Question 9 - A-Level Maths

WJEC Unit 4 2024 June — Question 9 11 marks

Exam Board	WJEC
Module	Unit 4 (Unit 4)
Year	2024
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Motion on a slope
Type	Motion down rough slope
Difficulty	Standard +0.3 This is a standard mechanics problem requiring application of Newton's second law to motion on a slope with friction and resistance. Students must resolve forces parallel to the slope (weight component, friction, variable resistance) and apply F=ma. The 'show that' format and straightforward force resolution make this easier than average, though it requires careful bookkeeping of multiple force terms.
Spec	1.07t Construct differential equations: in context 3.03v Motion on rough surface: including inclined planes

9. The diagram below shows a parcel, of mass $m \mathrm {~kg}$, sliding down a rough slope inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 7 } { 25 }$. \includegraphics[max width=\textwidth, alt={}, center]{8f47b2ff-f954-42ec-8ecc-fc64313a7b89-24_394_906_497_584} The coefficient of friction between the parcel and the slope is $\frac { 1 } { 12 }$. In addition to friction, the parcel experiences a variable resistive force of $m v \mathrm {~N}$, where $v \mathrm {~ms} ^ { - 1 }$ is the velocity of the parcel at time $t$ seconds.

Show that the motion of the parcel satisfies the differential equation $$5 \frac { \mathrm {~d} v } { \mathrm {~d} t } = g - 5 v$$
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Show mark scheme Show mark scheme source

Question 9:

Answer	Marks	Guidance
9	11
Total	80
School	College	Employment
Boys	33	49
Girls	40	40
Total	73	89
R2 =	0·9412

0·

Answer	Marks
R2 = 0	·8881

Question 9:
9 | 11
Total | 80
School | College | Employment | Other | Total
Boys | 33 | 49 | 8 | 2 | 92
Girls | 40 | 40 | 7 | 1 | 88
Total | 73 | 89 | 15 | 3 | 180
R2 = | 0·9412
0·
R2 = 0 | ·8881

Show LaTeX source

9. The diagram below shows a parcel, of mass $m \mathrm {~kg}$, sliding down a rough slope inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 7 } { 25 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{8f47b2ff-f954-42ec-8ecc-fc64313a7b89-24_394_906_497_584}

The coefficient of friction between the parcel and the slope is $\frac { 1 } { 12 }$. In addition to friction, the parcel experiences a variable resistive force of $m v \mathrm {~N}$, where $v \mathrm {~ms} ^ { - 1 }$ is the velocity of the parcel at time $t$ seconds.\\
(a) Show that the motion of the parcel satisfies the differential equation

$$5 \frac { \mathrm {~d} v } { \mathrm {~d} t } = g - 5 v$$

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Q1 3 Q2 8 Q3 8 Q4 21 Q5 7 Q6 8 Q7 7 Q8 7 Q9 11

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