| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: speed at given height |
| Difficulty | Standard +0.3 This is a straightforward mechanics question testing energy conservation and SUVAT equations. Part (i) involves standard kinetic/potential energy calculations with given values, requiring only direct substitution. Part (ii) uses the same principles with a ratio condition, needing one additional algebraic step. These are routine A-level mechanics exercises with clear methods and no novel problem-solving required. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| (i)(a) \(PE \text{ loss} = 0.2g(3 - h)\) | B1 | |
| \([0.2g(3 - h) = 1.6]\) | M1 | For using \(PE \text{ loss} = KE \text{ gain}\) |
| \(h = 2.2\) | A1 | [3] |
| (i)(b) \(KE \text{ is } 6J\) | B1 | [1] |
| (i)(c) \([v_G / v_B = (3/(3-2.2))^{\frac{1}{2}}\) or \(v_G / v_B = \sqrt{6/1.6}]\) | M1 | For using \(v^2 \propto (3 - ht)\) or \((v_G / v_B)^2 = \text{Ans. (i)(b)} \div 1.6\) |
| Ratio is 1.94 | A1 | [2] |
| (ii) For using \(v^2 \propto (H - ht)\) or using \(\frac{1}{2}m(2.55v_B)^2 = mgH\) and \(\frac{1}{2}mv_B^2 = mg(H - 2.2)\) and eliminating \(v_B^2\) | M1 | |
| \(H/(H - 2.2) = 2.55^2\) | A1 | |
| \(H = 2.6\) | A1 | [3] |
**(i)(a)** $PE \text{ loss} = 0.2g(3 - h)$ | B1 |
$[0.2g(3 - h) = 1.6]$ | M1 | For using $PE \text{ loss} = KE \text{ gain}$
$h = 2.2$ | A1 | [3]
**(i)(b)** $KE \text{ is } 6J$ | B1 | [1]
**(i)(c)** $[v_G / v_B = (3/(3-2.2))^{\frac{1}{2}}$ or $v_G / v_B = \sqrt{6/1.6}]$ | M1 | For using $v^2 \propto (3 - ht)$ or $(v_G / v_B)^2 = \text{Ans. (i)(b)} \div 1.6$
Ratio is 1.94 | A1 | [2] | Accept $\sqrt{60} \div 4$ or $\sqrt{15} \div 2$
**(ii)** For using $v^2 \propto (H - ht)$ or using $\frac{1}{2}m(2.55v_B)^2 = mgH$ and $\frac{1}{2}mv_B^2 = mg(H - 2.2)$ and eliminating $v_B^2$ | M1 |
$H/(H - 2.2) = 2.55^2$ | A1 |
$H = 2.6$ | A1 | [3]6\\
\includegraphics[max width=\textwidth, alt={}, center]{5125fab5-0be5-4904-afdf-93e91b16e773-3_476_1305_1519_420}
A smooth slide $A B$ is fixed so that its highest point $A$ is 3 m above horizontal ground. $B$ is $h \mathrm {~m}$ above the ground. A particle $P$ of mass 0.2 kg is released from rest at a point on the slide. The particle moves down the slide and, after passing $B$, continues moving until it hits the ground (see diagram). The speed of $P$ at $B$ is $v _ { B }$ and the speed at which $P$ hits the ground is $v _ { G }$.\\
(i) In the case that $P$ is released at $A$, it is given that the kinetic energy of $P$ at $B$ is 1.6 J . Find
\begin{enumerate}[label=(\alph*)]
\item the value of $h$,
\item the kinetic energy of the particle immediately before it reaches the ground,
\item the ratio $v _ { G } : v _ { B }$.\\
(ii) In the case that $P$ is released at the point $X$ of the slide, which is $H \mathrm {~m}$ above the ground (see diagram), it is given that $v _ { G } : v _ { B } = 2.55$. Find the value of $H$ correct to 2 significant figures.\\
\includegraphics[max width=\textwidth, alt={}, center]{5125fab5-0be5-4904-afdf-93e91b16e773-4_384_679_258_733}
Particles $P$ and $Q$, of masses 0.2 kg and 0.5 kg respectively, are connected by a light inextensible string. The string passes over a smooth pulley at the edge of a rough horizontal table. $P$ hangs freely and $Q$ is in contact with the table. A force of magnitude 3.2 N acts on $Q$, upwards and away from the pulley, at an angle of $30 ^ { \circ }$ to the horizontal (see diagram).
\begin{enumerate}[label=(\roman*)]
\item The system is in limiting equilibrium with $P$ about to move upwards. Find the coefficient of friction between $Q$ and the table.
The force of magnitude 3.2 N is now removed and $P$ starts to move downwards.
\item Find the acceleration of the particles and the tension in the string.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2010 Q6 [9]}}