| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Heavier particle hits ground, lighter continues upward - inclined plane involved |
| Difficulty | Standard +0.3 This is a standard M1 pulley system question requiring Newton's second law for connected particles, followed by kinematics with a change in motion. Part (i) involves routine resolution of forces and simultaneous equations. Part (ii) requires careful tracking of motion in two phases (while P falls, then after P stops), using standard SUVAT equations. While multi-step, all techniques are standard M1 fare with no novel insight required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - 0.85g \sin30 = 0.85a\), \(0.55g - T = 0.55a\), \(a = 1.225/1.4\), \(a = 0.875\) | B1, B1, M1, A1 | Either equation correct. Both eqns correct and consistent 'a' direction. Solves 2 sim eqn |
| \(T = 4.91\) | A1 | 4.908 or better – has to be positive |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = 2T\cos30\), \(F = 8.5(02...)\) | M1, A1ft | Or Pythagoras or cosine rule. \(cv(4.91)x\sqrt{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v^2 = 1.3^2 + 2x0.875x1.5\) (=4.315), \(a = +/-g\sin30\), \(v = 2.077...(v^2 = 1.69+3xcv(0.875))\), \(a = +/-4.9\) | M1, A1ft, B1, M1, A1, A1 | Uses \(v^2 = u^2 + 2a(1.5)\), u non-zero, a from (i). Uses \(v = u^2 +/-2as\), with a not g or (i), u not1.3. Maybe be implied – need not be 3sf |
| Answer | Marks | Guidance |
|---|---|---|
| \(S = 1.94\) | A1 | [6] |
**Part ia**
| $T - 0.85g \sin30 = 0.85a$, $0.55g - T = 0.55a$, $a = 1.225/1.4$, $a = 0.875$ | B1, B1, M1, A1 | Either equation correct. Both eqns correct and consistent 'a' direction. Solves 2 sim eqn |
|---|---|---|
| $T = 4.91$ | A1 | 4.908 or better – has to be positive |
**Part ib**
| $F = 2T\cos30$, $F = 8.5(02...)$ | M1, A1ft | Or Pythagoras or cosine rule. $cv(4.91)x\sqrt{3}$ |
**Part ii**
| $v^2 = 1.3^2 + 2x0.875x1.5$ (=4.315), $a = +/-g\sin30$, $v = 2.077...(v^2 = 1.69+3xcv(0.875))$, $a = +/-4.9$ | M1, A1ft, B1, M1, A1, A1 | Uses $v^2 = u^2 + 2a(1.5)$, u non-zero, a from (i). Uses $v = u^2 +/-2as$, with a not g or (i), u not1.3. Maybe be implied – need not be 3sf |
**Final answer**
| $S = 1.94$ | A1 | [6] |
---6 A block $B$ of mass 0.85 kg lies on a smooth slope inclined at $30 ^ { \circ }$ to the horizontal. $B$ is attached to one end of a light inextensible string which is parallel to the slope. At the top of the slope, the string passes over a smooth pulley. The other end of the string hangs vertically and is attached to a particle $P$ of mass 0.55 kg . The string is taut at the instant when $P$ is projected vertically downwards.\\
(i) Calculate
\begin{enumerate}[label=(\alph*)]
\item the acceleration of $B$ and the tension in the string,
\item the magnitude of the force exerted by the string on the pulley.
The initial speed of $P$ is $1.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and after moving $1.5 \mathrm {~m} P$ reaches the ground, where it remains at rest. $B$ continues to move up the slope and does not reach the pulley.\\
(ii) Calculate the total distance $B$ moves up the slope before coming instantaneously to rest.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2010 Q6 [13]}}