| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Elastic string – conical pendulum (string inclined to vertical) |
| Difficulty | Standard +0.3 This is a standard circular motion problem with elastic strings requiring resolution of forces, Hooke's law, and energy calculations. While it involves multiple steps (tension from extension, vertical/horizontal force resolution, angular speed calculation, then energy comparison), these are routine M2 techniques with no novel insight required. The multi-part structure and 9 total marks indicate moderate length, but the methods are textbook-standard, placing it slightly above average difficulty. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| 6(i) | T = 12 × 0.1 / 0.4 ( = 3 N) | B1 |
| 3sinθ = 0.15ω2(0.5sinθ) | M1 | Uses Newton's Second Law horizontally |
| ω = 6.32 rads−1 | A1 | |
| Tcosθ = 0.15g (cosθ = 0.5) | M1 | Resolves vertically |
| θ = 60 | A1 | |
| Total: | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(ii) | v = 6.32 × 0.5sin60 | B1 FT |
| KE = 0.15(6.32 × 0.5sin60)2 / 2 (=0.5625J) | B1 | |
| Difference = 0.5625 – 12 × 0.12 / (2 × 0.4) | M1 | Uses EE = λx2 / (2L) |
| Difference = 0.4125 J | A1 | |
| Total: | 4 |
Question 6:
--- 6(i) ---
6(i) | T = 12 × 0.1 / 0.4 ( = 3 N) | B1 | Uses T = λx / L
3sinθ = 0.15ω2(0.5sinθ) | M1 | Uses Newton's Second Law horizontally
ω = 6.32 rads−1 | A1
Tcosθ = 0.15g (cosθ = 0.5) | M1 | Resolves vertically
θ = 60 | A1
Total: | 5
--- 6(ii) ---
6(ii) | v = 6.32 × 0.5sin60 | B1 FT | Uses v = rω and r = 0.5sin60
KE = 0.15(6.32 × 0.5sin60)2 / 2 (=0.5625J) | B1
Difference = 0.5625 – 12 × 0.12 / (2 × 0.4) | M1 | Uses EE = λx2 / (2L)
Difference = 0.4125 J | A1
Total: | 4A particle $P$ of mass $0.15$ kg is attached to one end of a light elastic string of natural length $0.4$ m and modulus of elasticity $12$ N. The other end of the string is attached to a fixed point $A$. The particle $P$ moves in a horizontal circle which has its centre vertically below $A$, with the string inclined at $\theta°$ to the vertical and $AP = 0.5$ m.
\begin{enumerate}[label=(\roman*)]
\item Find the angular speed of $P$ and the value of $\theta$. [5]
\item Calculate the difference between the elastic potential energy stored in the string and the kinetic energy of $P$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2017 Q6 [9]}}