| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Variable mass or unknown mass |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem with multiple parts requiring systematic application of Newton's second law, kinematics equations, and energy considerations. While it involves several steps and careful tracking of the system after Q hits the ground, each individual part uses routine mechanics techniques without requiring novel insight or complex problem-solving strategies. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For P: \(T - 0.45g = 0.45 \times 0.98\) | M1 | Newton's 2nd law for either particle |
| \(T = 0.45 \times 9.8 + 0.45 \times 0.98 = 4.41 + 0.441 = 4.85\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For Q: \(mg - T = m \times 0.98\) | M1 | Newton's 2nd law for Q |
| \(mg - 4.851 = 0.98m\) | M1 | Substituting T |
| \(m(9.8 - 0.98) = 4.851\), \(m = 0.55\) kg | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v^2 = 2 \times 0.98 \times 0.36\) | M1 | Use of \(v^2 = u^2 + 2as\) |
| \(v = 0.840\) m s\(^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| After Q hits ground, P decelerates under gravity alone | M1 | |
| Additional height \(= \frac{v^2}{2g} = \frac{0.840^2}{2 \times 9.8} = 0.036\) m | M1 | |
| Greatest height \(= 0.36 + 0.036 = 0.396\) m | A1 |
# Question 2:
**(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| For P: $T - 0.45g = 0.45 \times 0.98$ | M1 | Newton's 2nd law for either particle |
| $T = 0.45 \times 9.8 + 0.45 \times 0.98 = 4.41 + 0.441 = 4.85$ N | A1 | |
**(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| For Q: $mg - T = m \times 0.98$ | M1 | Newton's 2nd law for Q |
| $mg - 4.851 = 0.98m$ | M1 | Substituting T |
| $m(9.8 - 0.98) = 4.851$, $m = 0.55$ kg | A1 | |
**(iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v^2 = 2 \times 0.98 \times 0.36$ | M1 | Use of $v^2 = u^2 + 2as$ |
| $v = 0.840$ m s$^{-1}$ | A1 | |
**(iv)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| After Q hits ground, P decelerates under gravity alone | M1 | |
| Additional height $= \frac{v^2}{2g} = \frac{0.840^2}{2 \times 9.8} = 0.036$ m | M1 | |
| Greatest height $= 0.36 + 0.036 = 0.396$ m | A1 | |
---2 Particles $P$ and $Q$, of masses 0.45 kg and $m \mathrm {~kg}$ respectively, are attached to the ends of a light inextensible string which passes over a small smooth pulley. The particles are released from rest with the string taut and both particles 0.36 m above a horizontal surface. $Q$ descends with acceleration $0.98 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. When $Q$ strikes the surface, it remains at rest.\\
(i) Calculate the tension in the string while both particles are in motion.\\
(ii) Find the value of $m$.\\
(iii) Calculate the speed at which $Q$ strikes the surface.\\
(iv) Calculate the greatest height of $P$ above the surface. (You may assume that $P$ does not reach the pulley.)
\hfill \mbox{\textit{OCR M1 2011 Q2 [10]}}