| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Difficulty | Standard +0.3 This is a straightforward application of the work-energy principle with clearly stated values. Part (i) requires calculating work against resistance and gravity at constant speed (so driving force = resistance + component down slope). Part (ii) uses work-energy theorem with given values to find distance. Both parts follow standard M1 procedures with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gain in \(PE = 15000g \times 16\) | B1 | |
| \(WD\) against resistance \(= 1800 \times 1440\) | B1 | |
| M1 | For using: \(WD\) by driving force \(=\) Gain in PE \(+\) WD against resistance | |
| Work done is \(4.99 \times 10^6\) J | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using: \(WD\) by engine \(=\) Increase in KE \(+\) WD against resistance | |
| \(5030\ 000 = \frac{1}{2} \times 15\ 000(24^2 - 15^2) + 1600d\) | A1 | |
| Distance is \(1500\) m | A1 [3] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gain in $PE = 15000g \times 16$ | B1 | |
| $WD$ against resistance $= 1800 \times 1440$ | B1 | |
| | M1 | For using: $WD$ by driving force $=$ Gain in PE $+$ WD against resistance |
| Work done is $4.99 \times 10^6$ J | A1 [4] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using: $WD$ by engine $=$ Increase in KE $+$ WD against resistance |
| $5030\ 000 = \frac{1}{2} \times 15\ 000(24^2 - 15^2) + 1600d$ | A1 | |
| Distance is $1500$ m | A1 [3] | |
---5 A lorry of mass 15000 kg climbs from the bottom to the top of a straight hill, of length 1440 m , at a constant speed of $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The top of the hill is 16 m above the level of the bottom of the hill. The resistance to motion is constant and equal to 1800 N .\\
(i) Find the work done by the driving force.
On reaching the top of the hill the lorry continues on a straight horizontal road and passes through a point $P$ with speed $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The resistance to motion is constant and is now equal to 1600 N . The work done by the lorry's engine from the top of the hill to the point $P$ is 5030 kJ .\\
(ii) Find the distance from the top of the hill to the point $P$.
\hfill \mbox{\textit{CAIE M1 2013 Q5 [7]}}