| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Motion on rough inclined plane |
| Difficulty | Standard +0.3 This is a standard multi-part work-energy question requiring systematic application of familiar mechanics principles: calculating friction work using μR×distance, applying work-energy theorem, and using energy conservation on the smooth section. All steps are routine for M1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i) | R = 0.25g × 0.6 [= 1.5] | B1 |
| [F = 0.5 × 0.25g × 0.6] [F = 0.75] | M1 | Use F = µR |
| [WD against friction = F × 8] | M1 | Using WD = Force × distance moved in direction of force |
| WD = 6 J | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(ii) | [½ × 0.25 ×152 = | |
| ½ × 0.25 × v2 + 6 + 0.25g × 8 × 0.8] | M1 | Work-energy equation in the form |
| Answer | Marks |
|---|---|
| A1ft | Correct Work–Energy equation for the motion to Q. ft on |
| Answer | Marks |
|---|---|
| M1 | Solving the work-energy equation for v |
| v = 7 m s–1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| [–F – 0.25g sin α = 0.25a] | M1 | Applying Newton’s second law to the particle along the plane |
| a = –11 m s–2 | A1ft | ft on friction found in (i) |
| M1 | Finding the speed of the particle at Q by applying v2 = u2 + |
| Answer | Marks |
|---|---|
| v = 7 m s–1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 7(iii) | [½ × 0.25 × 72 = 0.25 × g × H] |
| Answer | Marks | Guidance |
|---|---|---|
| [½ × m × 72 = m × g × H] | M1 | KE lost from Q to R = PE gain from Q to R |
| Answer | Marks |
|---|---|
| H = 72/2g = 2.45 m | A1 |
| Total height h = 6.4 + H = 8.85 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| [½ × 0.25 ×152 = 6 + 0.25g × h] | M1 | Work-energy from P to R |
| A1 | Correct Work-energy equation from P to R | |
| h = 8.85 | A1 |
Question 7:
--- 7(i) ---
7(i) | R = 0.25g × 0.6 [= 1.5] | B1
[F = 0.5 × 0.25g × 0.6] [F = 0.75] | M1 | Use F = µR
[WD against friction = F × 8] | M1 | Using WD = Force × distance moved in direction of force
WD = 6 J | A1
4
--- 7(ii) ---
7(ii) | [½ × 0.25 ×152 =
½ × 0.25 × v2 + 6 + 0.25g × 8 × 0.8] | M1 | Work-energy equation in the form
Initial KE = Final KE + WD against F + PE gain
A1ft | Correct Work–Energy equation for the motion to Q. ft on
WD
M1 | Solving the work-energy equation for v
v = 7 m s–1 | A1
Alternative method for question 7(ii)
[–F – 0.25g sin α = 0.25a] | M1 | Applying Newton’s second law to the particle along the plane
a = –11 m s–2 | A1ft | ft on friction found in (i)
M1 | Finding the speed of the particle at Q by applying v2 = u2 +
2as with u = 15, s = 8 or equivalent complete method
v = 7 m s–1 | A1
4
Question | Answer | Marks | Guidance
--- 7(iii) ---
7(iii) | [½ × 0.25 × 72 = 0.25 × g × H]
Or
[½ × m × 72 = m × g × H] | M1 | KE lost from Q to R = PE gain from Q to R
H is the height of R above Q
H = 72/2g = 2.45 m | A1
Total height h = 6.4 + H = 8.85 | A1
Alternative method for question 7(iii)
[½ × 0.25 ×152 = 6 + 0.25g × h] | M1 | Work-energy from P to R
A1 | Correct Work-energy equation from P to R
h = 8.85 | A1
3\includegraphics{figure_7}
The diagram shows the vertical cross-section $PQR$ of a slide. The part $PQ$ is a straight line of length $8$ m inclined at angle $α$ to the horizontal, where $\sin α = 0.8$. The straight part $PQ$ is tangential to the curved part $QR$ at $Q$, and $R$ is $h$ m above the level of $P$. The straight part $PQ$ of the slide is rough and the curved part $QR$ is smooth. A particle of mass $0.25$ kg is projected with speed $15$ m s$^{-1}$ from $P$ towards $Q$ and comes to rest at $R$. The coefficient of friction between the particle and $PQ$ is $0.5$.
\begin{enumerate}[label=(\roman*)]
\item Find the work done by the friction force during the motion of the particle from $P$ to $Q$. [4]
\item Hence find the speed of the particle at $Q$. [4]
\item Find the value of $h$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2019 Q7 [11]}}