AQA M2 — Question 5

Exam BoardAQA
ModuleM2 (Mechanics 2)
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TopicVariable acceleration (1D)
TypeAcceleration as function of velocity (separation of variables)
DifficultyStandard +0.3 This is a standard M2 differential equation problem requiring separation of variables to solve dv/dt = -40v/1600. The setup is straightforward (F=ma with resistance force), and the integration is routine, making it slightly easier than average but still requiring proper technique.
Spec3.03c Newton's second law: F=ma one dimension6.06a Variable force: dv/dt or v*dv/dx methods
5 A car, of mass 1600 kg , is travelling along a straight horizontal road at a speed of \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when the driving force is removed. The car then freewheels and experiences a resistance force. The resistance force has magnitude \(40 v\) newtons, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of the car after it has been freewheeling for \(t\) seconds. Find an expression for \(v\) in terms of \(t\).
Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\omega = 900 \times \frac{2\pi}{60} = 30\pi\) rad s\(^{-1}\)B1 Correct conversion
\(\omega^2 r = (30\pi)^2 \times 0.3 = 2664\) ms\(^{-2}\)M1 A1 Centripetal acceleration
At top: \(mg + N_{min} = m\omega^2 r\), so \(N_{min} = m(\omega^2 r - g)\)M1
\(N_{min} = 0.8(2664 - 9.8) = 2123\) NA1
At bottom: \(N_{max} - mg = m\omega^2 r\), so \(N_{max} = m(\omega^2 r + g)\)M1
\(N_{max} = 0.8(2664 + 9.8) = 2139\) NA1 
# Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\omega = 900 \times \frac{2\pi}{60} = 30\pi$ rad s$^{-1}$ | B1 | Correct conversion |
| $\omega^2 r = (30\pi)^2 \times 0.3 = 2664$ ms$^{-2}$ | M1 A1 | Centripetal acceleration |
| At top: $mg + N_{min} = m\omega^2 r$, so $N_{min} = m(\omega^2 r - g)$ | M1 | |
| $N_{min} = 0.8(2664 - 9.8) = 2123$ N | A1 | |
| At bottom: $N_{max} - mg = m\omega^2 r$, so $N_{max} = m(\omega^2 r + g)$ | M1 | |
| $N_{max} = 0.8(2664 + 9.8) = 2139$ N | A1 | |

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5 A car, of mass 1600 kg , is travelling along a straight horizontal road at a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when the driving force is removed. The car then freewheels and experiences a resistance force. The resistance force has magnitude $40 v$ newtons, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of the car after it has been freewheeling for $t$ seconds.

Find an expression for $v$ in terms of $t$.

\hfill \mbox{\textit{AQA M2  Q5}}
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