| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Acceleration from power and speed |
| Difficulty | Standard +0.3 This is a straightforward multi-part work-energy question requiring standard formula application (KE = ½mv², PE = mgh, Work = Force × distance, Power = Force × velocity). Part (iii) requires combining P = Fv to find driving force, then F = ma, but all steps are routine M2 techniques with no novel problem-solving or geometric insight needed. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Gain in KE is \(\frac{1}{2} \times 80 \times 5^2 = 1000\) J. Gain in PE is \(80 \times 9.8 \times 4 = 3136\) J | M1, M1, A1 (3 marks) | For use of formula \(\frac{1}{2}mv^2\); For use of formula \(mgh\); For both answers 1000 and 3136 correct |
| (ii) \(8000 = 1000 + 3136 + 70d\). Hence distance AB is 55.2 m | M1, M1, A1 (3 marks) | For equating work done to energy change; For relevant use of force × distance; For correct answer 55.2 |
| (iii) \(\frac{720}{5} - 70 = 80a\). Hence acceleration is \(0.925\) m s\(^{-2}\) | B1, M1, A1, A1 (4 marks) | For driving force \(\frac{720}{5}\); For use of Newton II with 3-term equation; For a completely correct equation; For correct answer 0.925 |
(i) Gain in KE is $\frac{1}{2} \times 80 \times 5^2 = 1000$ J. Gain in PE is $80 \times 9.8 \times 4 = 3136$ J | M1, M1, A1 (3 marks) | For use of formula $\frac{1}{2}mv^2$; For use of formula $mgh$; For both answers 1000 and 3136 correct
(ii) $8000 = 1000 + 3136 + 70d$. Hence distance AB is 55.2 m | M1, M1, A1 (3 marks) | For equating work done to energy change; For relevant use of force × distance; For correct answer 55.2
(iii) $\frac{720}{5} - 70 = 80a$. Hence acceleration is $0.925$ m s$^{-2}$ | B1, M1, A1, A1 (4 marks) | For driving force $\frac{720}{5}$; For use of Newton II with 3-term equation; For a completely correct equation; For correct answer 0.925
---5 A cyclist and his machine have a combined mass of 80 kg . The cyclist ascends a straight hill $A B$ of constant slope, starting from rest at $A$ and reaching a speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $B$. The level of $B$ is 4 m above the level of $A$.\\
(i) Find the gain in kinetic energy and the gain in gravitational potential energy of the cyclist and his machine.
During the ascent the resistance to motion is constant and has magnitude 70 N .\\
(ii) Given that the work done by the cyclist in ascending the hill is 8000 J , find the distance $A B$.
At $B$ the cyclist is working at 720 watts and starts to move in a straight line along horizontal ground. The resistance to motion has the same magnitude of 70 N as before.\\
(iii) Find the acceleration with which the cyclist starts to move horizontally.
\hfill \mbox{\textit{OCR M2 Q5 [10]}}