign errors |
| \(U\sin 45°\cdot t - \frac{1}{2}gt^2 = -25\) | A1 | Correct equation (\(g\) does not need to be substituted) |
| Solve by eliminating \(t\) and solving for \(U\) | M1 | Must have earned previous two M marks; may solve for \(t\) first from \(100 - \frac{1}{2}gt^2 = -25\) then find \(U\) |
| \(U = 28\)* | A1* | Exact given answer correctly obtained with no wrong working (e.g. \(g = 9.81\) used) or approximation seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Using vertical motion | M1 | Complete method to give equation in \(h\) only (allow if \(U\) not substituted), condone sin/cos confusion and sign errors |
| \(0^2 = (28\sin 45°)^2 - 2gh\) | A1 | Correct equation (\(g\) does not need to be substituted; A0 if \(U\) used instead of 28) |
| Greatest height \(= 45\) m | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| New value \(> 28\) | B1 | Clear statement; penalise incorrect extras |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| e.g. wind effects, more accurate value of \(g\), spin of ball, include size of the ball, not model as a particle, shape of ball | B1 | Penalise incorrect extras (B0); ground being horizontal, cliff being vertical are not part of the model so B0; include weight/mass of the ball B0 |
## Question 5(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Using horizontal motion | M1 | Complete method to give equation in $U$ and $t$ only, condone sin/cos confusion and sign errors |
| $U\cos 45° \cdot t = 100$ | A1 | Correct equation |
| Using vertical motion | M1 | Complete method to give equation in $U$ and $t$ only, condone sin/cos confusion and sign errors |
| $U\sin 45°\cdot t - \frac{1}{2}gt^2 = -25$ | A1 | Correct equation ($g$ does not need to be substituted) |
| Solve by eliminating $t$ and solving for $U$ | M1 | Must have earned previous two M marks; may solve for $t$ first from $100 - \frac{1}{2}gt^2 = -25$ then find $U$ |
| $U = 28$* | A1* | Exact given answer correctly obtained with no wrong working (e.g. $g = 9.81$ used) or approximation seen |
---
## Question 5(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Using vertical motion | M1 | Complete method to give equation in $h$ only (allow if $U$ not substituted), condone sin/cos confusion and sign errors |
| $0^2 = (28\sin 45°)^2 - 2gh$ | A1 | Correct equation ($g$ does not need to be substituted; A0 if $U$ used instead of 28) |
| Greatest height $= 45$ m | A1 | cao |
---
## Question 5(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| New value $> 28$ | B1 | Clear statement; penalise incorrect extras |
---
## Question 5(d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| e.g. wind effects, more accurate value of $g$, spin of ball, include size of the ball, not model as a particle, shape of ball | B1 | Penalise incorrect extras (B0); ground being horizontal, cliff being vertical are not part of the model so B0; include weight/mass of the ball B0 |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d1989e18-1a4a-47e9-9f12-3beb8985ed87-16_532_1002_237_533}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A small ball is projected with speed $U \mathrm {~ms} ^ { - 1 }$ from a point $O$ at the top of a vertical cliff. The point $O$ is 25 m vertically above the point $N$ which is on horizontal ground.
The ball is projected at an angle of $45 ^ { \circ }$ above the horizontal.\\
The ball hits the ground at a point $A$, where $A N = 100 \mathrm {~m}$, as shown in Figure 2 .\\
The motion of the ball is modelled as that of a particle moving freely under gravity.\\
Using this initial model,
\begin{enumerate}[label=(\alph*)]
\item show that $U = 28$
\item find the greatest height of the ball above the horizontal ground $N A$.
In a refinement to the model of the motion of the ball from $O$ to $A$, the effect of air resistance is included.
This refined model is used to find a new value of $U$.
\item How would this new value of $U$ compare with 28, the value given in part (a)?
\item State one further refinement to the model that would make the model more realistic.
\section*{" " $_ { \text {" } } ^ { \text {" } }$ " "}
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 2020 Q5 [11]}}