| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force up incline, find KE/PE changes as sub-parts |
| Difficulty | Standard +0.3 This is a straightforward multi-part work-energy question requiring standard formulas (KE = ½mv², PE = mgh, work = force × distance) and the work-energy principle. Part (iv) involves resolving forces but follows directly from equating work done to energy changes. Slightly above average due to the four-part structure and final angle calculation, but all steps are routine applications of standard mechanics principles. |
| Spec | 3.03a Force: vector nature and diagrams3.03e Resolve forces: two dimensions6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{1}{2} \times 50(t - 3^2)\) | M1 | For using loss in KE \(= \frac{1}{2}mv^2 - \frac{1}{2}mv^2\) |
| Loss is \(1000\) J | A1 | 2 |
| (ii) \(50 \times 10 \times 15\) | M1 | For using 'Gain in PE \(= mgh\)' |
| Gain is \(7500\) J | A1 | 2 |
| (iii) WD by pulling force \(= 7500 + 1500 - 1000\) | M1 | For using WD against resistance \(= 7.5 \times 200\) and attempting to find WD by the pulling force as a linear combination of the three relevant components |
| (iv) Work done is \(8000\) J | A1 ft | 2 |
| \(8000 = 45 \times 200\cos \alpha\) | M1 | Any correct numerical form of \(\cos \alpha\) |
| \(\cos \alpha = \frac{8000}{45 \times 200}\) | ||
| \(\alpha = 27.3\) | A1 | 3 |
**(i)** $\frac{1}{2} \times 50(t - 3^2)$ | M1 | For using loss in KE $= \frac{1}{2}mv^2 - \frac{1}{2}mv^2$ |
Loss is $1000$ J | A1 | 2 |
**(ii)** $50 \times 10 \times 15$ | M1 | For using 'Gain in PE $= mgh$' |
Gain is $7500$ J | A1 | 2 |
**(iii)** WD by pulling force $= 7500 + 1500 - 1000$ | M1 | For using WD against resistance $= 7.5 \times 200$ and attempting to find WD by the pulling force as a linear combination of the three relevant components |
**(iv)** Work done is $8000$ J | A1 ft | 2 | For using WD $= F\cos \alpha$ |
$8000 = 45 \times 200\cos \alpha$ | M1 | Any correct numerical form of $\cos \alpha$ |
$\cos \alpha = \frac{8000}{45 \times 200}$ | | |
$\alpha = 27.3$ | A1 | 3 |
---6 A block of mass 50 kg is pulled up a straight hill and passes through points $A$ and $B$ with speeds $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively. The distance $A B$ is 200 m and $B$ is 15 m higher than $A$. For the motion of the block from $A$ to $B$, find\\
(i) the loss in kinetic energy of the block,\\
(ii) the gain in potential energy of the block.
The resistance to motion of the block has magnitude 7.5 N.\\
(iii) Find the work done by the pulling force acting on the block.
The pulling force acting on the block has constant magnitude 45 N and acts at an angle $\alpha ^ { \circ }$ upwards from the hill.\\
(iv) Find the value of $\alpha$.
\hfill \mbox{\textit{CAIE M1 2006 Q6 [9]}}