| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Easy -1.2 This is a standard SUVAT kinematics question requiring straightforward application of constant acceleration equations across multiple parts. While it involves several calculations, each part follows routine procedures typical of M1 mechanics with no novel problem-solving or geometric insight required. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \([T_A - 2.5 = 0.25 \times a]\) and \([7.5 - T_B = 0.75 \times a]\) | M1 | For applying Newton's 2nd law to either particle A or particle B |
| \(T_A = 2.5 + 0.25a\) | A1 | |
| \(T_B = 7.5 - 0.75a\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(F = 0.4 \times 5\) | B1 | |
| \([T_B - T_A - F = 0.5a]\) | M1 | For using Newton's 2nd law for P with friction and both tensions represented (4 terms) |
| \(7.5 - 0.75a - (2.5 + 0.25a) - 2 = 0.5a \Rightarrow a = 2\) | A1 | 3 marks; AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = 0.4 \times 5\) | B1 | |
| \(a = 2\) used to find \(T_A = 3\), \(T_B = 6\) and used in \(T_B - T_A - F = 0.5 \times a\) | M1 | Assume given value of \(a\), find \(T_A\) and \(T_B\) and use the values in 4 term Newton's 2nd law |
| \(a = 2\) | A1 | Justify the value \(a = 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Total |
| \([v^2 = 2 \times 2 \times 0.36]\) | M1 | |
| Speed is \(1.2 \text{ ms}^{-1}\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Total |
| \(-T_A - 2 = 0.5a\) and \(T_A - 2.5 = 0.25a\) | M1 | |
| Deceleration is \(6 \text{ ms}^{-2}\) | A1 | 2 |
## Question 7:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $[T_A - 2.5 = 0.25 \times a]$ and $[7.5 - T_B = 0.75 \times a]$ | M1 | For applying Newton's 2nd law to either particle A or particle B |
| $T_A = 2.5 + 0.25a$ | A1 | |
| $T_B = 7.5 - 0.75a$ | A1 | 3 marks |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $F = 0.4 \times 5$ | B1 | |
| $[T_B - T_A - F = 0.5a]$ | M1 | For using Newton's 2nd law for P with friction and both tensions represented (4 terms) |
| $7.5 - 0.75a - (2.5 + 0.25a) - 2 = 0.5a \Rightarrow a = 2$ | A1 | 3 marks; AG |
## Question (ii) [Alternative Method]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = 0.4 \times 5$ | B1 | |
| $a = 2$ used to find $T_A = 3$, $T_B = 6$ and used in $T_B - T_A - F = 0.5 \times a$ | M1 | Assume given value of $a$, find $T_A$ and $T_B$ and use the values in 4 term Newton's 2nd law |
| $a = 2$ | A1 | Justify the value $a = 2$ |
## Question (iii):
| Answer/Working | Mark | Total | Guidance |
|---|---|---|---|
| $[v^2 = 2 \times 2 \times 0.36]$ | M1 | | For using $v^2 = 2as$ with $s = 1 - \frac{1}{2}(5.28 - 4)$ |
| Speed is $1.2 \text{ ms}^{-1}$ | A1 | 2 | |
## Question (iv):
| Answer/Working | Mark | Total | Guidance |
|---|---|---|---|
| $-T_A - 2 = 0.5a$ and $T_A - 2.5 = 0.25a$ | M1 | | For applying Newton's 2nd law to particle P **and** substituting for $T_A$ |
| Deceleration is $6 \text{ ms}^{-2}$ | A1 | 2 | $a = -6$ or $d = 6$ |7\\
\includegraphics[max width=\textwidth, alt={}, center]{77976dad-c055-45fd-93fe-e37fa8e9ae22-4_333_1001_264_573}
A light inextensible string of length 5.28 m has particles $A$ and $B$, of masses 0.25 kg and 0.75 kg respectively, attached to its ends. Another particle $P$, of mass 0.5 kg , is attached to the mid-point of the string. Two small smooth pulleys $P _ { 1 }$ and $P _ { 2 }$ are fixed at opposite ends of a rough horizontal table of length 4 m and height 1 m . The string passes over $P _ { 1 }$ and $P _ { 2 }$ with particle $A$ held at rest vertically below $P _ { 1 }$, the string taut and $B$ hanging freely below $P _ { 2 }$. Particle $P$ is in contact with the table halfway between $P _ { 1 }$ and $P _ { 2 }$ (see diagram). The coefficient of friction between $P$ and the table is 0.4 . Particle $A$ is released and the system starts to move with constant acceleration of magnitude $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The tension in the part $A P$ of the string is $T _ { A } \mathrm {~N}$ and the tension in the part $P B$ of the string is $T _ { B } \mathrm {~N}$.\\
(i) Find $T _ { A }$ and $T _ { B }$ in terms of $a$.\\
(ii) Show by considering the motion of $P$ that $a = 2$.\\
(iii) Find the speed of the particles immediately before $B$ reaches the floor.\\
(iv) Find the deceleration of $P$ immediately after $B$ reaches the floor.
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\hfill \mbox{\textit{CAIE M1 2014 Q7}}