| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion down smooth slope |
| Difficulty | Moderate -0.3 This is a straightforward two-part mechanics question requiring standard application of Newton's second law and equations of motion. Part (i) involves resolving forces on a smooth slope and using v=u+at (routine calculation). Part (ii) requires finding speed at bottom using energy or kinematics, then applying friction to find stopping distance. All techniques are standard M1 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension6.02a Work done: concept and definition6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = g\sin 30 = 5\) | B1 | |
| \(2.5 = 0 + 5t\) | M1 | Using \(v = u + at\) |
| \(t = 0.5\), Time \(= 0.5\) s | A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v^2 = 0 + 2 \times 5 \times 3 = 30\) | B1 | |
| \(-1 = 0.5a \rightarrow a = -2\) | For applying Newton's second law to the particle and using \(v^2 = u^2 + 2as\) | |
| \(0 = 30 + 2 \times (-2) \times s\) | M1 | |
| Distance \(= 7.5\) m | A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v^2 = 0 + 2 \times 5 \times 3 = 30\) | B1 | |
| \(0.5 \times 0.5 \times 30 = 1 \times \text{distance}\) | M1 | KE lost \(=\) WD against Friction |
| Distance \(= 7.5\) m | A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| PE lost \(= 0.5 \times 10 \times 3\sin 30 = 7.5\) | B1 | Using PE lost \(= mgh\) |
| \(7.5 = 1 \times \text{distance}\) | M1 | PE lost \(=\) WD against Friction |
| Distance \(= 7.5\) m | A1 (3) |
## Question 2:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = g\sin 30 = 5$ | B1 | |
| $2.5 = 0 + 5t$ | M1 | Using $v = u + at$ |
| $t = 0.5$, Time $= 0.5$ s | A1 (3) | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v^2 = 0 + 2 \times 5 \times 3 = 30$ | B1 | |
| $-1 = 0.5a \rightarrow a = -2$ | | For applying Newton's second law to the particle and using $v^2 = u^2 + 2as$ |
| $0 = 30 + 2 \times (-2) \times s$ | M1 | |
| Distance $= 7.5$ m | A1 (3) | |
**First alternative method for 2(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v^2 = 0 + 2 \times 5 \times 3 = 30$ | B1 | |
| $0.5 \times 0.5 \times 30 = 1 \times \text{distance}$ | M1 | KE lost $=$ WD against Friction |
| Distance $= 7.5$ m | A1 (3) | |
**Second alternative method for 2(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| PE lost $= 0.5 \times 10 \times 3\sin 30 = 7.5$ | B1 | Using PE lost $= mgh$ |
| $7.5 = 1 \times \text{distance}$ | M1 | PE lost $=$ WD against Friction |
| Distance $= 7.5$ m | A1 (3) | |
---2 A particle of mass 0.5 kg starts from rest and slides down a line of greatest slope of a smooth plane. The plane is inclined at an angle of $30 ^ { \circ }$ to the horizontal.\\
(i) Find the time taken for the particle to reach a speed of $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
When the particle has travelled 3 m down the slope from its starting point, it reaches rough horizontal ground at the bottom of the slope. The frictional force acting on the particle is 1 N .\\
(ii) Find the distance that the particle travels along the ground before it comes to rest.
\hfill \mbox{\textit{CAIE M1 2015 Q2 [6]}}