| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force on horizontal road |
| Difficulty | Moderate -0.3 This is a straightforward work-energy problem requiring standard application of formulas (KE = ½mv², work = force × distance) and the work-energy principle. All three parts follow directly from given information with no conceptual difficulty or novel insight required, making it slightly easier than average for A-level mechanics. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times 50 \times 2^2 = 100\) J | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(250 \times 20 = 5000\) J | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| WD by the force \(= 5100\) J or \(a = \frac{1}{10}\) | B1 | |
| \(5100 = 400 \times 20\cos\alpha\) or \(400\cos\alpha - 250 = 50 \times \frac{1}{10}\) | M1 | For using WD by force \(= Fd\cos\alpha\), or for using Newton's second law (3 terms required) |
| \(\alpha = 50.4°\) | A1 |
## Question 2:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 50 \times 2^2 = 100$ J | B1 | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $250 \times 20 = 5000$ J | B1 | |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| WD by the force $= 5100$ J or $a = \frac{1}{10}$ | B1 | |
| $5100 = 400 \times 20\cos\alpha$ or $400\cos\alpha - 250 = 50 \times \frac{1}{10}$ | M1 | For using WD by force $= Fd\cos\alpha$, or for using Newton's second law (3 terms required) |
| $\alpha = 50.4°$ | A1 | |
---2 A crate of mass 50 kg is dragged along a horizontal floor by a constant force of magnitude 400 N acting at an angle $\alpha ^ { \circ }$ upwards from the horizontal. The total resistance to motion of the crate has constant magnitude 250 N . The crate starts from rest at the point $O$ and passes the point $P$ with a speed of $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The distance $O P$ is 20 m . For the crate's motion from $O$ to $P$, find\\
(i) the increase in kinetic energy of the crate,\\
(ii) the work done against the resistance to the motion of the crate,\\
(iii) the value of $\alpha$.
\hfill \mbox{\textit{CAIE M1 2005 Q2 [5]}}