| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Uniform beam on two supports |
| Difficulty | Moderate -0.8 This is a straightforward moments equilibrium problem with a uniform rod on two supports. Part (a) requires standard application of taking moments about a point and resolving vertically—routine textbook exercise. Part (b) is conceptual reasoning about whether the rod would tip, requiring minimal calculation. Below average difficulty for A-level mechanics. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks |
|---|---|
| (ii) | 550v2 |
| Answer | Marks |
|---|---|
| v2 = 2x – 3200 | B1 |
| Answer | Marks |
|---|---|
| A1 | [4] |
| Answer | Marks |
|---|---|
| [4] | v2 |
Question 5:
--- 5 (i)
(ii) ---
5 (i)
(ii) | 550v2
KE gain =
PE gain = 1000x
[1800x = 550v2 + 1000x + 700x]
k = 5.5
At A 5.5v2 = 1760 → v2 = 320
550(v2
– 320) =
1800(x – 1760) – 700(x – 1760)
v2 = 2x – 3200 (cwo)
Alternative for part (ii)
[1800 – 700 = 1100a and 5.5v2 = 1760]
a = 1 and v2 = 320
[v2 = 320 + 2 × 1 × (x – 1760)]
v2 = 2x – 3200 | B1
B1
M1
A1
B1
M1
A1
A1
M1
A1
M1
A1 | [4]
[4]
[4] | v2
ft for incorrect coeff(s) of and/or
of x
For using from A, KEgain= WD by
DF –WD against R
AG
For applying Newton’s 2nd Law to find
acceleration along AB and for using
kv2 = x to find v2 at A
For using v2 = u2 + 2as for motion from
A to B\includegraphics{figure_5}
A uniform rod AB has length $2$ m and weight $20$ N. The rod rests horizontally in equilibrium on two supports at points C and D, where AC = $0.4$ m and BD = $0.6$ m.
\begin{enumerate}[label=(\alph*)]
\item Find the reaction at each support. [4]
\item State what happens if the support at D is removed. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2014 Q5 [6]}}