| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Bearing and compass direction problems |
| Difficulty | Moderate -0.3 This is a standard M1 vector resolution problem requiring students to resolve forces into components and use Pythagoras/trigonometry. While it involves multiple steps (resolving the resultant, finding components of F2, calculating magnitude and bearing), these are routine techniques practiced extensively in M1. The problem is slightly easier than average A-level because it's a straightforward application of well-drilled methods with no conceptual surprises or proof elements. |
| Spec | 3.03a Force: vector nature and diagrams3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
| Answer | Marks |
|---|---|
| 4(i) | M1 A1 |
| Solve for = 8.1 (N) or better | M1 A1 (4) |
| OR: | M1 A1 |
| Solve for = 8.1 (N) or better | M1 A1 (4) |
| 4(i) | First M1 for use of cos rule with 30o |
| Answer | Marks | Guidance |
|---|---|---|
| 4(ii) | or | M 1 A 1 |
| Answer | Marks |
|---|---|
| or | M1 A1 |
| Bearing is 149o (nearest degree) | A1 (5) |
| OR: | M1 A1 |
| Solve: | M1 A1 |
| Bearing is 149o (nearest degree) | A1 (5) |
| Answer | Marks |
|---|---|
| 4(ii) | First M1 for use of sin rule with 30o |
| Answer | Marks |
|---|---|
| 5a | M1A1 |
| Answer | Marks |
|---|---|
| 5b | M1A1 |
| Answer | Marks |
|---|---|
| 5c | M1 A1 |
| Answer | Marks |
|---|---|
| 5a | N.B. If they use g = 9.81, lose first A mark (once for whole question) |
| Answer | Marks |
|---|---|
| 5b | First M1 for a complete method to find the required time (they may find |
| Answer | Marks |
|---|---|
| 5c | First M1 for a complete method to find the speed / velocity(Could |
| Answer | Marks |
|---|---|
| 6a | V |
| O 270 | B1 shape |
| Answer | Marks | Guidance |
|---|---|---|
| 6b | Given answer | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6c | Time decelerating is 5V | B1 |
| OR: | M1 A2 | |
| Given answer | DM1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6d | or | M1 solving |
| Answer | Marks |
|---|---|
| 6a | First B1 for a trapezium with line starting at the origin |
| Answer | Marks |
|---|---|
| 6b | M1 for (t =) ; N.B. M1A0 for V=0.6t then answer |
| Answer | Marks |
|---|---|
| 6c | B1 for 5V identified appropriately |
| Answer | Marks |
|---|---|
| 6d | First M1 for solving their 3 term quadratic equation for V |
| Answer | Marks |
|---|---|
| 7a | M1A1 |
| Answer | Marks |
|---|---|
| 7b | B1; M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Given answer | A1 (5) | |
| 7c | Particles have same acceleration | B1 (1) |
| 7d | (= 0.8g) | M1 A1 |
| Answer | Marks |
|---|---|
| Total distance = 0.5 + 1.75 = 2.25 (m) Accept 2.3 (m) | A1 (7) |
| Answer | Marks |
|---|---|
| 7a | First M1 for equation of motion for A with usual rules |
| Answer | Marks |
|---|---|
| 7b | B1 for seen e.g. on diagram |
| Answer | Marks |
|---|---|
| 7c | B1 for particles have same acceleration (B0 for same velocity or if |
| Answer | Marks |
|---|---|
| 7d | First M1 for attempt to find speed (or speed2) when B hits the ground |
Question 4:
--- 4(i) ---
4(i) | M1 A1
Solve for = 8.1 (N) or better | M1 A1 (4)
OR: | M1 A1
Solve for = 8.1 (N) or better | M1 A1 (4)
4(i) | First M1 for use of cos rule with 30o
First A1 for a correct equation
OR: First M1 for ‘resolving’ in 2 directions with 30o / 60o (N.B. M0
here if cos/sin confused)
First A1 for TWO correct equations
Second M1 for solving for independent but must be solving a
,
‘correct cosine formula but with wrong angle’ if using method 1
OR for eliminating from two equations, independent but equations
must have the correct structure if using method 2
Second A1 for 8.1 (N) or better
--- 4(ii) ---
4(ii) | or | M 1 A 1
Solve:
or | M1 A1
Bearing is 149o (nearest degree) | A1 (5)
OR: | M1 A1
Solve: | M1 A1
Bearing is 149o (nearest degree) | A1 (5)
Notes
4(ii) | First M1 for use of sin rule with 30o
First A1 for a correct equation (allow 8.12 or better)
OR: First M1 for ‘resolving’ in 2 directions with 30o / 60o
First A1 for TWO correct equations (allow 4.12 or better)
Second M1, independent, for solving a ‘correct sine formula’ for or
OR independent for solving two equations, with correct structure, for
Second A1 for or
OR
Third A1 for Bearing is 149o (nearest degree)
N.B. First M1A1 Could use cos rule to find an angle
N.B. If the resolving method is used and there are no (i) or (ii) labels,
only award M1A1 in both cases when an answer is reached.
5a | M1A1
A1
A1 (4)
5b | M1A1
DM1
A1 (4)
5c | M1 A1
A1 (3)
11
Notes
5a | N.B. If they use g = 9.81, lose first A mark (once for whole question)
but all other A marks can be scored.
First M1 for a complete method to find the height (Could involve two
suvat equations) condone sign errors.
First A1 for a correct equation (or equations)
Second A1 for h = 11 (may be unsimplified) or better (For other
methods, give this A1 for any correct (may be unsimplified)
intermediate answer)
Third A1 for 13.5 or 14 (m)
5b | First M1 for a complete method to find the required time (they may find
the time up (1.5 s) and then add on the time down. Condone sign errors
First A1 for a correct equation or equations
Second DM1, dependent, for solving to find required time
Second A1 for 3.1 or 3.10 (s)
5c | First M1 for a complete method to find the speed / velocity(Could
involve two suvat equations) Condone sign errors but must have correct
numbers in their equation(s)
First A1 for a correct equation (or equations)
Second A1 for 16 or 16.3 (m s-1) Must be positive (speed)
6a | V
O 270 | B1 shape
B1 270, V
(2)
6b | Given answer | M1A1
(2)
6c | Time decelerating is 5V | B1
OR: | M1 A2
Given answer | DM1A1
(6)
6d | or | M1 solving
A1 A1
B1 (4)
14
Notes
6a | First B1 for a trapezium with line starting at the origin
Second B1 for 270 and V correctly marked
6b | M1 for (t =) ; N.B. M1A0 for V=0.6t then answer
Must see division or intermediate step from V=0.6t e.g. Changing 0.6
into 3/5.
A1 for t Given answer
6c | B1 for 5V identified appropriately
First M1 for clear attempt to equate the total area under graph to 1500.
(Must include all 3 parts (if not using the trapezium rule) with seen at
least once to give equation in V only; may use (1 triangle + 1 trapezium)
or (rectangle - trapezium)
(May use suvat for one or more parts of the area)
A2 for a correct equation, -1 e.e.o.o.
Second DM1 dependent on first M1 for multiplying out and collecting
terms and putting into appropriate form
Third A1 for correct equation. Given answer
6d | First M1 for solving their 3 term quadratic equation for V
N.B. This M1 can be implied by two correct roots but if either answer incorrect
then an explicit method must be shown for this M mark.
First A1 for V = 6
Second A1 for V = 75
B1 on ePEN but treat as DM1, dependent on both previous A marks, for
either reason
7a | M1A1
M1A1 (4)
7b | B1; M1A1
M1
Given answer | A1 (5)
7c | Particles have same acceleration | B1 (1)
7d | (= 0.8g) | M1 A1
M1
A1
M1 A1
Total distance = 0.5 + 1.75 = 2.25 (m) Accept 2.3 (m) | A1 (7)
17
Notes
7a | First M1 for equation of motion for A with usual rules
First A1 for a correct equation
Second M1 for equation of motion for B with usual rules
Second A1 for a correct equation
N.B. If using different tension in second equation, M0 for that equation
7b | B1 for seen e.g. on diagram
First M1 for resolving for A perp to the plane
First A1 for correct equation
N.B. These first 3 marks can be earned in (a).
Second M1 (Hence) for substituting for R and F and trig. and solving
for a (must be some evidence of this) their equations of motion from
part (a)
Second A1 for given answer (Not available if not using exact values
for trig ratios)
7c | B1 for particles have same acceleration (B0 for same velocity or if
incorrect extras given)
7d | First M1 for attempt to find speed (or speed2) when B hits the ground
(M0 if uses g)
First A1 for a correct expression
Second M1 for attempt to find deceleration of A
Second A1 for correct deceleration
Third M1 for using deceleration (must have found a deceleration) with v
= 0 to find distance (M0 if uses g)
Third A1 for a correct equation
Fourth A1 for 2.25 (m)
Two forces $\mathbf{F_1}$ and $\mathbf{F_2}$ act on a particle. The force $\mathbf{F_1}$ has magnitude 8 N and acts due east. The resultant of $\mathbf{F_1}$ and $\mathbf{F_2}$ is a force of magnitude 14 N acting in a direction whose bearing is $120°$.
Find
\begin{enumerate}[label=(\roman*)]
\item the magnitude of $\mathbf{F_2}$,
[4]
\item the direction of $\mathbf{F_2}$, giving your answer as a bearing to the nearest degree.
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2017 Q4 [9]}}