| 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 | Moderate -0.5 This is a standard velocity-time graph question requiring calculation of distance (area under curve) and possibly acceleration (gradient). These are routine M1 skills with straightforward geometric area calculations, making it slightly easier than average but still requiring proper method application. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(v^2 = 2 \times g \times 7.2 \Rightarrow\) speed at surface is \(12\text{ ms}^{-1}\) | B1 | |
| \([6^2 = 12^2 + 2a \times 0.8]\) | M1 | For using \(6^2 = v^2 + 2as\) and finding \(a\) |
| Deceleration is \(67.5\text{ ms}^{-2}\) | A1 | |
| \([0.2g - R = -0.2 \times 67.5]\) | M1 | For using Newton's 2nd law with three terms for P in the liquid |
| \(R = 15.5\) | A1 | 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \([3.6 = \frac{1}{2}a \times 4^2]\) | M1 | For using \(s = 0 + \frac{1}{2}at^2\) and finding \(a\) |
| \(a = 0.45\text{ ms}^{-2}\) | A1 | |
| \([T - R - 0.2g = 0.2 \times 0.45]\) | M1 | For using Newton's 2nd law with P in the liquid |
| Tension is \(17.6\text{ N}\) (17.59 exact) | A1\(\checkmark\) | 4 marks; ft incorrect R |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| M1 | For using PE lost \(=\) WD by R in liquid \(+\) KE gain | |
| \(0.2g \times 8 = R(0.8) + \frac{1}{2}(0.2)6^2\) | A1 | |
| \(R = 15.5\) | A1 | Finding R |
| \(0.2g - 15.5 = 0.2a\) | M1 | For using Newton's 2nd law in the liquid |
| \(a = -67.5\) | A1 | 5 marks |
| M1 | For using \(s = (0+v)/2 \times t\) to find \(v\) at surface of liquid | |
| \(3.6 = v/2 \times 4 \Rightarrow v = 1.8\) | A1 | |
| \(T(3.6) = R(3.6) + 0.2g(3.6) + \frac{1}{2}(0.2)(1.8)^2\) | M1 | For using WD by T \(=\) WD by R \(+\) PE gain \(+\) KE gain |
| \(T = 17.6\text{ N}\) | A1 | 4 marks |
## Question 6:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $v^2 = 2 \times g \times 7.2 \Rightarrow$ speed at surface is $12\text{ ms}^{-1}$ | B1 | |
| $[6^2 = 12^2 + 2a \times 0.8]$ | M1 | For using $6^2 = v^2 + 2as$ and finding $a$ |
| Deceleration is $67.5\text{ ms}^{-2}$ | A1 | |
| $[0.2g - R = -0.2 \times 67.5]$ | M1 | For using Newton's 2nd law with three terms for P in the liquid |
| $R = 15.5$ | A1 | 5 marks |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $[3.6 = \frac{1}{2}a \times 4^2]$ | M1 | For using $s = 0 + \frac{1}{2}at^2$ and finding $a$ |
| $a = 0.45\text{ ms}^{-2}$ | A1 | |
| $[T - R - 0.2g = 0.2 \times 0.45]$ | M1 | For using Newton's 2nd law with P in the liquid |
| Tension is $17.6\text{ N}$ (17.59 exact) | A1$\checkmark$ | 4 marks; ft incorrect R |
**Alternative Energy Method:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| | M1 | For using PE lost $=$ WD by R in liquid $+$ KE gain |
| $0.2g \times 8 = R(0.8) + \frac{1}{2}(0.2)6^2$ | A1 | |
| $R = 15.5$ | A1 | Finding R |
| $0.2g - 15.5 = 0.2a$ | M1 | For using Newton's 2nd law in the liquid |
| $a = -67.5$ | A1 | 5 marks |
| | M1 | For using $s = (0+v)/2 \times t$ to find $v$ at surface of liquid |
| $3.6 = v/2 \times 4 \Rightarrow v = 1.8$ | A1 | |
| $T(3.6) = R(3.6) + 0.2g(3.6) + \frac{1}{2}(0.2)(1.8)^2$ | M1 | For using WD by T $=$ WD by R $+$ PE gain $+$ KE gain |
| $T = 17.6\text{ N}$ | A1 | 4 marks |
---6 A particle $P$ of mass 0.2 kg is released from rest at a point 7.2 m above the surface of the liquid in a container. $P$ falls through the air and into the liquid. There is no air resistance and there is no instantaneous change of speed as $P$ enters the liquid. When $P$ is at a distance of 0.8 m below the surface of the liquid, $P$ 's speed is $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The only force on $P$ due to the liquid is a constant resistance to motion of magnitude $R \mathrm {~N}$.\\
(i) Find the deceleration of $P$ while it is falling through the liquid, and hence find the value of $R$.
The depth of the liquid in the container is $3.6 \mathrm {~m} . P$ is taken from the container and attached to one end of a light inextensible string. $P$ is placed at the bottom of the container and then pulled vertically upwards with constant acceleration. The resistance to motion of $R \mathrm {~N}$ continues to act. The particle reaches the surface 4 s after leaving the bottom of the container.\\
(ii) Find the tension in the string.
\hfill \mbox{\textit{CAIE M1 2014 Q6}}