Moderate -0.5 This is a standard M1 mechanics question requiring resolution of forces into components and finding the resultant. While it involves three forces at different angles requiring careful trigonometry and Pythagoras, it follows a routine procedure taught in all M1 courses with no novel problem-solving insight needed. The 6 marks reflect computational work rather than conceptual difficulty, making it slightly easier than average.
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\includegraphics[max width=\textwidth, alt={}, center]{e060fc3b-ae93-46b5-b476-dcecb14d6d06-2_684_257_1114_945}
Coplanar forces of magnitudes \(250 \mathrm {~N} , 100 \mathrm {~N}\) and 300 N act at a point in the directions shown in the diagram. The resultant of the three forces has magnitude \(R \mathrm {~N}\), and acts at an angle \(\alpha ^ { \circ }\) anticlockwise from the force of magnitude 100 N . Find \(R\) and \(\alpha\). [0pt]
[6]
ft only if one B1 scored, or if expressions for candidate's \(X\) and \(Y\) are those of the equilibrant
\(\tan\alpha = 65.1/185.5\)
M1
For using \(\tan\alpha = Y/X\)
\(\alpha = 19.3\)
A1 ft (×6)
ft only if one B1 is scored; SR for sin/cos mix (max 4/6): \(X = 100 + 250\sin70°\) and \(Y = 300 - 250\cos70°\) (334.9 and 214.5) B1; Method marks as scheme M1 M1; \(R = 398\) N and \(\alpha = 32.6\) A1
OR method:
Answer
Marks
Guidance
Answer/Working
Marks
Guidance
316(.227766..) or 107(.4528..) or 299(.3343..)
B1
Magnitude of resultant of two of the forces
\(71.565...°\) or \(37.2743...°\) or \(-51.7039...°\)
# Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X = 100 + 250\cos70°$ | B1 | |
| $Y = 300 - 250\sin70°$ | B1 | |
| $R^2 = 185.5^2 + 65.1^2$ | M1 | For using $R^2 = X^2 + Y^2$ |
| $R = 197$ | A1 ft | ft only if one B1 scored, or if expressions for candidate's $X$ and $Y$ are those of the equilibrant |
| $\tan\alpha = 65.1/185.5$ | M1 | For using $\tan\alpha = Y/X$ |
| $\alpha = 19.3$ | A1 ft (×6) | ft only if one B1 is scored; SR for sin/cos mix (max 4/6): $X = 100 + 250\sin70°$ and $Y = 300 - 250\cos70°$ (334.9 and 214.5) B1; Method marks as scheme M1 M1; $R = 398$ N and $\alpha = 32.6$ A1 |
## OR method:
| Answer/Working | Marks | Guidance |
|---|---|---|
| 316(.227766..) or 107(.4528..) or 299(.3343..) | B1 | Magnitude of resultant of two of the forces |
| $71.565...°$ or $37.2743...°$ or $-51.7039...°$ | B1 | Direction of resultant of two of the forces |
| $R^2 = 316.2^2 + 250^2 - 2\times316.2\times250\cos38.4°$ | M1 | For using cosine rule to find $R$ |
| $R^2 = 107.5^2 + 100^2 - 2\times107.5\times100\cos142.7°$ | | |
| $R^2 = 299.3^2 + 300^2 - 2\times299.3\times300\cos38.3°$ | | |
| $R = 197$ | A1 ft | ft only if one B1 is scored |
| $\sin(71.6-\alpha) = 250\sin38.4 \div 197$ | M1 | For using sine rule to find $\alpha$ |
| $\sin(37.3-\alpha) = 100\sin142.7 \div 197$ | | |
| $\sin(51.7+\alpha) = 300\sin38.3 \div 197$ | | |
| $\alpha = 19.3°$ | A1 ft | ft only if one B1 is scored |
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\includegraphics[max width=\textwidth, alt={}, center]{e060fc3b-ae93-46b5-b476-dcecb14d6d06-2_684_257_1114_945}
Coplanar forces of magnitudes $250 \mathrm {~N} , 100 \mathrm {~N}$ and 300 N act at a point in the directions shown in the diagram. The resultant of the three forces has magnitude $R \mathrm {~N}$, and acts at an angle $\alpha ^ { \circ }$ anticlockwise from the force of magnitude 100 N . Find $R$ and $\alpha$.\\[0pt]
[6]
\hfill \mbox{\textit{CAIE M1 2004 Q2 [6]}}