| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Vertical motion under gravity |
| Difficulty | Moderate -0.3 This is a straightforward two-stage kinematics problem requiring standard SUVAT equations and Newton's second law. Part (i) uses v²=u²+2as with free fall, part (ii) applies F=ma to find deceleration in water then uses v=u+at, and part (iii) requires sketching two linear segments. While it involves multiple steps across 9 marks, each step follows routine mechanics procedures without requiring problem-solving insight or novel approaches—slightly easier than average due to its predictable structure. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02h Motion under gravity: vector form3.03c Newton's second law: F=ma one dimension |
| Answer | Marks |
|---|---|
| 6(i) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | KE gain=PE lost |
| v=6ms–1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| v2=02+2 × g × 1.8 | M1 | Use constant acceleration equation(s) with a=g to find v |
| v=6ms–1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(ii) | 0.4g–5.6 = 0.4a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| a=–4ms–2 | A1 | |
| 0=6–4t | M1 | Use of constant acceleration equation(s) such as v=u+at to |
| Answer | Marks |
|---|---|
| t=1.5s | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(iii) | Straight line starting at (0,0) with positive gradient | B1 |
| Answer | Marks |
|---|---|
| gradient and ending with v = 0 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| i.e. (0.6, 6) and finishing at (2.1, 0) | B1FT | FT on their v from (i) and/or their t from (ii) |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
--- 6(i) ---
6(i) | 1
0.4g×1.8= ×0.4×v2
2 | M1 | KE gain=PE lost
v=6ms–1 | A1
Alternative method for question 6(i)
v2=02+2 × g × 1.8 | M1 | Use constant acceleration equation(s) with a=g to find v
v=6ms–1 | A1
2
--- 6(ii) ---
6(ii) | 0.4g–5.6 = 0.4a | M1 | Use Newton’s second law for the particle in the vertical
(3 terms)
a=–4ms–2 | A1
0=6–4t | M1 | Use of constant acceleration equation(s) such as v=u+at to
find t
t=1.5s | A1
4
--- 6(iii) ---
6(iii) | Straight line starting at (0,0) with positive gradient | B1
Second straight line starting at end of the first line with negative
gradient and ending with v = 0 | B1
All correct, start at (0, 0) with max velocity v=6 at t=0.6
i.e. (0.6, 6) and finishing at (2.1, 0) | B1FT | FT on their v from (i) and/or their t from (ii)
3
Question | Answer | Marks | GuidanceA particle of mass 0.4 kg is released from rest at a height of 1.8 m above the surface of the water in a tank. There is no instantaneous change of speed when the particle enters the water. The water exerts an upward force of 5.6 N on the particle when it is in the water.
\begin{enumerate}[label=(\roman*)]
\item Find the velocity of the particle at the instant when it reaches the surface of the water. [2]
\end{enumerate}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the time that it takes from the instant when the particle enters the water until it comes to instantaneous rest in the water. You may assume that the tank is deep enough so that the particle does not reach the bottom of the tank. [4]
\end{enumerate}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Sketch a velocity-time graph for the motion of the particle from the instant at which it is released until it comes to instantaneous rest in the water. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2019 Q6 [9]}}