| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Motion with exponential force |
| Difficulty | Standard +0.3 This is a standard M2 variable force question requiring Newton's second law in the form F=ma=mv(dv/dx), followed by integration with given boundary conditions. Part (i) is straightforward algebraic manipulation, and part (ii) involves separating variables and integrating exponential and polynomial terms—routine techniques for this module with no novel insight required. |
| Spec | 1.08h Integration by substitution3.03d Newton's second law: 2D vectors6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.8v\,dv/dx = 4e^{-x} - 2.4x^2\) | M1 | N2L, terms different signs |
| \(v\,dv/dx = 5e^{-x} - 3x^2\) AG | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int v\,dv = \int(5e^{-x} - 3x^2)\,dx\) | M1 | Attempts integration |
| \(v^2/2 = -5e^{-x} - 3x^3/3\ (+c)\) | A1 | Accept \(c\) omitted |
| \(x=0,\ v=6\), hence \(c = 23\) | B1 | Or uses limits 0 and 2 |
| \(v^2/2 = -5e^{-2} - 3x2^3/3 + 23\) | M1 | Puts \(x=2\) in \(v(x)\) expression |
| \(v = 5.35\ \text{ms}^{-1}\) | A1 [5] | \(v = 5.352...\) |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.8v\,dv/dx = 4e^{-x} - 2.4x^2$ | M1 | N2L, terms different signs |
| $v\,dv/dx = 5e^{-x} - 3x^2$ AG | A1 [2] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int v\,dv = \int(5e^{-x} - 3x^2)\,dx$ | M1 | Attempts integration |
| $v^2/2 = -5e^{-x} - 3x^3/3\ (+c)$ | A1 | Accept $c$ omitted |
| $x=0,\ v=6$, hence $c = 23$ | B1 | Or uses limits 0 and 2 |
| $v^2/2 = -5e^{-2} - 3x2^3/3 + 23$ | M1 | Puts $x=2$ in $v(x)$ expression |
| $v = 5.35\ \text{ms}^{-1}$ | A1 [5] | $v = 5.352...$ |
---3 A particle $P$ of mass 0.8 kg moves along the $x$-axis on a horizontal surface. When the displacement of $P$ from the origin $O$ is $x \mathrm {~m}$ the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the positive $x$-direction. Two horizontal forces act on $P$. One force has magnitude $4 \mathrm { e } ^ { - x } \mathrm {~N}$ and acts in the positive $x$-direction. The other force has magnitude $2.4 x ^ { 2 } \mathrm {~N}$ and acts in the negative $x$-direction.\\
(i) Show that $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 5 \mathrm { e } ^ { - x } - 3 x ^ { 2 }$.\\
(ii) The velocity of $P$ as it passes through $O$ is $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the velocity of $P$ when $x = 2$.
\hfill \mbox{\textit{CAIE M2 2013 Q3 [7]}}