| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of velocity (separation of variables) |
| Difficulty | Standard +0.8 This is a Further Maths mechanics question requiring integration with variable acceleration expressed as a function of velocity. Part (a) needs separation of variables (dv/dt = 4/(2+v)) and integration with initial conditions, while part (b) requires using v dv/dx and integrating again. The techniques are standard for FM2 but require careful algebraic manipulation and understanding of different forms of acceleration, making it moderately challenging but within expected Further Maths scope. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = \dfrac{4}{2+v} \Rightarrow \int(2+v)\,dv = \int 4\,dt\) | M1 | Form differential equation in \(v\) and \(t\) and prepare to integrate |
| \(\dfrac{(2+v)^2}{2} = 4t + C_1\) | M1 | Integrate to obtain \(k(2+v)^2\) or equivalent |
| A1 | Correct integration. Condone missing constant of integration. | |
| \(t=0, v=2 \Rightarrow C_1 = 8\) | M1 | Use the model to find the value of constant of integration |
| \(\dfrac{(2+v)^2}{2} = 4t + 8\) | A1 | Correct solution in any form |
| \((2+v)^2 = 8t+16, \quad v = \sqrt{8t+16} - 2\) * | A1* | Obtain given solution from correct working. Allow use of quadratic formula. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = \dfrac{4}{2+v} \Rightarrow \int(2+v)\,dv = \int 4\,dt\) | M1 | Form differential equation in \(v\) and \(t\) and prepare to integrate |
| \(2v + \dfrac{v^2}{2} = 4t + C_2\) | M1 | Integrate to obtain \(k(2+v)^2\) or equivalent |
| A1 | Correct integration. Condone missing constant of integration. | |
| \(t=0, v=2 \Rightarrow C_2 = 6\) | M1 | Use the model to find value of constant of integration |
| \(2v + \dfrac{v^2}{2} = 4t + 6\) | A1 | Correct solution in any form |
| \(4v + v^2 = 8t+12, \quad (v+2)^2 = 8t+16 \Rightarrow v = \sqrt{8t+16}-2\) * | A1* | Obtain given solution from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v=4 \Rightarrow 36 = 8t+16 \Rightarrow t = 2.5\) | B1 | Use result from (a) to find \(t\) when \(v=4\): seen or implied |
| \(\dfrac{dx}{dt} = \sqrt{8t+16} - 2\) | M1 | Form differential equation in \(x\) and \(t\) |
| \(x = k(8t+16)^{\frac{3}{2}} - 2t + C\) | M1 | Integrate to obtain terms of correct form. Condone missing constant of integration. |
| \(x = \dfrac{1}{12}(8t+16)^{\frac{3}{2}} - 2t + C\) | A1 | Correct integration. Condone missing constant of integration. |
| \(t=0, x=0 \Rightarrow \dfrac{64}{12} + C = 0, \quad C = -\dfrac{16}{3}\) | M1 | Use boundary conditions to find constant. Note independent M mark. M0 if \(t=4\) used. |
| \(AB = \dfrac{1}{12}(36)^{\frac{3}{2}} - 5 - \dfrac{16}{3} = \dfrac{23}{3}\) (m) | A1 | Correct answer only. 7.7 (m) or better |
# Question 2:
## Part 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = \dfrac{4}{2+v} \Rightarrow \int(2+v)\,dv = \int 4\,dt$ | M1 | Form differential equation in $v$ and $t$ and prepare to integrate |
| $\dfrac{(2+v)^2}{2} = 4t + C_1$ | M1 | Integrate to obtain $k(2+v)^2$ or equivalent |
| | A1 | Correct integration. Condone missing constant of integration. |
| $t=0, v=2 \Rightarrow C_1 = 8$ | M1 | Use the model to find the value of constant of integration |
| $\dfrac{(2+v)^2}{2} = 4t + 8$ | A1 | Correct solution in any form |
| $(2+v)^2 = 8t+16, \quad v = \sqrt{8t+16} - 2$ * | A1* | Obtain **given solution** from correct working. Allow use of quadratic formula. |
**Total: (6)**
## Part 2(a) alt:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = \dfrac{4}{2+v} \Rightarrow \int(2+v)\,dv = \int 4\,dt$ | M1 | Form differential equation in $v$ and $t$ and prepare to integrate |
| $2v + \dfrac{v^2}{2} = 4t + C_2$ | M1 | Integrate to obtain $k(2+v)^2$ or equivalent |
| | A1 | Correct integration. Condone missing constant of integration. |
| $t=0, v=2 \Rightarrow C_2 = 6$ | M1 | Use the model to find value of constant of integration |
| $2v + \dfrac{v^2}{2} = 4t + 6$ | A1 | Correct solution in any form |
| $4v + v^2 = 8t+12, \quad (v+2)^2 = 8t+16 \Rightarrow v = \sqrt{8t+16}-2$ * | A1* | Obtain **given solution** from correct working |
**Total: (6)**
## Part 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v=4 \Rightarrow 36 = 8t+16 \Rightarrow t = 2.5$ | B1 | Use result from (a) to find $t$ when $v=4$: seen or implied |
| $\dfrac{dx}{dt} = \sqrt{8t+16} - 2$ | M1 | Form differential equation in $x$ and $t$ |
| $x = k(8t+16)^{\frac{3}{2}} - 2t + C$ | M1 | Integrate to obtain terms of correct form. Condone missing constant of integration. |
| $x = \dfrac{1}{12}(8t+16)^{\frac{3}{2}} - 2t + C$ | A1 | Correct integration. Condone missing constant of integration. |
| $t=0, x=0 \Rightarrow \dfrac{64}{12} + C = 0, \quad C = -\dfrac{16}{3}$ | M1 | Use boundary conditions to find constant. Note independent M mark. M0 if $t=4$ used. |
| $AB = \dfrac{1}{12}(36)^{\frac{3}{2}} - 5 - \dfrac{16}{3} = \dfrac{23}{3}$ (m) | A1 | Correct answer only. 7.7 (m) or better |
**Total: (6)**
---\begin{enumerate}
\item A car moves in a straight line along a horizontal road. The car is modelled as a particle. At time $t$ seconds, where $t \geqslant 0$, the speed of the car is $v \mathrm {~ms} ^ { - 1 }$
\end{enumerate}
At the instant when $t = 0$, the car passes through the point $A$ with speed $2 \mathrm {~ms} ^ { - 1 }$\\
The acceleration, $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, of the car is modelled by
$$a = \frac { 4 } { 2 + v }$$
in the direction of motion of the car.\\
(a) Use algebraic integration to show that $v = \sqrt { 8 t + 16 } - 2$
At the instant when the car passes through the point $B$, the speed of the car is $4 \mathrm {~ms} ^ { - 1 }$\\
(b) Use algebraic integration to find the distance $A B$.
\hfill \mbox{\textit{Edexcel FM2 AS 2019 Q2 [12]}}