CAIE M1 2020 November — Question 3 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2020
SessionNovember
Marks5
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Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeParticle suspended by strings
DifficultyModerate -0.8 This is a straightforward two-string equilibrium problem requiring resolution of forces in two perpendicular directions and solving simultaneous equations. The angles are standard (45° and 60°) with known exact trig values, and one tension is given. This is a routine mechanics exercise with no conceptual challenges beyond basic force resolution.
Spec1.05g Exact trigonometric values: for standard angles3.03m Equilibrium: sum of resolved forces = 0
3 \includegraphics[max width=\textwidth, alt={}, center]{fcc3d739-5c36-48ad-9c34-f69b28a06dba-05_446_851_260_646} A block of mass \(m \mathrm {~kg}\) is held in equilibrium below a horizontal ceiling by two strings, as shown in the diagram. One of the strings is inclined at \(45 ^ { \circ }\) to the horizontal and the tension in this string is \(T \mathrm {~N}\). The other string is inclined at \(60 ^ { \circ }\) to the horizontal and the tension in this string is 20 N . Find \(T\) and \(m\).
Question 3:
Method 1 (Resolving forces):
AnswerMarks Guidance
AnswerMark Guidance
\(20\cos 60 = T\cos 45\)M1 Resolve forces horizontally, 2 terms
\(T = 10\sqrt{2}\) or \(T = 14.1\)A1
\(20\sin 60 + T\sin 45 = mg\) or \(W\)M1 Resolve forces vertically, 3 terms
\(20\sin 60 + T\sin 45 = mg\)A1
\(m = 2.73\ [=\sqrt{3}+1]\)A1
Alternative Method 1 (Lami's Method):
AnswerMarks Guidance
AnswerMark Guidance
\(\left[\frac{T}{\sin 150} = \frac{mg \text{ or } W}{\sin 75} = \frac{20}{\sin 135}\right]\)M1 Attempt at one pair of terms using Lami's Method
\(\frac{T}{\sin 150} = \frac{mg}{\sin 75} = \frac{20}{\sin 135}\)A1 All terms correct in Lami's Method
Attempt to solve for either \(T\) or \(m\) or \(W\)M1
\(T = 10\sqrt{2}\) or \(T = 14.1\)A1
\(m = 2.73\ [=\sqrt{3}+1]\)A1
Alternative Method 2 (Triangle of forces):
AnswerMarks Guidance
AnswerMark Guidance
\(\left[\frac{T}{\sin 30} = \frac{mg \text{ or } W}{\sin 105} = \frac{20}{\sin 45}\right]\)M1 Attempt the triangle of forces method and state one equation which involves any two of the forces \(T\), \(m\) and 20
\(\frac{T}{\sin 30} = \frac{mg}{\sin 105} = \frac{20}{\sin 45}\)A1 All correct
Attempt to solve for either \(T\) or \(m\) or \(W\)M1
\(T = 10\sqrt{2}\) or \(T = 14.1\)A1
\(m = 2.73\ [=\sqrt{3}+1]\)A1
5 
## Question 3:

**Method 1 (Resolving forces):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $20\cos 60 = T\cos 45$ | M1 | Resolve forces horizontally, 2 terms |
| $T = 10\sqrt{2}$ or $T = 14.1$ | A1 | |
| $20\sin 60 + T\sin 45 = mg$ or $W$ | M1 | Resolve forces vertically, 3 terms |
| $20\sin 60 + T\sin 45 = mg$ | A1 | |
| $m = 2.73\ [=\sqrt{3}+1]$ | A1 | |

**Alternative Method 1 (Lami's Method):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[\frac{T}{\sin 150} = \frac{mg \text{ or } W}{\sin 75} = \frac{20}{\sin 135}\right]$ | M1 | Attempt at one pair of terms using Lami's Method |
| $\frac{T}{\sin 150} = \frac{mg}{\sin 75} = \frac{20}{\sin 135}$ | A1 | All terms correct in Lami's Method |
| Attempt to solve for either $T$ or $m$ or $W$ | M1 | |
| $T = 10\sqrt{2}$ or $T = 14.1$ | A1 | |
| $m = 2.73\ [=\sqrt{3}+1]$ | A1 | |

**Alternative Method 2 (Triangle of forces):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[\frac{T}{\sin 30} = \frac{mg \text{ or } W}{\sin 105} = \frac{20}{\sin 45}\right]$ | M1 | Attempt the triangle of forces method and state one equation which involves any two of the forces $T$, $m$ and 20 |
| $\frac{T}{\sin 30} = \frac{mg}{\sin 105} = \frac{20}{\sin 45}$ | A1 | All correct |
| Attempt to solve for either $T$ or $m$ or $W$ | M1 | |
| $T = 10\sqrt{2}$ or $T = 14.1$ | A1 | |
| $m = 2.73\ [=\sqrt{3}+1]$ | A1 | |
| | **5** | |

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3\\
\includegraphics[max width=\textwidth, alt={}, center]{fcc3d739-5c36-48ad-9c34-f69b28a06dba-05_446_851_260_646}

A block of mass $m \mathrm {~kg}$ is held in equilibrium below a horizontal ceiling by two strings, as shown in the diagram. One of the strings is inclined at $45 ^ { \circ }$ to the horizontal and the tension in this string is $T \mathrm {~N}$. The other string is inclined at $60 ^ { \circ }$ to the horizontal and the tension in this string is 20 N .

Find $T$ and $m$.\\

\hfill \mbox{\textit{CAIE M1 2020 Q3 [5]}}
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