Standard +0.3 This is a straightforward mechanics problem requiring resolution of forces, calculation of friction (F = μR), application of F = ma, and use of a SUVAT equation. All steps are standard M1 techniques with no conceptual subtlety—slightly easier than average due to the direct application of well-practiced methods.
3 A ring of mass 0.3 kg is threaded on a horizontal rough rod. The coefficient of friction between the ring and the rod is 0.8 . A force of magnitude 8 N acts on the ring. This force acts at an angle of \(10 ^ { \circ }\) above the horizontal in the vertical plane containing the rod.
Find the time taken for the ring to move, from rest, 0.6 m along the rod.
Complete method leading to an equation in \(t\), such as \(s = ut + \frac{1}{2}at^2\) with \(s = 0.6\), \(u = 0\) and using *their* value of \(a\) found from Newton's second law with 3 terms, namely, component of 8 N, any friction and \(0.3a\).
\(t = 0.234\) seconds
A1
Allow use of \(a = 22\) for M1 and A1
Alternative method for Question 3:
Answer
Marks
Guidance
Answer
Marks
Guidance
Resolving perpendicular to the rod
M1
\(8\sin 10 + R = 0.3g\)
A1
\(F = 0.8R \quad [R = 1.61081..., F = 1.28865...]\)
M1
Using \(F = \mu R\), where \(R\) must involve \(0.3g\) and a component of 8 N.
Work energy equation to find \(v\) after 0.6 metres.
\(0.6 = \frac{1}{2}(0 + 5.134) \times t\)
M1
Using \(s = \frac{1}{2}(u+v)t\) to find \(t\).
\(t = 0.234\) seconds
A1
6
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolving along or perpendicular to the rod | M1 | 3 terms in either direction |
| $8\sin 10 + R = 0.3g$ | A1 | |
| $8\cos 10 - F = 0.3a$ | A1 | |
| $F = 0.8R \quad [R = 1.61081..., F = 1.28865...]$ | M1 | Using $F = \mu R$, where $R$ is 2 terms involving weight and a component of 8 N. |
| $[a = 21.966...]$ $0.6 = \frac{1}{2} \times 21.966 \times t^2$ | M1 | Complete method leading to an equation in $t$, such as $s = ut + \frac{1}{2}at^2$ with $s = 0.6$, $u = 0$ and using *their* value of $a$ found from Newton's second law with 3 terms, namely, component of 8 N, any friction and $0.3a$. |
| $t = 0.234$ seconds | A1 | Allow use of $a = 22$ for M1 and A1 |
**Alternative method for Question 3:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolving perpendicular to the rod | M1 | |
| $8\sin 10 + R = 0.3g$ | A1 | |
| $F = 0.8R \quad [R = 1.61081..., F = 1.28865...]$ | M1 | Using $F = \mu R$, where $R$ must involve $0.3g$ and a component of 8 N. |
| $8\cos 10 \times 0.6 = F \times 0.6 + \frac{1}{2} \times 0.3v^2 \quad [v = 5.134]$ | B1 | Work energy equation to find $v$ after 0.6 metres. |
| $0.6 = \frac{1}{2}(0 + 5.134) \times t$ | M1 | Using $s = \frac{1}{2}(u+v)t$ to find $t$. |
| $t = 0.234$ seconds | A1 | |
| | **6** | |
---
3 A ring of mass 0.3 kg is threaded on a horizontal rough rod. The coefficient of friction between the ring and the rod is 0.8 . A force of magnitude 8 N acts on the ring. This force acts at an angle of $10 ^ { \circ }$ above the horizontal in the vertical plane containing the rod.
Find the time taken for the ring to move, from rest, 0.6 m along the rod.\\
\hfill \mbox{\textit{CAIE M1 2021 Q3 [6]}}