| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Read and interpret velocity-time graph |
| Difficulty | Standard +0.3 This is a standard M1 velocity-time graph question requiring area calculations for displacement, gradient for acceleration, and basic algebraic manipulation. Part (iv) involves comparing two particles' displacements but follows routine procedures. All parts use familiar SUVAT/graph techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\text{Area} = \frac{1}{2}(10 + 4) \times 6 = 42\ \text{m}]\) Displacement \(= 42\) m | B1 | |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{v}{2} = \frac{6}{4}\) or \([\text{gradient} = 1.5,\ v = 6 + 1.5 \times 6]\) | M1 | Using similar triangles or using acceleration \(=\) gradient and \(v = u + at\) |
| \(v = 3\ \text{ms}^{-1}\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Total distance travelled \(= 42 + \frac{1}{2}(T - 10) \times 3\) | B1 FT | Area found with FT distance from (i) and FT speed from (ii) |
| \([42 + \frac{1}{2}(T - 10) \times 3 = 49.5] \rightarrow T = \ldots\) | M1 | For equation and solving for \(T\) |
| \(T = 15\) s | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(V = 1.75 \times 4 = 7 \text{ ms}^{-1}\) | B1 | |
| \(Q\) travels \([\frac{1}{2}(13+6) \times 7 = 66.5 \text{ m}]\), Distance apart \(= [66.5 + 42 - 7.5]\) | M1 | Finding area for \(Q\) and interpreting total distance between particles |
| Distance between \(P\) and \(Q = 101 \text{ m}\) | A1 | |
| Total: 3 |
# Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\text{Area} = \frac{1}{2}(10 + 4) \times 6 = 42\ \text{m}]$ Displacement $= 42$ m | **B1** | |
| | **1** | |
# Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{v}{2} = \frac{6}{4}$ or $[\text{gradient} = 1.5,\ v = 6 + 1.5 \times 6]$ | **M1** | Using similar triangles or using acceleration $=$ gradient and $v = u + at$ |
| $v = 3\ \text{ms}^{-1}$ | **A1** | |
| | **2** | |
# Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Total distance travelled $= 42 + \frac{1}{2}(T - 10) \times 3$ | **B1 FT** | Area found with FT distance from **(i)** and FT speed from **(ii)** |
| $[42 + \frac{1}{2}(T - 10) \times 3 = 49.5] \rightarrow T = \ldots$ | **M1** | For equation and solving for $T$ |
| $T = 15$ s | **A1** | |
| | **3** | |
## Question 6(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $V = 1.75 \times 4 = 7 \text{ ms}^{-1}$ | **B1** | |
| $Q$ travels $[\frac{1}{2}(13+6) \times 7 = 66.5 \text{ m}]$, Distance apart $= [66.5 + 42 - 7.5]$ | **M1** | Finding area for $Q$ and interpreting total distance between particles |
| Distance between $P$ and $Q = 101 \text{ m}$ | **A1** | |
| **Total: 3** | | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{f08a4870-9466-4f8b-bd0f-431fb1803514-08_661_1244_262_452}
The diagram shows the velocity-time graphs for two particles, $P$ and $Q$, which are moving in the same straight line. The graph for $P$ consists of four straight line segments. The graph for $Q$ consists of three straight line segments. Both particles start from the same initial position $O$ on the line. $Q$ starts 2 seconds after $P$ and both particles come to rest at time $t = T$. The greatest velocity of $Q$ is $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Find the displacement of $P$ from $O$ at $t = 10$.\\
(ii) Find the velocity of $P$ at $t = 12$.\\
(iii) Given that the total distance covered by $P$ during the $T$ seconds of its motion is 49.5 m , find the value of $T$.\\
(iv) Given also that the acceleration of $Q$ from $t = 2$ to $t = 6$ is $1.75 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, find the value of $V$ and hence find the distance between the two particles when they both come to rest at $t = T$.\\
\includegraphics[max width=\textwidth, alt={}, center]{f08a4870-9466-4f8b-bd0f-431fb1803514-10_392_529_262_808}
A particle $P$ of mass 0.2 kg rests on a rough plane inclined at $30 ^ { \circ }$ to the horizontal. The coefficient of friction between the particle and the plane is 0.3 . A force of magnitude $T \mathrm {~N}$ acts upwards on $P$ at $15 ^ { \circ }$ above a line of greatest slope of the plane (see diagram).\\
(i) Find the least value of $T$ for which the particle remains at rest.\\
The force of magnitude $T \mathrm {~N}$ is now removed. A new force of magnitude 0.25 N acts on $P$ up the plane, parallel to a line of greatest slope of the plane. Starting from rest, $P$ slides down the plane. After moving a distance of $3 \mathrm {~m} , P$ passes through the point $A$.\\
(ii) Use an energy method to find the speed of $P$ at $A$.\\
\hfill \mbox{\textit{CAIE M1 2017 Q6 [9]}}