| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Projectile on rough surface |
| Difficulty | Standard +0.3 This is a standard M1 mechanics problem involving constant acceleration with friction, energy loss on impact, and motion in both directions. It requires systematic application of SUVAT equations and energy principles across three phases of motion, but follows a predictable structure with no novel insights needed. The multi-part nature and graph sketching add some complexity, but this is typical fare for M1 students who have practiced similar problems. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \([-0.12 = 0.15a]\) | M1 | For using Newton's 2nd law |
| \(a = -0.8 \text{ ms}^{-2}\) | A1 | |
| \([v = 3 - 0.8 \times 2]\) | M1 | For using \(v = u + at\) to find speed of approach |
| \(v_{\text{approach}} = 1.4\) | A1 | |
| \([\frac{1}{2}\ 0.15(1.4^2 - v_r^2)]\) | M1 | For using KE loss \(= \frac{1}{2}m(v_a^2 - v_r^2)\) |
| \(v_{\text{return}} = -1\) | A1 | |
| M1 | For using \(0 = v_{\text{return}} + a(t-2)\) | |
| \(t = 3.25\) s when block comes to rest | A1 | |
| B1 | Alternative: \(t_{YZ} = 1.25\), \(t = 3.25\) s when block is at rest (B1ft) | |
| For correct sketch | B1ft [9] | ft incorrect values of \(v\) and \(t\) (although \(v_{\text{return}}\) must be negative) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \([XY = \frac{1}{2}(3 + 1.4) \times 2,\ YZ = \frac{1}{2}\ 1.25 \times 1]\) | M1 | For using area property (or equivalent) to find distances XY and YZ |
| \(s = 4.4\) at Y and \(3.775\) at Z, stated or on graph | A1 | (accept 3.77 or 3.78) |
| Curve starts at origin, \(s\) increases, slope decreases (convex upwards) for \(0 < t < 2\), value of \(s(2)\) shown | B1ft | ft incorrect value for \(s(2)\) |
| Curve starts at \((2,\ 4.4)\), \(s\) decreases, magnitude of slope decreases to zero at \((3.25,\ 3.775)\) | B1ft [4] | ft incorrect values of \(s\) and \(t\) |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $[-0.12 = 0.15a]$ | M1 | For using Newton's 2nd law |
| $a = -0.8 \text{ ms}^{-2}$ | A1 | |
| $[v = 3 - 0.8 \times 2]$ | M1 | For using $v = u + at$ to find speed of approach |
| $v_{\text{approach}} = 1.4$ | A1 | |
| $[\frac{1}{2}\ 0.15(1.4^2 - v_r^2)]$ | M1 | For using KE loss $= \frac{1}{2}m(v_a^2 - v_r^2)$ |
| $v_{\text{return}} = -1$ | A1 | |
| | M1 | For using $0 = v_{\text{return}} + a(t-2)$ |
| $t = 3.25$ s when block comes to rest | A1 | |
| | B1 | Alternative: $t_{YZ} = 1.25$, $t = 3.25$ s when block is at rest (B1ft) |
| For correct sketch | B1ft [9] | ft incorrect values of $v$ and $t$ (although $v_{\text{return}}$ must be negative) |
### Part (ii):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $[XY = \frac{1}{2}(3 + 1.4) \times 2,\ YZ = \frac{1}{2}\ 1.25 \times 1]$ | M1 | For using area property (or equivalent) to find distances XY and YZ |
| $s = 4.4$ at Y and $3.775$ at Z, stated or on graph | A1 | (accept 3.77 or 3.78) |
| Curve starts at origin, $s$ increases, slope decreases (convex upwards) for $0 < t < 2$, value of $s(2)$ shown | B1ft | ft incorrect value for $s(2)$ |
| Curve starts at $(2,\ 4.4)$, $s$ decreases, magnitude of slope decreases to zero at $(3.25,\ 3.775)$ | B1ft [4] | ft incorrect values of $s$ and $t$ |7\\
\includegraphics[max width=\textwidth, alt={}, center]{fa0e0e0d-b0a6-44e0-8b4f-4923e235c6c6-3_168_803_1909_671}
The frictional force acting on a small block of mass 0.15 kg , while it is moving on a horizontal surface, has magnitude 0.12 N . The block is set in motion from a point $X$ on the surface, with speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It hits a vertical wall at a point $Y$ on the surface 2 s later. The block rebounds from the wall and moves directly towards $X$ before coming to rest at the point $Z$ (see diagram). At the instant that the block hits the wall it loses 0.072 J of its kinetic energy. The velocity of the block, in the direction from $X$ to $Y$, is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t \mathrm {~s}$ after it leaves $X$.\\
(i) Find the values of $v$ when the block arrives at $Y$ and when it leaves $Y$, and find also the value of $t$ when the block comes to rest at $Z$. Sketch the velocity-time graph.\\
(ii) The displacement of the block from $X$, in the direction from $X$ to $Y$, is $s \mathrm {~m}$ at time $t \mathrm {~s}$. Sketch the displacement-time graph. Show on your graph the values of $s$ and $t$ when the block is at $Y$ and when it comes to rest at $Z$.
\hfill \mbox{\textit{CAIE M1 2012 Q7 [13]}}