| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Inverse power force - non-gravitational context |
| Difficulty | Standard +0.8 This M3 variable force question requires setting up F=ma with v dv/dx, integrating a rational function with substitution, then separating variables and integrating again to find distance. While mechanically demanding with multiple integration steps, the techniques are standard for M3 and the algebraic manipulation is straightforward once the method is identified. Harder than average due to the two-stage integration process, but not requiring novel insight. |
| Spec | 1.08h Integration by substitution6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| M1A1 | Replace \(v\) with \(\frac{dx}{dt}\) in either the answer in (a) (with or without taking square root first) or in an equivalent expression drawn from their working in (a) |
| dM1A1 | Separate the variables ready to integrate. Must reach an integrand which can be integrated. All correct so far |
| ddM1 | Attempt the integration by any valid means (e.g. inspection or substitution). Must include \(\sqrt{x+1} = A\sqrt{(x+1)^2 - 1} + B\sqrt{x^2 + 2x}\) or equivalent |
| A1cso | Correct integration (not ft). Constant may be omitted. |
| Answer | Marks |
|---|---|
| M1 | Use the given values of \(t\) to obtain the distances OA and OB and hence the distance AB. The function must have been integrated but third M mark may have been lost. |
| M1A1 | Correct distance. Exact or decimal accepted. Constant must have been included and shown to be 0. |
| Answer | Marks |
|---|---|
| M1 | Replace \(v\) with \(\frac{dx}{dt}\) in either the answer in (a) (with or without taking square root first) or in an equivalent expression drawn from their working in (a) |
| M1 | Separate the variables ready to integrate. Must reach an integrand which can be integrated. |
| A1 | All correct so far (ignore limits) |
| M1A1 | Attempt the integration by any valid means. Must include \(\sqrt{x+1} = A\sqrt{(x+1)^2 - 1} + B\sqrt{x^2 + 2x}\) or equivalent (integration only, ignore limits) |
| M1A1 | Correct integration (not ft). Constant may be omitted. |
| M1A1 | Substitute correct limits |
| A1cso | Correct completion. Distance \(= 8\sqrt{11} - 2\sqrt{11} = 9 - 2\sqrt{1}\) or \(4.417\) m (accept \(4.4\) or better) |
**5(a)**
M1A1 | Replace $v$ with $\frac{dx}{dt}$ in either the answer in (a) (with or without taking square root first) or in an equivalent expression drawn from their working in (a)
dM1A1 | Separate the variables ready to integrate. Must reach an integrand which can be integrated. All correct so far
ddM1 | Attempt the integration by any valid means (e.g. inspection or substitution). Must include $\sqrt{x+1} = A\sqrt{(x+1)^2 - 1} + B\sqrt{x^2 + 2x}$ or equivalent
A1cso | Correct integration (not ft). Constant may be omitted.
[6]
**5(b)**
M1 | Use the given values of $t$ to obtain the distances OA and OB and hence the distance AB. The function must have been integrated but third M mark may have been lost.
M1A1 | Correct distance. Exact or decimal accepted. Constant must have been included and shown to be 0.
Distance $= 8\sqrt{11} - 2\sqrt{11} = 9 - 2\sqrt{1}$ or $4.417$ m (accept $4.4$ or better)
[7]
**5(b) ALT: By definite integration**
M1 | Replace $v$ with $\frac{dx}{dt}$ in either the answer in (a) (with or without taking square root first) or in an equivalent expression drawn from their working in (a)
M1 | Separate the variables ready to integrate. Must reach an integrand which can be integrated.
A1 | All correct so far (ignore limits)
M1A1 | Attempt the integration by any valid means. Must include $\sqrt{x+1} = A\sqrt{(x+1)^2 - 1} + B\sqrt{x^2 + 2x}$ or equivalent (integration only, ignore limits)
M1A1 | Correct integration (not ft). Constant may be omitted.
M1A1 | Substitute correct limits
A1cso | Correct completion. Distance $= 8\sqrt{11} - 2\sqrt{11} = 9 - 2\sqrt{1}$ or $4.417$ m (accept $4.4$ or better)
[7]
[13]
---\begin{enumerate}
\item A particle $P$ of mass 0.8 kg moves along the $x$-axis in the positive $x$ direction under the action of a resultant force. This force acts in the direction of $x$ increasing. At time $t$ seconds, $t \geqslant 0 , P$ is $x$ metres from the origin $O , P$ is moving with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the force has magnitude $\frac { 4 } { ( x + 1 ) ^ { 3 } } \mathrm {~N}$.
\end{enumerate}
When $t = 0 , P$ is at rest at $O$.\\
(a) Show that $v ^ { 2 } = 5 \left( \frac { ( x + 1 ) ^ { 2 } - 1 } { ( x + 1 ) ^ { 2 } } \right)$
When $t = 2 , P$ is at the point $A$. When $t = 4 , P$ is at the point $B$.\\
(b) Using algebraic integration, find the distance $A B$.
\hfill \mbox{\textit{Edexcel M3 2018 Q5 [13]}}