| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Particle moving through liquid or resistance |
| Difficulty | Standard +0.3 This is a straightforward two-stage SUVAT problem requiring basic application of equations of motion and Newton's second law. Part (i) uses F=ma with given acceleration to find resistance force. Part (ii) requires sketching a standard v-t graph with two linear sections of different gradients, using SUVAT to calculate the required values—routine mechanics with no conceptual challenges beyond standard M1 content. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03f Weight: W=mg |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([3g - R = 3 \times 5.5]\) | M1 | For using Newton's 2nd law |
| Resistance is \(13.5\) N | A1 | [Total: 2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Graph consists of two line segments; the first starts at the origin and has a positive gradient. | B1 | |
| The second starts where first one ends and has positive but less steep gradient. | B1 | [Total: 2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([v_S^2 = 2 \times 10 \times 5 = 100]\) or \([v_B^2 = v_T^2 + 2 \times 5.5 \times 4]\) | M1 | For using \(v^2 = u^2 + 2as\) (for either stage) |
| \(v_S = 10\) ms\(^{-1}\) at surface and \(v_B = 12\) ms\(^{-1}\) at bottom – both shown on sketch | A1 | |
| \([10 = 0 + 10t_1\) or \(12 = 10 + 5.5(t_2 - t_1)]\) | M1 | For using \(v = u + at\) (for either stage) |
| \(t_1 = 1\) s at surface and shown on sketch | A1 | |
| \(t_2 = 1.36\) s at bottom and shown on sketch | A1 | [Total: 5] |
## Question 6 (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[3g - R = 3 \times 5.5]$ | M1 | For using Newton's 2nd law |
| Resistance is $13.5$ N | A1 | **[Total: 2]** |
## Question 6 (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Graph consists of two line segments; the first starts at the origin and has a positive gradient. | B1 | |
| The second starts where first one ends and has positive but less steep gradient. | B1 | **[Total: 2]** |
## Question 6 (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[v_S^2 = 2 \times 10 \times 5 = 100]$ **or** $[v_B^2 = v_T^2 + 2 \times 5.5 \times 4]$ | M1 | For using $v^2 = u^2 + 2as$ (for either stage) |
| $v_S = 10$ ms$^{-1}$ at surface **and** $v_B = 12$ ms$^{-1}$ at bottom – both shown on sketch | A1 | |
| $[10 = 0 + 10t_1$ or $12 = 10 + 5.5(t_2 - t_1)]$ | M1 | For using $v = u + at$ (for either stage) |
| $t_1 = 1$ s at surface and shown on sketch | A1 | |
| $t_2 = 1.36$ s at bottom and shown on sketch | A1 | **[Total: 5]** |
---6 A particle of mass 3 kg falls from rest at a point 5 m above the surface of a liquid which is in a container. There is no instantaneous change in speed of the particle as it enters the liquid. The depth of the liquid in the container is 4 m . The downward acceleration of the particle while it is moving in the liquid is $5.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Find the resistance to motion of the particle while it is moving in the liquid.\\
(ii) Sketch the velocity-time graph for the motion of the particle, from the time it starts to move until the time it reaches the bottom of the container. Show on your sketch the velocity and the time when the particle enters the liquid, and when the particle reaches the bottom of the container.
\hfill \mbox{\textit{CAIE M1 2014 Q6 [9]}}