CAIE M1 2016 November — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2016
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeResultant of three coplanar forces
DifficultyModerate -0.8 This is a straightforward 2D vector resolution problem requiring basic trigonometry to resolve forces into components and find the resultant. The angles are given clearly, and the method is standard M1 content with no conceptual challenges—simpler than average A-level questions which typically involve more steps or problem-solving insight.
Spec3.03p Resultant forces: using vectors

3 \includegraphics[max width=\textwidth, alt={}, center]{221f995d-2ef2-4be8-935b-49f8acf1cbe0-2_317_1104_1050_518} A boat is being pulled along a river by two people. One of the people walks along a path on one side of the river and the other person walks along a path on the opposite side of the river. The first person exerts a horizontal force of 60 N at an angle of \(25 ^ { \circ }\) to the direction of the river. The second person exerts a horizontal force of 50 N at an angle of \(15 ^ { \circ }\) to the direction of the river (see diagram).
  1. Find the total force exerted by the two people in the direction of the river.
  2. Find the magnitude and direction of the resultant force exerted by the two people.

Question 3:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\([X = 60\cos25 + 50\cos15]\)M1 For resolving both forces in the direction of river
\(= 103 \text{ N}\)A1 [2] Value of \(X\) is \(102.7 \text{ N}\)
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(Y = 60\sin25 - 50\sin15\ [= 12.4]\)B1 Component perpendicular to the direction of the river
\([R^2 = X^2 + Y^2]\) or \([\alpha = \arctan(Y/X)]\)M1 For using Pythagoras or for using arctan to find the resultant force or its direction
Magnitude is \(103 \text{ N}\) (or \(\alpha = 6.9°\) with direction specified unambiguously)A1 Magnitude is \(103.4 \text{ N}\)
\(\alpha = 6.9°\) with direction specified unambiguously (or Magnitude \(= 103 \text{ N}\))B1 [4]
## Question 3:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $[X = 60\cos25 + 50\cos15]$ | M1 | For resolving both forces in the direction of river |
| $= 103 \text{ N}$ | A1 [2] | Value of $X$ is $102.7 \text{ N}$ |

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $Y = 60\sin25 - 50\sin15\ [= 12.4]$ | B1 | Component perpendicular to the direction of the river |
| $[R^2 = X^2 + Y^2]$ or $[\alpha = \arctan(Y/X)]$ | M1 | For using Pythagoras or for using arctan to find the resultant force or its direction |
| Magnitude is $103 \text{ N}$ (or $\alpha = 6.9°$ with direction specified unambiguously) | A1 | Magnitude is $103.4 \text{ N}$ |
| $\alpha = 6.9°$ with direction specified unambiguously (or Magnitude $= 103 \text{ N}$) | B1 [4] | |

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\includegraphics[max width=\textwidth, alt={}, center]{221f995d-2ef2-4be8-935b-49f8acf1cbe0-2_317_1104_1050_518}

A boat is being pulled along a river by two people. One of the people walks along a path on one side of the river and the other person walks along a path on the opposite side of the river. The first person exerts a horizontal force of 60 N at an angle of $25 ^ { \circ }$ to the direction of the river. The second person exerts a horizontal force of 50 N at an angle of $15 ^ { \circ }$ to the direction of the river (see diagram).\\
(i) Find the total force exerted by the two people in the direction of the river.\\
(ii) Find the magnitude and direction of the resultant force exerted by the two people.

\hfill \mbox{\textit{CAIE M1 2016 Q3 [6]}}