| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: same start time, different heights |
| Difficulty | Standard +0.3 This is a straightforward two-particle SUVAT problem requiring students to set up equations for motion on an incline and vertical free fall, then solve simultaneously. The setup is clear, the mathematics is routine (solving quadratic equations and applying SUVAT formulas), and it's a standard M1 textbook exercise with no novel insight required. Slightly easier than average due to the structured nature and clear diagram. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a_P = g\sin 30°\) | B1 | |
| \(3.2 = \frac{1}{2}gt_q^2\) | B1 | |
| \([6.4 = u(0.8) + \frac{1}{2} \times 5 \times (0.8)^2]\) | M1 | For applying \(s = ut + \frac{1}{2}at^2\) to P |
| \(u = 6\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([v = 6 + 5 \times 0.8\) or \(v^2 = 36 + 2\times5\times6.4]\) | M1 | For using \(v = u + at\) or \(v^2 = u^2 + 2as\) for P |
| Speed of P is \(10\ \text{ms}^{-1}\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3.2 = \frac{1}{2}gt_q^2\) | B1 | |
| For using KE gain = PE loss to obtain an equation in \(u\) and \(v\): \([\frac{1}{2}(v^2 - u^2) = 6.4g\sin 30°]\) | M1 | |
| For using \(s = \frac{1}{2}(u+v)t\) to obtain a second equation in \(u\) and \(v\): \([6.4 = \frac{1}{2}(u+v)\times 0.8]\) | DM1 | |
| \(u = 6\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitutes for \(u\) to find \(v\) | M1 | |
| Speed is \(10\ \text{ms}^{-1}\) | A1 [2] |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a_P = g\sin 30°$ | B1 | |
| $3.2 = \frac{1}{2}gt_q^2$ | B1 | |
| $[6.4 = u(0.8) + \frac{1}{2} \times 5 \times (0.8)^2]$ | M1 | For applying $s = ut + \frac{1}{2}at^2$ to P |
| $u = 6$ | A1 **[4]** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[v = 6 + 5 \times 0.8$ or $v^2 = 36 + 2\times5\times6.4]$ | M1 | For using $v = u + at$ or $v^2 = u^2 + 2as$ for P |
| Speed of P is $10\ \text{ms}^{-1}$ | A1 **[2]** | |
### Alternative for Parts (i) and (ii) when $a$ is not used:
**Part (i):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3.2 = \frac{1}{2}gt_q^2$ | B1 | |
| For using KE gain = PE loss to obtain an equation in $u$ and $v$: $[\frac{1}{2}(v^2 - u^2) = 6.4g\sin 30°]$ | M1 | |
| For using $s = \frac{1}{2}(u+v)t$ to obtain a second equation in $u$ and $v$: $[6.4 = \frac{1}{2}(u+v)\times 0.8]$ | DM1 | |
| $u = 6$ | A1 **[4]** | |
**Part (ii):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitutes for $u$ to find $v$ | M1 | |
| Speed is $10\ \text{ms}^{-1}$ | A1 **[2]** | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{8d64372d-0b9a-4b93-8c41-7096c813f714-2_443_825_755_661}
A particle $P$ is projected from the top of a smooth ramp with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and travels down a line of greatest slope. The ramp has length 6.4 m and is inclined at $30 ^ { \circ }$ to the horizontal. Another particle $Q$ is released from rest at a point 3.2 m vertically above the bottom of the ramp, at the same instant that $P$ is projected (see diagram). Given that $P$ and $Q$ reach the bottom of the ramp simultaneously, find\\
(i) the value of $u$,\\
(ii) the speed with which $P$ reaches the bottom of the ramp.\\
\includegraphics[max width=\textwidth, alt={}, center]{8d64372d-0b9a-4b93-8c41-7096c813f714-3_609_1539_255_303}
The diagram shows the velocity-time graphs for the motion of two particles $P$ and $Q$, which travel in the same direction along a straight line. $P$ and $Q$ both start at the same point $X$ on the line, but $Q$ starts to move $T$ s later than $P$. Each particle moves with speed $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for the first 20 s of its motion. The speed of each particle changes instantaneously to $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ after it has been moving for 20 s and the particle continues at this speed.\\
\hfill \mbox{\textit{CAIE M1 2011 Q3 [6]}}