Edexcel M4 — Question 2 8 marks

Exam BoardEdexcel
ModuleM4 (Mechanics 4)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeAir resistance kv² - projected vertically upward
DifficultyChallenging +1.8 This M4 question requires setting up and solving a differential equation for motion with air resistance proportional to v², then integrating to find distance. It involves non-trivial manipulation of the equation of motion (F=ma with resistance mkv² and weight mg), separation of variables, and careful integration with substitution. The specific initial condition and target (speed halving) require precise algebraic manipulation. While systematic, it demands strong calculus skills and multi-step reasoning beyond standard mechanics questions.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods

2. At time \(t = 0\), a particle \(P\) of mass \(m\) is projected vertically upwards with speed \(\sqrt { \frac { g } { k } }\), where \(k\) is a constant. At time \(t\) the speed of \(P\) is \(v\). The particle \(P\) moves against air resistance whose magnitude is modelled as being \(m k v ^ { 2 }\) when the speed of \(P\) is \(v\). Find, in terms of \(k\), the distance travelled by \(P\) until its speed first becomes half of its initial speed.

Question 2:
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
Keith swims perpendicular to bank for minimum time; velocity component across river \(= \frac{10}{9}\) m s\(^{-1}\)M1 Recognising perpendicular crossing gives least time
\(t = \frac{50}{10/9} = 45\) sA1 Correct time
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(BC = \frac{2}{3} \times 45 = 30\) mM1 A1 Drift = river speed × time
Part (c):
AnswerMarks Guidance
Working/AnswerMark Guidance
For straight line to \(B\), swimmer must aim upstream; component along river: \(\frac{10}{9}\sin\theta = \frac{2}{3}\)M1 Setting up velocity triangle
\(\sin\theta = \frac{2/3}{10/9} = \frac{3}{5}\), \(\cos\theta = \frac{4}{5}\)A1 Correct angle
Speed across river \(= \frac{10}{9}\cos\theta = \frac{10}{9} \times \frac{4}{5} = \frac{8}{9}\) m s\(^{-1}\)A1 Correct component
\(t = \frac{50}{8/9} = \frac{450}{8} = 56.25\) sA1 Correct time
# Question 2:

## Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Keith swims perpendicular to bank for minimum time; velocity component across river $= \frac{10}{9}$ m s$^{-1}$ | M1 | Recognising perpendicular crossing gives least time |
| $t = \frac{50}{10/9} = 45$ s | A1 | Correct time |

## Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $BC = \frac{2}{3} \times 45 = 30$ m | M1 A1 | Drift = river speed × time |

## Part (c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| For straight line to $B$, swimmer must aim upstream; component along river: $\frac{10}{9}\sin\theta = \frac{2}{3}$ | M1 | Setting up velocity triangle |
| $\sin\theta = \frac{2/3}{10/9} = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$ | A1 | Correct angle |
| Speed across river $= \frac{10}{9}\cos\theta = \frac{10}{9} \times \frac{4}{5} = \frac{8}{9}$ m s$^{-1}$ | A1 | Correct component |
| $t = \frac{50}{8/9} = \frac{450}{8} = 56.25$ s | A1 | Correct time |

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2. At time $t = 0$, a particle $P$ of mass $m$ is projected vertically upwards with speed $\sqrt { \frac { g } { k } }$, where $k$ is a constant. At time $t$ the speed of $P$ is $v$. The particle $P$ moves against air resistance whose magnitude is modelled as being $m k v ^ { 2 }$ when the speed of $P$ is $v$. Find, in terms of $k$, the distance travelled by $P$ until its speed first becomes half of its initial speed.

\hfill \mbox{\textit{Edexcel M4  Q2 [8]}}