| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Constant power on inclined plane |
| Difficulty | Standard +0.3 This is a standard constant power mechanics problem requiring P=Fv and F=ma with resistance forces. Part (i) is a straightforward 'show that' using Newton's second law. Part (ii) involves setting driving force equal to resistance plus component of weight at steady speed. Part (iii) requires calculating net force at a different speed. All steps are routine applications of standard mechanics formulas with no novel insight required, making it slightly easier than average. |
| Spec | 3.03c Newton's second law: F=ma one dimension6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Driving force \(= 160/5\ (= 32\text{ N})\) | B1 | |
| \([160/5 - 20 = m \times 0.15]\) | M1 | For using Newton's Second Law |
| Total mass is \(80\text{ kg}\) | AG A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([300/v - 20 - 80g\sin 2° = 0]\) | M1 | For resolving up hill |
| Speed is \(6.26\text{ ms}^{-1}\) | AG A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Driving force \(= 300/(0.9 \times 6.26)\ (= 53.2\text{ N})\) | B1 | |
| M1 | For using Newton's Second Law | |
| \(300/(0.9 \times 6.26) - 20 - 80g\sin 2° = 80a\) | A1 | |
| Acceleration is \(0.0666\text{ ms}^{-2}\) | A1 | [4] |
# Question 6:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Driving force $= 160/5\ (= 32\text{ N})$ | B1 | |
| $[160/5 - 20 = m \times 0.15]$ | M1 | For using Newton's Second Law |
| Total mass is $80\text{ kg}$ | AG A1 | [3] |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[300/v - 20 - 80g\sin 2° = 0]$ | M1 | For resolving up hill |
| Speed is $6.26\text{ ms}^{-1}$ | AG A1 | [2] |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Driving force $= 300/(0.9 \times 6.26)\ (= 53.2\text{ N})$ | B1 | |
| | M1 | For using Newton's Second Law |
| $300/(0.9 \times 6.26) - 20 - 80g\sin 2° = 80a$ | A1 | |
| Acceleration is $0.0666\text{ ms}^{-2}$ | A1 | [4] |6 A cyclist is cycling with constant power of 160 W along a horizontal straight road. There is a constant resistance to motion of 20 N . At an instant when the cyclist's speed is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, his acceleration is $0.15 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Show that the total mass of the cyclist and bicycle is 80 kg .
The cyclist comes to a hill inclined at $2 ^ { \circ }$ to the horizontal. When the cyclist starts climbing the hill, he increases his power to a constant 300 W . The resistance to motion remains 20 N .\\
(ii) Show that the steady speed up the hill which the cyclist can maintain when working at this power is $6.26 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, correct to 3 significant figures.\\
(iii) Find the acceleration at an instant when the cyclist is travelling at $90 \%$ of the speed in part (ii).
\hfill \mbox{\textit{CAIE M1 2016 Q6 [9]}}