| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Standard +0.8 This M3 question requires chain rule differentiation (v dv/dx = a), algebraic manipulation with surds, and integration of a reciprocal square root function. While the techniques are standard for M3, the combination of steps and careful algebraic handling of the surd expressions elevates it above routine exercises. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{1}{\sqrt{(2x+1)}} = \frac{1}{3} \Rightarrow x = 4\) | M1A1 | M1 for putting \(v = \frac{1}{3}\) and solving for \(x\) e.g. \((2x+1)=9\) |
| \(a = v\frac{dv}{dx} = \frac{1}{\sqrt{(2x+1)}} \times -\frac{1}{2} \times 2(2x+1)^{-\frac{3}{2}}\) \(\left(= -(2x+1)^{-2}\right)\) | M1A1 | M1 for use of \(v\frac{dv}{dx}\) with clear attempt at differentiation; power \(-\frac{3}{2}\) needed |
| \(a = (-)\frac{1}{81}\) | A1 | Correct unsimplified expression for \(a\) in terms of \(x\) only |
| \(\frac{2}{81}\) (N), 0.025 or better | A1 (6) | cao (must be positive) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(t = \int \sqrt{(2x+1)}\, dx\) | M1 | Use of \(\frac{dx}{dt} = \frac{1}{\sqrt{(2x+1)}}\), separate variables and attempt to integrate; \(k(2x+1)^{\frac{3}{2}}\) should be seen |
| \(t = \frac{2}{3} \times \frac{1}{2}(2x+1)^{\frac{3}{2}} + (C)\) | A1 | Correct unsimplified integration, \(C\) not needed |
| Use of \(t=0\), \(x=0\) to obtain \(C = -\frac{1}{3}\) | M1 | M1 for use of initial conditions to find value for \(C\) |
| Substitute \(x = 7.5\) | dM1 | Dependent on first M1; for definite integration, award M1 for substitution of lower limits and DM1 for substitution of upper limits |
| \(t = 21\) | A1 (5) | cao |
# Question 2:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{1}{\sqrt{(2x+1)}} = \frac{1}{3} \Rightarrow x = 4$ | M1A1 | M1 for putting $v = \frac{1}{3}$ and solving for $x$ e.g. $(2x+1)=9$ |
| $a = v\frac{dv}{dx} = \frac{1}{\sqrt{(2x+1)}} \times -\frac{1}{2} \times 2(2x+1)^{-\frac{3}{2}}$ $\left(= -(2x+1)^{-2}\right)$ | M1A1 | M1 for use of $v\frac{dv}{dx}$ with clear attempt at differentiation; power $-\frac{3}{2}$ needed |
| $a = (-)\frac{1}{81}$ | A1 | Correct unsimplified expression for $a$ in terms of $x$ only |
| $\frac{2}{81}$ (N), 0.025 or better | A1 (6) | cao (must be positive) |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $t = \int \sqrt{(2x+1)}\, dx$ | M1 | Use of $\frac{dx}{dt} = \frac{1}{\sqrt{(2x+1)}}$, separate variables and attempt to integrate; $k(2x+1)^{\frac{3}{2}}$ should be seen |
| $t = \frac{2}{3} \times \frac{1}{2}(2x+1)^{\frac{3}{2}} + (C)$ | A1 | Correct unsimplified integration, $C$ not needed |
| Use of $t=0$, $x=0$ to obtain $C = -\frac{1}{3}$ | M1 | M1 for use of initial conditions to find value for $C$ |
| Substitute $x = 7.5$ | dM1 | Dependent on first M1; for definite integration, award M1 for substitution of lower limits and DM1 for substitution of upper limits |
| $t = 21$ | A1 (5) | cao |
---2. In this question solutions relying on calculator technology are not acceptable.
A particle $P$ of mass 2 kg is moving along the positive $x$-axis.\\
At time $t$ seconds, where $t \geqslant 0 , P$ is $x$ metres from the origin $O$ and is moving away from $O$ with speed $v \mathrm {~ms} ^ { - 1 }$ where $v = \frac { 1 } { \sqrt { ( 2 x + 1 ) } }$
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the resultant force acting on $P$ when its speed is $\frac { 1 } { 3 } \mathrm {~ms} ^ { - 1 }$
When $t = 0 , P$ is at $O$
\item Find the value of $t$ when $P$ is 7.5 m from $O$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2021 Q2 [11]}}