| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Constant speed up/down incline |
| Difficulty | Standard +0.3 This is a straightforward work-energy-power question requiring standard formulas and bookwork methods. Part (i) involves calculating power from work done against resistance plus gravitational PE gain. Part (ii) applies work-energy principle with given values. Both parts are routine applications with clear problem structure and no novel insight required, making it slightly easier than average. |
| Spec | 6.02a Work done: concept and definition6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks |
|---|---|
| 5(i) | EITHER: |
| Answer | Marks |
|---|---|
| 25 | (B1 |
| Answer | Marks | Guidance |
|---|---|---|
| = 32 N | B1 | For correct unsimplified numerical form |
| Answer | Marks | Guidance |
|---|---|---|
| [Power = 56 × 4] | M1 | For use of P = Fv where F is from two |
| Answer | Marks |
|---|---|
| Power = 224 W | A1) |
| Answer | Marks | Guidance |
|---|---|---|
| = 800 | (B1 | For a correct unsimplified numerical |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | B1 | |
| [WD by cyclist = P × 6.25 = 800 + 600] | M1 | For using WD = P × t where WD is from |
| Answer | Marks |
|---|---|
| Power = 224 W | A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | Work done by cyclist | |
| = 224 × 10 ( = 2240J) | B1 FT | For stating WD = power × time |
| Answer | Marks | Guidance |
|---|---|---|
| Initial KE = ½ × 80 × 42 [= 640 J] | B1 | |
| [½ × 80v2 = 640 + P×10 –1200] | M1 | For using Work/Energy equation |
| Speed = 6.48 m s–1 | A1 | Allow speed = √42 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | EITHER:
600
Resistance force = = 24 N
25 | (B1
Weight component = 80 g (0.04)
= 32 N | B1 | For correct unsimplified numerical form
of the weight component
[Power = 56 × 4] | M1 | For use of P = Fv where F is from two
relevant force terms
Power = 224 W | A1)
4
OR:
PE gain = 80g × 25 (0.04)
= 800 | (B1 | For a correct unsimplified numerical
expression for PE
25
Time taken = = 6.25
4 | B1
[WD by cyclist = P × 6.25 = 800 + 600] | M1 | For using WD = P × t where WD is from
two relevant terms
Power = 224 W | A1)
4
Question | Answer | Marks | Guidance
--- 5(ii) ---
5(ii) | Work done by cyclist
= 224 × 10 ( = 2240J) | B1 FT | For stating WD = power × time
FT on P value found in 5(i)
Initial KE = ½ × 80 × 42 [= 640 J] | B1
[½ × 80v2 = 640 + P×10 –1200] | M1 | For using Work/Energy equation
Speed = 6.48 m s–1 | A1 | Allow speed = √42
4
Question | Answer | Marks | GuidanceA cyclist is riding up a straight hill inclined at an angle $α$ to the horizontal, where $\sin α = 0.04$. The total mass of the bicycle and rider is 80 kg. The cyclist is riding at a constant speed of 4 m s$^{-1}$. There is a force resisting the motion. The work done by the cyclist against this resistance force over a distance of 25 m is 600 J.
\begin{enumerate}[label=(\roman*)]
\item Find the power output of the cyclist. [4]
\end{enumerate}
The cyclist reaches the top of the hill, where the road becomes horizontal, with speed 4 m s$^{-1}$. The cyclist continues to work at the same rate on the horizontal part of the road.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the speed of the cyclist 10 seconds after reaching the top of the hill, given that the work done by the cyclist during this period against the resistance force is 1200 J. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2017 Q5 [8]}}