| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on slope then horizontal |
| Difficulty | Standard +0.3 This is a multi-part work-energy question requiring application of conservation of energy and work-energy principles across different surface sections. While it involves several steps and geometric reasoning (finding heights, understanding the path), the techniques are standard M1 fare: using KE + PE conservation on smooth surfaces and work done against friction. The question guides students through each stage clearly, making it slightly easier than average for A-level mechanics. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| PE gain \(= mg(2.5\sin60°)\) | B1 |
| For using KE \(= \frac{1}{2}mv^2\) | M1 |
| For using the principle of conservation of energy \(\left(\frac{1}{2}m8^2 - \frac{1}{2}mv^2 = mg(2.5\sin60°)\right)\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| For using Newton's Second Law or stating \(a = -g\sin60°\) | M1* | |
| \(a = -8.66\) (may be implied) | A1 | |
| For using \(v^2 = u^2 + 2as\) \((v^2 = 64 - 2 \times 8.66 \times 2.5)\) | M1dep* | |
| Speed is 4.55 ms\(^{-1}\) | A1 | Accept 4.64 from 9.8 or 9.81 |
| Answer | Marks |
|---|---|
| For using \(\frac{1}{2}mu^2 > mgh_{\max}\) \(\left(\frac{1}{2}8^2 > 10h_{\max}\right)\) | M1 |
| For obtaining 3.2 m A.G. | A1 |
| Answer | Marks |
|---|---|
| Energy is conserved or absence of friction or curve \(BC\) is smooth (or equivalent) and \(B\) and \(C\) are at the same height or the PE is the same at \(A\) and \(B\) (or equivalent) | B1 |
| Answer | Marks |
|---|---|
| WD against friction is \(1.4 \times 5.2\) | B1 |
| For WD \(=\) KE loss (or equivalent) used | M1 |
| \(1.4 \times 5.2 = \frac{1}{2}0.4(8^2 - v^2)\) or \(1.4 \times 5.2 = \frac{1}{2}0.4((i)^2 - v^2) + 0.4 \times 10(2.5\sin60°)\) \((12.8\) or \(4.14 + 8.66)\) | A1 |
| Answer | Marks |
|---|---|
| For using Newton's Second Law \(0.4g(2.5\sin60° \div 5.2) - 1.4 = 0.4a\) \((a = 0.6636)\) | M1* |
| A1 | |
| For using \(v^2 = u^2 + 2as\) with \(u \neq 0\) \((v^2 = 4.55^2 + 2 \times 0.6636 \times 5.2)\) | M1dep* |
| Speed is 5.25 ms\(^{-1}\) | A1 |
# Question 7:
## Part (i)
| PE gain $= mg(2.5\sin60°)$ | B1 | |
| For using KE $= \frac{1}{2}mv^2$ | M1 | |
| For using the principle of conservation of energy $\left(\frac{1}{2}m8^2 - \frac{1}{2}mv^2 = mg(2.5\sin60°)\right)$ | M1 | |
**Alternative** for the above 3 marks:
| For using Newton's Second Law or stating $a = -g\sin60°$ | M1* | |
| $a = -8.66$ (may be implied) | A1 | |
| For using $v^2 = u^2 + 2as$ $(v^2 = 64 - 2 \times 8.66 \times 2.5)$ | M1dep* | |
| Speed is 4.55 ms$^{-1}$ | A1 | Accept 4.64 from 9.8 or 9.81 |
## Part (ii)
| For using $\frac{1}{2}mu^2 > mgh_{\max}$ $\left(\frac{1}{2}8^2 > 10h_{\max}\right)$ | M1 | |
| For obtaining 3.2 m A.G. | A1 | |
## Part (iii)
| Energy is conserved or absence of friction or curve $BC$ is smooth (or equivalent) and $B$ and $C$ are at the same height or the PE is the same at $A$ and $B$ (or equivalent) | B1 | |
## Part (iv)
| WD against friction is $1.4 \times 5.2$ | B1 | |
| For WD $=$ KE loss (or equivalent) used | M1 | |
| $1.4 \times 5.2 = \frac{1}{2}0.4(8^2 - v^2)$ or $1.4 \times 5.2 = \frac{1}{2}0.4((i)^2 - v^2) + 0.4 \times 10(2.5\sin60°)$ $(12.8$ or $4.14 + 8.66)$ | A1 | |
**Alternative** for the above 3 marks:
| For using Newton's Second Law $0.4g(2.5\sin60° \div 5.2) - 1.4 = 0.4a$ $(a = 0.6636)$ | M1* | |
| | A1 | |
| For using $v^2 = u^2 + 2as$ with $u \neq 0$ $(v^2 = 4.55^2 + 2 \times 0.6636 \times 5.2)$ | M1dep* | |
| Speed is 5.25 ms$^{-1}$ | A1 | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{cb04a09c-af23-4e9d-b3da-da9e351fe879-4_257_988_267_580}
The diagram shows a vertical cross-section $A B C D$ of a surface. The parts $A B$ and $C D$ are straight and have lengths 2.5 m and 5.2 m respectively. $A D$ is horizontal, and $A B$ is inclined at $60 ^ { \circ }$ to the horizontal. The points $B$ and $C$ are at the same height above $A D$. The parts of the surface containing $A B$ and $B C$ are smooth. A particle $P$ is given a velocity of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $A$, in the direction $A B$, and it subsequently reaches $D$. The particle does not lose contact with the surface during this motion.\\
(i) Find the speed of $P$ at $B$.\\
(ii) Show that the maximum height of the cross-section, above $A D$, is less than 3.2 m .\\
(iii) State briefly why $P$ 's speed at $C$ is the same as its speed at $B$.\\
(iv) The frictional force acting on the particle as it travels from $C$ to $D$ is 1.4 N . Given that the mass of $P$ is 0.4 kg , find the speed with which $P$ reaches $D$.
\hfill \mbox{\textit{CAIE M1 2003 Q7 [11]}}