Standard +0.3 This is a standard variable force problem requiring integration of F = ma with force inversely proportional to distance cubed. The setup is straightforward (released from rest, smooth surface), and the solution follows a routine method: use v dv/dx = a, integrate the force law, and apply boundary conditions. Slightly above average difficulty due to the integration technique required, but this is a textbook M2/Further Mechanics exercise with no novel insight needed.
3
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A particle \(P\) of mass 0.6 kg moves in a straight line on a smooth horizontal surface. A force of magnitude \(\frac { 3 } { x ^ { 3 } }\) newtons acts on the particle in the direction from \(P\) to \(O\), where \(O\) is a fixed point of the surface and \(x \mathrm {~m}\) is the distance \(O P\) (see diagram). The particle \(P\) is released from rest at the point where \(x = 10\). Find the speed of \(P\) when \(x = 2.5\).
For applying Newton's 2nd law and using \(a = v\dfrac{dv}{dx}\)
M1
\(0.6v\dfrac{dv}{dx} = -\dfrac{3}{x^3}\)
A1
For separating the variables and integrating
M1
\(0.3v^2 = -\dfrac{3x^{-2}}{(-2)}\) \((+C)\)
A1 ft
ft omission of minus sign in line 2 only
For using \(v = 0\) when \(x = 10\)
M1
\(v^2 = \dfrac{5}{x^2} - \dfrac{1}{20}\) (aef)
A1 ft
ft wrong sign in line 4 only
Speed is \(\dfrac{\sqrt{3}}{2}\) ms\(^{-1}\) \((= 0.866)\)
A1
7 marks total
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| For applying Newton's 2nd law and using $a = v\dfrac{dv}{dx}$ | M1 | |
| $0.6v\dfrac{dv}{dx} = -\dfrac{3}{x^3}$ | A1 | |
| For separating the variables and integrating | M1 | |
| $0.3v^2 = -\dfrac{3x^{-2}}{(-2)}$ $(+C)$ | A1 ft | ft omission of minus sign in line 2 only |
| For using $v = 0$ when $x = 10$ | M1 | |
| $v^2 = \dfrac{5}{x^2} - \dfrac{1}{20}$ (aef) | A1 ft | ft wrong sign in line 4 only |
| Speed is $\dfrac{\sqrt{3}}{2}$ ms$^{-1}$ $(= 0.866)$ | A1 | **7 marks total** |
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\includegraphics[max width=\textwidth, alt={}, center]{835616aa-0b2b-4e8c-bbbf-60b72dc5ea3e-2_145_792_1656_680}
A particle $P$ of mass 0.6 kg moves in a straight line on a smooth horizontal surface. A force of magnitude $\frac { 3 } { x ^ { 3 } }$ newtons acts on the particle in the direction from $P$ to $O$, where $O$ is a fixed point of the surface and $x \mathrm {~m}$ is the distance $O P$ (see diagram). The particle $P$ is released from rest at the point where $x = 10$. Find the speed of $P$ when $x = 2.5$.
\hfill \mbox{\textit{CAIE M2 2004 Q3 [7]}}