| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance with other powers |
| Difficulty | Standard +0.8 This question requires setting up and solving a differential equation with a non-standard resistance force (proportional to √v rather than v or v²), then integrating to find distance. The separation of variables and algebraic manipulation needed to reach the given form v = ¼(t-6)² is non-trivial, requiring careful handling of square roots and integration. While methodical, it goes beyond routine mechanics problems and demands strong calculus skills alongside mechanics understanding. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(dv/dt = -2.5k\sqrt{v}\); \(\int v^{-0.5}dv = -2.5k\int dt\); \(v^{0.5}/0.5 = -2.5kt\) (+ c); \(t = 0, v = 9\) hence c = 6 and \(2 = 4\) hence k = 0.4; \(v = (6 - t)^2/4 = (t - 6)^2/4\) | B1, M1, A1, M1, A1 [5] | \(0.4dv/dt = -k\sqrt{v}\); LHS = \(0.8\sqrt{v}\); \(\sqrt{v} = (6 - t)/2\); Uses correct limits; AG |
| (ii) \(x = \int(t - 6)^2/4dt\); \(x = (t - 6)^3/(3 \times 4)\) (+ c); \(t = 0, x = 0\) hence c = 18; \(x(3) = 18 - (3 - 6)^3/12\); \(x(3) = 15.75\) OR \(\int v^2dv = \int -dx\); \(\frac{2}{3}v^{3/2} = -x\) (+ c); \(x = 18 - \frac{2}{3}v^{1/2}\); \(x = 15.75\) | M1, A1, M1, A1 [4] | \(\int(6 - t)^2/4dt\); \(-(6 - t)^3/(3 \times 4)\) (+ c); Or uses limits 0, 3; Accept 15.7 or 15.8; From \(mvdv/dx = -k\sqrt{v}\); Using \(v = 9, x = 0\) so c = 18; Put t = 3 to find v = 2.25 |
(i) $dv/dt = -2.5k\sqrt{v}$; $\int v^{-0.5}dv = -2.5k\int dt$; $v^{0.5}/0.5 = -2.5kt$ (+ c); $t = 0, v = 9$ hence c = 6 and $2 = 4$ hence k = 0.4; $v = (6 - t)^2/4 = (t - 6)^2/4$ | B1, M1, A1, M1, A1 [5] | $0.4dv/dt = -k\sqrt{v}$; LHS = $0.8\sqrt{v}$; $\sqrt{v} = (6 - t)/2$; Uses correct limits; AG
(ii) $x = \int(t - 6)^2/4dt$; $x = (t - 6)^3/(3 \times 4)$ (+ c); $t = 0, x = 0$ hence c = 18; $x(3) = 18 - (3 - 6)^3/12$; $x(3) = 15.75$ OR $\int v^2dv = \int -dx$; $\frac{2}{3}v^{3/2} = -x$ (+ c); $x = 18 - \frac{2}{3}v^{1/2}$; $x = 15.75$ | M1, A1, M1, A1 [4] | $\int(6 - t)^2/4dt$; $-(6 - t)^3/(3 \times 4)$ (+ c); Or uses limits 0, 3; Accept 15.7 or 15.8; From $mvdv/dx = -k\sqrt{v}$; Using $v = 9, x = 0$ so c = 18; Put t = 3 to find v = 2.255 A particle $P$ of mass 0.4 kg moves in a straight line on a horizontal surface and has velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t \mathrm {~s}$. A horizontal force of magnitude $k \sqrt { } v \mathrm {~N}$ opposes the motion of $P$. When $t = 0 , v = 9$ and when $t = 2 , v = 4$.\\
(i) Express $\frac { \mathrm { d } v } { \mathrm {~d} t }$ in terms of $k$ and $v$, and hence show that $v = \frac { 1 } { 4 } ( t - 6 ) ^ { 2 }$.\\
(ii) Find the distance travelled by $P$ in the first 3 seconds of its motion.
\hfill \mbox{\textit{CAIE M2 2011 Q5 [9]}}