Edexcel M3 — Question 5 13 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeAir resistance kv² - falling from rest or projected downward
DifficultyChallenging +1.2 This is a standard M3 viscous resistance problem requiring setting up Newton's second law with a quadratic resistance term, then solving a separable differential equation using substitution and partial fractions. While it involves multiple techniques (differential equations, integration, partial fractions), these are routine M3 methods applied in a straightforward context with no novel insight required. The 13-mark allocation reflects the algebraic work rather than conceptual difficulty.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods

A small sphere \(S\), of mass \(m\) kg is released from rest at the surface of a liquid in a right circular cylinder whose axis is vertical. When \(S\) is moving downwards with speed \(v\) ms\(^{-1}\), the viscous resistive force acting upwards on it has magnitude \(v^2\) N.
  1. Write down a differential equation for the motion of \(S\), clearly defining any symbol(s) that you introduce. [4 marks]
  2. Find, in terms of \(m\), the distance \(S\) has fallen when its speed is \(\sqrt{\frac{mg}{2}}\) ms\(^{-1}\). [9 marks]

AnswerMarks
(a) When distance from surface \(= x\) m, \(mv\frac{dv}{dx} = mg - v^2\)B1 M1 A1 A1
(b) \(\int\frac{mv}{mg-v^2}dv = x + c\)M1 A1 M1 A1
\(\frac{1}{m}\ln(mg - v^2) = x + c\)M1 A1 A1
When \(v = 0, x = 0\): \(c = -\frac{1}{m}\ln mg\)M1 A1 A1
\(x = \frac{1}{m}\ln\frac{mv}{mg-v^2}\)M1 A1
When \(v = \sqrt{\frac{mg}{2}}\), \(x = \frac{1}{2}\ln 2\)M1 A1
Total: 13 marks
(a) When distance from surface $= x$ m, $mv\frac{dv}{dx} = mg - v^2$ | B1 M1 A1 A1 |

(b) $\int\frac{mv}{mg-v^2}dv = x + c$ | M1 A1 M1 A1 |

$\frac{1}{m}\ln(mg - v^2) = x + c$ | M1 A1 A1 |

When $v = 0, x = 0$: $c = -\frac{1}{m}\ln mg$ | M1 A1 A1 |

$x = \frac{1}{m}\ln\frac{mv}{mg-v^2}$ | M1 A1 |

When $v = \sqrt{\frac{mg}{2}}$, $x = \frac{1}{2}\ln 2$ | M1 A1 |

| **Total: 13 marks** |
A small sphere $S$, of mass $m$ kg is released from rest at the surface of a liquid in a right circular cylinder whose axis is vertical. When $S$ is moving downwards with speed $v$ ms$^{-1}$, the viscous resistive force acting upwards on it has magnitude $v^2$ N.

\begin{enumerate}[label=(\alph*)]
\item Write down a differential equation for the motion of $S$, clearly defining any symbol(s) that you introduce. [4 marks]
\item Find, in terms of $m$, the distance $S$ has fallen when its speed is $\sqrt{\frac{mg}{2}}$ ms$^{-1}$. [9 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3  Q5 [13]}}