| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Projectile on rough surface |
| Difficulty | Standard +0.3 This is a straightforward M1 mechanics question involving constant friction and kinematics with SUVAT equations. Parts (i)-(iii) are routine calculations (F=μR, F=ma, then s=ut+½at²), while parts (iv)-(v) require applying the same methods to the rebound motion. The multi-part structure and rebound adds slight complexity beyond a basic friction question, but all steps follow standard procedures with no novel problem-solving required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| For using \(F = \mu R\) and \(R = mg\) \((F = 0.025 \times 0.15 \times 10)\) | M1 | |
| Frictional force is 0.0375 N or 3/80 N | A1 | Accept 0.0368 from 9.8 or 9.81 |
| Answer | Marks |
|---|---|
| For using \(F = ma\) \((-0.0375 = 0.15a)\) or \(d = \mu g\) | M1 |
| Deceleration is 0.25 ms\(^{-2}\) (or \(a = -0.25\)) A.G. | A1 |
| Answer | Marks |
|---|---|
| For using \(s = ut + \frac{1}{2}at^2\) \(\left(s = 5.5 \times 4 + \frac{1}{2}(-0.25)16\right)\) | M1 |
| Distance \(AB\) is 20 m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| For using \(v^2 = u^2 + 2as\) \((v^2 = 3.5^2 - 2 \times 0.25 \times 20)\) | M1 | |
| Speed is 1.5 ms\(^{-1}\) | A1ft | \(\left(\text{ft } \sqrt{(24.5 - (iii))/2}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Return dist. \(= \frac{3.5^2}{2 \times 0.25}\) or distance beyond \(A = \frac{(iv)^2}{2 \times 0.25}\) | M1 | |
| Total distance is 44.5 m | A1ft | ft \(24.5 + \text{(iii)}\) or \(2(\mathbf{(iv)})^2 + \text{(iii)}\) |
# Question 6:
## Part (i)
| For using $F = \mu R$ and $R = mg$ $(F = 0.025 \times 0.15 \times 10)$ | M1 | |
| Frictional force is 0.0375 N or 3/80 N | A1 | Accept 0.0368 from 9.8 or 9.81 |
## Part (ii)
| For using $F = ma$ $(-0.0375 = 0.15a)$ or $d = \mu g$ | M1 | |
| Deceleration is 0.25 ms$^{-2}$ (or $a = -0.25$) A.G. | A1 | |
## Part (iii)
| For using $s = ut + \frac{1}{2}at^2$ $\left(s = 5.5 \times 4 + \frac{1}{2}(-0.25)16\right)$ | M1 | |
| Distance $AB$ is 20 m | A1 | |
## Part (iv)
| For using $v^2 = u^2 + 2as$ $(v^2 = 3.5^2 - 2 \times 0.25 \times 20)$ | M1 | |
| Speed is 1.5 ms$^{-1}$ | A1ft | $\left(\text{ft } \sqrt{(24.5 - (iii))/2}\right)$ |
## Part (v)
| Return dist. $= \frac{3.5^2}{2 \times 0.25}$ or distance beyond $A = \frac{(iv)^2}{2 \times 0.25}$ | M1 | |
| Total distance is 44.5 m | A1ft | ft $24.5 + \text{(iii)}$ or $2(\mathbf{(iv)})^2 + \text{(iii)}$ |
---6 A small block of mass 0.15 kg moves on a horizontal surface. The coefficient of friction between the block and the surface is 0.025 .\\
(i) Find the frictional force acting on the block.\\
(ii) Show that the deceleration of the block is $0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
The block is struck from a point $A$ on the surface and, 4 s later, it hits a boundary board at a point $B$. The initial speed of the block is $5.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(iii) Find the distance $A B$.
The block rebounds from the board with a speed of $3.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and moves along the line $B A$. Find\\
(iv) the speed with which the block passes through $A$,\\
(v) the total distance moved by the block, from the instant when it was struck at $A$ until the instant when it comes to rest.
\hfill \mbox{\textit{CAIE M1 2003 Q6 [10]}}