| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: relative motion |
| Difficulty | Standard +0.3 This is a standard two-particle SUVAT problem requiring systematic application of kinematic equations. Part (i) involves equating flight times using s = ut + ½at², while part (ii) requires setting up simultaneous equations for positions and solving for velocities. The problem is slightly above average difficulty due to the two-part structure and the need to handle the time delay carefully, but it follows predictable patterns for M1 relative motion questions with no novel insights required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([T = 2 \times 1.7 - 2 \times 0.7]\) [for P \(17t - 5t^2 = 0\) and for Q \(7t = 5t^2 = 0\)] | M1 | \(T =\) 2 × time to max height for P − 2 × time to max height for Q; or for using \(T =\) time for P to return to ground − time for Q to return to ground |
| \(T = 2\) | A1 [2] | SR (max 1/2) for candidates who find difference in time to maximum height: \(T = 1.7 - 0.7 = 1\) B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(h_P - h_Q = 5\) and \(s = ut - 5t^2\) for both P and Q | |
| \(17(t+2) - 5(t+2)^2 - (7t - 5t^2) = 5\) or \(17t - 5t^2 - 7(t-2) + 5(t-2)^2 = 5\) | ||
| A1 | ft \(T\) from part (i) | |
| \(t = 0.9\) or \(t = 2.9\) | A1 | |
| M1 | For using \(v = u - 10t\) for P and Q | |
| \(v_P = 17 - 10(0.9 + 2)\), \(v_Q = 7 - 10 \times 0.9 \Rightarrow\) Magnitudes are \(12\ \text{m s}^{-1}\) & \(2\ \text{m s}^{-1}\) | A1 | ft using \(t_P\) and \(t_P - T\) or using \(t_Q\) and \(t_Q + T\) |
| The direction for both is vertically downwards | A1 [6] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[T = 2 \times 1.7 - 2 \times 0.7]$ [for P $17t - 5t^2 = 0$ and for Q $7t = 5t^2 = 0$] | M1 | $T =$ 2 × time to max height for P − 2 × time to max height for Q; or for using $T =$ time for P to return to ground − time for Q to return to ground |
| $T = 2$ | A1 [2] | SR (max 1/2) for candidates who find difference in time to maximum height: $T = 1.7 - 0.7 = 1$ B1 |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $h_P - h_Q = 5$ and $s = ut - 5t^2$ for both P and Q |
| $17(t+2) - 5(t+2)^2 - (7t - 5t^2) = 5$ or $17t - 5t^2 - 7(t-2) + 5(t-2)^2 = 5$ | | |
| | A1 | ft $T$ from part (i) |
| $t = 0.9$ or $t = 2.9$ | A1 | |
| | M1 | For using $v = u - 10t$ for P and Q |
| $v_P = 17 - 10(0.9 + 2)$, $v_Q = 7 - 10 \times 0.9 \Rightarrow$ Magnitudes are $12\ \text{m s}^{-1}$ & $2\ \text{m s}^{-1}$ | A1 | ft using $t_P$ and $t_P - T$ or using $t_Q$ and $t_Q + T$ |
| The direction for both is vertically downwards | A1 [6] | |
---5 A particle $P$ is projected vertically upwards from a point on the ground with speed $17 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Another particle $Q$ is projected vertically upwards from the same point with speed $7 \mathrm {~ms} ^ { - 1 }$. Particle $Q$ is projected $T$ seconds later than particle $P$.\\
(i) Given that the particles reach the ground at the same instant, find the value of $T$.\\
(ii) At a certain instant when both $P$ and $Q$ are in motion, $P$ is 5 m higher than $Q$. Find the magnitude and direction of the velocity of each of the particles at this instant.
\hfill \mbox{\textit{CAIE M1 2013 Q5 [8]}}