| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find acceleration on incline given power |
| Difficulty | Moderate -0.3 This is a straightforward application of the power-force-velocity relationship (P=Fv) combined with Newton's second law. Part (i) requires simple rearrangement and division; part (ii) adds a standard component of weight down the slope. Both parts are routine mechanics calculations with no conceptual difficulty beyond standard M1 content. |
| Spec | 6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = 420/5\) \([= 84]\) | B1, M1 | For using Newton's second law |
| (i) Acceleration is 1.12\(ms^{-2}\) | A1ft | From \(F = 75a\) |
| (ii) \([420/5 - 750\sin 1.5° = 75a\) or \(a = 1.12 - g\sin 1.5°]\) | M1 | For including weight component in N2 equation or for \(a(ii) = a(i) - wt. comp./m\) |
| Acceleration is 0.858\(ms^{-2}\) | A1ft | 5 |
$F = 420/5$ $[= 84]$ | B1, M1 | For using Newton's second law
(i) Acceleration is 1.12$ms^{-2}$ | A1ft | From $F = 75a$
(ii) $[420/5 - 750\sin 1.5° = 75a$ or $a = 1.12 - g\sin 1.5°]$ | M1 | For including weight component in N2 equation or for $a(ii) = a(i) - wt. comp./m$
Acceleration is 0.858$ms^{-2}$ | A1ft | 5 | $ft$ ans(i) – 0.2623 A cyclist travels along a straight road working at a constant rate of 420 W . The total mass of the cyclist and her cycle is 75 kg . Ignoring any resistance to motion, find the acceleration of the cyclist at an instant when she is travelling at $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$,\\
(i) given that the road is horizontal,\\
(ii) given instead that the road is inclined at $1.5 ^ { \circ }$ to the horizontal and the cyclist is travelling up the slope.
\hfill \mbox{\textit{CAIE M1 2006 Q3 [5]}}