Edexcel M4 2018 June — Question 3 7 marks

Exam BoardEdexcel
ModuleM4 (Mechanics 4)
Year2018
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeApparent wind problems
DifficultyChallenging +1.2 This is a standard M4 apparent wind problem requiring vector addition and simultaneous equations. While it involves setting up two conditions and solving a quadratic, the technique is well-practiced in M4 and follows a familiar pattern. The 'two possible values' hint guides students toward the quadratic solution. More challenging than basic mechanics but routine for this module.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry
3. When a man walks due West at a constant speed of \(4 \mathrm {~km} \mathrm {~h} ^ { - 1 }\), the wind appears to be blowing from due South. When he runs due North at a constant speed of \(8 \mathrm {~km} \mathrm {~h} ^ { - 1 }\), the speed of the wind appears to be \(5 \mathrm {~km} \mathrm {~h} ^ { - 1 }\).
The velocity of the wind relative to the Earth is constant with magnitude \(w \mathrm {~km} \mathrm {~h} ^ { - 1 }\).
Find the two possible values of \(w\).
AnswerMarks
Walking: \(v_w = u_vm + v_m\) walking: \(v_w = a\vec{i} - 4\vec{i}\)B1
Running: \(v_w = b\vec{i} + (c+8)\vec{j}\) \((b^2 + c^2 = 25)\)B1
Compare components and use \(b^2 + c^2 = 25\):M1
\(b = -4\)A1
\(a = c + 8\), \(c^2 = 25 - 16 = 9\), \(c = \pm 3\)A1
Correct method to obtain a value of \(w\): \(w = \sqrt{4^2 + 5^2} = \sqrt{41}(= 6.40)\)M1
Second value correct: \(w = \sqrt{4^2 + 11^2} = \sqrt{137}(= 11.7)\)A1
Alternative:
AnswerMarks
Triangle of velocities for walkingB1
Either form of triangle of velocities for running using their \(v_w\)B1
Two triangles combined using their common velocityM1
Either correct diagram seen or impliedA1
Both possibilities shownA1
Correct method to obtain a value of \(w\): \(w = \sqrt{4^2 + 5^2} = \sqrt{41}(= 6.40)\)M1
Second value correct: \(w = \sqrt{4^2 + 11^2} = \sqrt{137}(= 11.7)\)A1
Question 4a
AnswerMarks
Equation of motion: \(\frac{1}{2}\frac{d^2x}{dt^2}(-28e^{-4t} + 80re^{-4t}) = -kx - \lambda v\)M1A2
Differentiate: \(\frac{dx}{dt} = -4(1.5 + 10r)e^{-4t} + 10e^{-4t} = 4e^{-4t} - 40ue^{-4t}\)M1
\(\frac{d^2x}{dt^2} = -16e^{-4t} - 40e^{-4t} + 160re^{-4t} = -56e^{-4t} + 160re^{-4t}\)A1
Substitute and compare coefficients: \(-28e^{-4t} + 80re^{-4t} = e^{-4t}(-1.5k - 10kt - 4\lambda + 40\lambda t)\)M1
\(1.5k + 4\lambda = 28\)
\(-10k + 40\lambda = 80\)
\(k = 8\), \(\lambda = 4\)A1, A1
Alt
AnswerMarks
Equation of motion: \(\frac{1}{2}\frac{d^2x}{dt^2}(-28e^{-4t} + 80re^{-4t}) = -kx - \lambda v\)M1A2
\(\ddot{x} + 2\lambda\dot{x} + 2kx = 0\)
\(m^2 + 24m + 2k = 0 \Rightarrow m = \frac{-2\lambda \pm\sqrt{4\lambda^2 - 8k}}{2} = -\lambda \pm\sqrt{\lambda^2 - 2k}\)M1A1
\(\Rightarrow \lambda = 4\), and \(\lambda^2 - 2k = 0 \Rightarrow k = 8\)M1A1, A1
Alternative for last 5 marks:
AnswerMarks
AE has a repeated root \(m = -4\)M1A1
\(\Rightarrow m^2 + 2\lambda m + 2k = m^2 + 8m + 16\)M1
\(k = 8\), \(\lambda = 4\)A1, A1
Question 4b
AnswerMarks
\(\frac{dx}{dt} = 0\) when \(t = \frac{1}{10}\)B1
\(x = 2.5e^{-0.4} = 1.68\)M1A1 
| Walking: $v_w = u_vm + v_m$ walking: $v_w = a\vec{i} - 4\vec{i}$ | B1 | |
| Running: $v_w = b\vec{i} + (c+8)\vec{j}$ $(b^2 + c^2 = 25)$ | B1 | |
| Compare components and use $b^2 + c^2 = 25$: | M1 | |
| $b = -4$ | A1 | |
| $a = c + 8$, $c^2 = 25 - 16 = 9$, $c = \pm 3$ | A1 | |
| Correct method to obtain a value of $w$: $w = \sqrt{4^2 + 5^2} = \sqrt{41}(= 6.40)$ | M1 | |
| Second value correct: $w = \sqrt{4^2 + 11^2} = \sqrt{137}(= 11.7)$ | A1 | |

### Alternative:
| Triangle of velocities for walking | B1 | |
| Either form of triangle of velocities for running using their $v_w$ | B1 | |
| Two triangles combined using their common velocity | M1 | |
| Either correct diagram seen or implied | A1 | |
| Both possibilities shown | A1 | |
| Correct method to obtain a value of $w$: $w = \sqrt{4^2 + 5^2} = \sqrt{41}(= 6.40)$ | M1 | |
| Second value correct: $w = \sqrt{4^2 + 11^2} = \sqrt{137}(= 11.7)$ | A1 | |

## Question 4a
| Equation of motion: $\frac{1}{2}\frac{d^2x}{dt^2}(-28e^{-4t} + 80re^{-4t}) = -kx - \lambda v$ | M1A2 | |
| Differentiate: $\frac{dx}{dt} = -4(1.5 + 10r)e^{-4t} + 10e^{-4t} = 4e^{-4t} - 40ue^{-4t}$ | M1 | |
| $\frac{d^2x}{dt^2} = -16e^{-4t} - 40e^{-4t} + 160re^{-4t} = -56e^{-4t} + 160re^{-4t}$ | A1 | |
| Substitute and compare coefficients: $-28e^{-4t} + 80re^{-4t} = e^{-4t}(-1.5k - 10kt - 4\lambda + 40\lambda t)$ | M1 | |
| $1.5k + 4\lambda = 28$ | | |
| $-10k + 40\lambda = 80$ | | |
| $k = 8$, $\lambda = 4$ | A1, A1 | |

### Alt
| Equation of motion: $\frac{1}{2}\frac{d^2x}{dt^2}(-28e^{-4t} + 80re^{-4t}) = -kx - \lambda v$ | M1A2 | |
| $\ddot{x} + 2\lambda\dot{x} + 2kx = 0$ | | |
| $m^2 + 24m + 2k = 0 \Rightarrow m = \frac{-2\lambda \pm\sqrt{4\lambda^2 - 8k}}{2} = -\lambda \pm\sqrt{\lambda^2 - 2k}$ | M1A1 | |
| $\Rightarrow \lambda = 4$, and $\lambda^2 - 2k = 0 \Rightarrow k = 8$ | M1A1, A1 | |

### Alternative for last 5 marks:
| AE has a repeated root $m = -4$ | M1A1 | |
| $\Rightarrow m^2 + 2\lambda m + 2k = m^2 + 8m + 16$ | M1 | |
| $k = 8$, $\lambda = 4$ | A1, A1 | |

## Question 4b
| $\frac{dx}{dt} = 0$ when $t = \frac{1}{10}$ | B1 | |
| $x = 2.5e^{-0.4} = 1.68$ | M1A1 | |
3. When a man walks due West at a constant speed of $4 \mathrm {~km} \mathrm {~h} ^ { - 1 }$, the wind appears to be blowing from due South. When he runs due North at a constant speed of $8 \mathrm {~km} \mathrm {~h} ^ { - 1 }$, the speed of the wind appears to be $5 \mathrm {~km} \mathrm {~h} ^ { - 1 }$.\\
The velocity of the wind relative to the Earth is constant with magnitude $w \mathrm {~km} \mathrm {~h} ^ { - 1 }$.\\
Find the two possible values of $w$.

\hfill \mbox{\textit{Edexcel M4 2018 Q3 [7]}}
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