A particle \(P\) of mass \(0.5\) kg moves in a straight line. At time \(t\) s the velocity of \(P\) is \(v\) m s\(^{-1}\) and its displacement from a fixed point \(O\) on the line is \(x\) m. The only forces acting on \(P\) are a force of magnitude \(\frac{150}{(x+1)^2}\) N in the direction of increasing displacement and a resistive force of magnitude \(\frac{450}{(x+1)^3}\) N. When \(t = 0\), \(x = 0\) and \(v = 20\). Find \(v\) in terms of \(x\), giving your answer in the form \(v = \frac{Ax + B}{(x + 1)}\), where \(A\) and \(B\) are constants to be determined. [6]

Question 2:
2 | dv 150 450
0.5v = −
dx (x+1)2 (x+1)3 | M1 | Allow sign errors.
300 450
Integrate: 0.5v2 =− + + A
x+1 (x+1)2 | M1A1 | Correct powers, allow sign errors.
x=0, v=20; A=50 | M1 | Use initial condition.
100(x2 −4x+4)
Rearrange: v2 =
(x+1)2 | A1 | AEF
100(x−2)2 10(x−2)
v2 = so v=
(x+1)2 (x+1)
20−10x
From initial condition, sign must be negative, v=
x+1 | A1 | Signs dealt with convincingly.
6
Question | Answer | Marks | Guidance

A particle $P$ of mass $0.5$ kg moves in a straight line. At time $t$ s the velocity of $P$ is $v$ m s$^{-1}$ and its displacement from a fixed point $O$ on the line is $x$ m. The only forces acting on $P$ are a force of magnitude $\frac{150}{(x+1)^2}$ N in the direction of increasing displacement and a resistive force of magnitude $\frac{450}{(x+1)^3}$ N. When $t = 0$, $x = 0$ and $v = 20$.

Find $v$ in terms of $x$, giving your answer in the form $v = \frac{Ax + B}{(x + 1)}$, where $A$ and $B$ are constants to be determined. [6]

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q2 [6]}}