| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | March |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - smooth inclined plane (no resistance) |
| Difficulty | Moderate -0.8 This is a straightforward application of standard kinetic energy and work-energy principle formulas. Part (i) is direct substitution into KE = ½mv², and part (ii) requires equating initial KE to gravitational PE gained (mgh) with basic trigonometry to find distance along the slope. No problem-solving insight needed, just routine application of memorized methods. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 1(i) | KE = ½ × 0.4 × 122 = 28.8J | B1 |
| Total: | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1(ii) | PE gain = 0.4gh [= 4d sin 30] | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4h = 28.8 | M1 | Using KE loss = PE gain |
| h = 7.2 h = d sin 30 d = 14.4 m | A1 | |
| Total: | 3 | |
| Question | Answer | Marks |
Question 1:
--- 1(i) ---
1(i) | KE = ½ × 0.4 × 122 = 28.8J | B1
Total: | 1
--- 1(ii) ---
1(ii) | PE gain = 0.4gh [= 4d sin 30] | B1 | h = height gained
d = distance travelled up the plane
4h = 28.8 | M1 | Using KE loss = PE gain
h = 7.2 h = d sin 30 d = 14.4 m | A1
Total: | 3
Question | Answer | Marks | GuidanceA particle of mass $0.4$ kg is projected with a speed of $12$ m s$^{-1}$ up a line of greatest slope of a smooth plane inclined at $30°$ to the horizontal.
\begin{enumerate}[label=(\roman*)]
\item Find the initial kinetic energy of the particle. [1]
\item Use an energy method to find the distance the particle moves up the plane before coming to instantaneous rest. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2017 Q1 [4]}}