| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on smooth curved surface |
| Difficulty | Standard +0.3 This is a straightforward energy conservation problem on a smooth surface. Part (i) uses basic PE to KE conversion with simple trigonometry, (ii) applies conservation of energy between equal heights, and (iii) uses the given maximum speed at the lowest point. All steps are standard M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([KE = \text{Loss of PE} = 0.8g(2.4\sin50°)\), \(KE = \frac{1}{2}(0.8) \times 2(g\sin50°)(2.4)]\) | M1 | For using \(KE = PE\) loss \(= mgh\) or \(KE = \frac{1}{2}mv^2\) and \(v^2 = 2as\) |
| Kinetic energy at A is 14.7 J | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([14.7 = \frac{1}{2}mv^2]\) | M1 | For using KE at C \(=\) KE at A \(= \frac{1}{2}mv^2\) |
| Speed at C is \(6.06 \text{ ms}^{-1}\) | A1ft [2] | ft \(v = (2.5\ KE)^{\frac{1}{2}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\frac{1}{2}m8^2 = mgH,\ \frac{1}{2}m8^2 - \frac{1}{2}m(6.06)^2 = mgh]\) | M1 | For using principle of conservation of energy |
| \(h = 3.2 - 2.4\sin50°\) or \(10h = \frac{1}{2}(8^2 - 6.06^2)\) | A1ft | ft \(10h = \frac{1}{2}(8^2 - v_c^2)\) |
| Depth is 1.36 m | A1 [3] | |
| SR in (iii) (max. mark 1/3): For depth \(= 1.36\) from \(v^2 = u^2 + 2gs\) | B1 |
## Question 4:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[KE = \text{Loss of PE} = 0.8g(2.4\sin50°)$, $KE = \frac{1}{2}(0.8) \times 2(g\sin50°)(2.4)]$ | M1 | For using $KE = PE$ loss $= mgh$ **or** $KE = \frac{1}{2}mv^2$ and $v^2 = 2as$ |
| Kinetic energy at A is 14.7 J | A1 [2] | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[14.7 = \frac{1}{2}mv^2]$ | M1 | For using KE at C $=$ KE at A $= \frac{1}{2}mv^2$ |
| Speed at C is $6.06 \text{ ms}^{-1}$ | A1ft [2] | ft $v = (2.5\ KE)^{\frac{1}{2}}$ |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\frac{1}{2}m8^2 = mgH,\ \frac{1}{2}m8^2 - \frac{1}{2}m(6.06)^2 = mgh]$ | M1 | For using principle of conservation of energy |
| $h = 3.2 - 2.4\sin50°$ **or** $10h = \frac{1}{2}(8^2 - 6.06^2)$ | A1ft | ft $10h = \frac{1}{2}(8^2 - v_c^2)$ |
| Depth is 1.36 m | A1 [3] | |
| SR in **(iii)** (max. mark 1/3): For depth $= 1.36$ from $v^2 = u^2 + 2gs$ | B1 | |
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$O A B C$ is a vertical cross-section of a smooth surface. The straight part $O A$ has length 2.4 m and makes an angle of $50 ^ { \circ }$ with the horizontal. $A$ and $C$ are at the same horizontal level and $B$ is the lowest point of the cross-section (see diagram). A particle $P$ of mass 0.8 kg is released from rest at $O$ and moves on the surface. $P$ remains in contact with the surface until it leaves the surface at $C$. Find\\
(i) the kinetic energy of $P$ at $A$,\\
(ii) the speed of $P$ at $C$.
The greatest speed of $P$ is $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(iii) Find the depth of $B$ below the horizontal through $A$ and $C$.
\hfill \mbox{\textit{CAIE M1 2008 Q4 [7]}}