| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2004 |
| Session | June |
| Marks | 11 |
| 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 SUVAT problem with two particles under gravity requiring comparison of heights and velocities. While it has three parts and requires careful setup of equations, the techniques are routine (solving quadratic equations, substituting times into velocity formulas). The relative motion aspect adds mild complexity but no novel insight is needed—it's a straightforward application of projectile motion formulas that's slightly above average due to the multi-part nature and bookkeeping required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(25 = 30t - 5t^2 \Rightarrow t^2 - 6t + 5 = 0 \Rightarrow (t-1)(t-5) = 0\) or \(v^2 = 30^2 - 500\); \(t_{up} = (20-0)/10\) | M1 | For using \(25 = ut - \frac{1}{2}gt^2\) and attempting to solve for \(t\), or for using \(v^2 = u^2 - 2g(25)\) and \(t_{up} = (v-0)/g\) |
| \(t = 1, 5\) or \(t_{up} = 2\) | A1 | |
| Time \(= 5 - 1 = 4\) s or Time \(= 2\times2 = 4\) s, \(1 < t < 5\) | A1 (×3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s_1 = 30t - 5t^2\) and \(s_2 = 10t - 5t^2\) | M1 | For using \(s = ut - \frac{1}{2}gt^2\) for \(P_1\) and \(P_2\) |
| \(30t - 10t = 25\) | M1 | For using \(s_1 = s_2 + 25\) and attempting to solve for \(t\) |
| \(t = 1.25\) | A1 | |
| \(v_1 = 30 - 10\times1.25\) or \(v_2 = 10 - 10\times1.25\) | M1 | For using \(v = u - gt\) (either case) or for calculating \(s_1\) and substituting into \(v_1^2 = 30^2 - 2\times10s_1\) or calculating \(s_2\) and substituting into \(v_2^2 = 10^2 - 2\times10s_2\) |
| Velocities 17.5 ms⁻¹ and \(-2.5\) ms⁻¹ | A1 (×5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v_1 = 30 - 10t\), \(v_2 = 10 - 10t\) \(\Rightarrow v_1 - v_2 = 20\) | M1 | For using \(v = u - gt\) for \(P_1\) and \(P_2\) and eliminating \(t\) |
| \((30^2 - v_1^2)\div20 = (10^2 - v_2^2)\div20 + 25\) | M1, A1 | For using \(v^2 = u^2 - 2gs\) for \(P_1\) and \(P_2\) and then \(s_1 = s_2 + 25\) |
| \(v_1 - v_2 = 20\), \(v_1^2 - v_2^2 = 300\) | M1 | For solving simultaneous equations in \(v_1\) and \(v_2\) |
| Velocities 17.5 ms⁻¹ and \(-2.5\) ms⁻¹ | A1 (×5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(t_{up} = 3\) | B1 | |
| \(3 - 1.25\) | M1 | For using \(t_{\text{up and above}} = t_{up} - t_{equal}\) |
| Time is 1.75 s or \(1.25 < t < 3\) | A1 (×3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0 = 17.5 - 10t\) | M2 | For using \(0 = u - gt\) with \(u\) equal to the answer found for \(v_1\) in (ii) |
| Time is 1.75 s or \(1.25 < t < 3\) | A1 | SR (max 1 out of 3): \(0 = 17.5 + 10t\) B1 ft |
# Question 7:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $25 = 30t - 5t^2 \Rightarrow t^2 - 6t + 5 = 0 \Rightarrow (t-1)(t-5) = 0$ or $v^2 = 30^2 - 500$; $t_{up} = (20-0)/10$ | M1 | For using $25 = ut - \frac{1}{2}gt^2$ and attempting to solve for $t$, or for using $v^2 = u^2 - 2g(25)$ and $t_{up} = (v-0)/g$ |
| $t = 1, 5$ or $t_{up} = 2$ | A1 | |
| Time $= 5 - 1 = 4$ s or Time $= 2\times2 = 4$ s, $1 < t < 5$ | A1 (×3) | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s_1 = 30t - 5t^2$ and $s_2 = 10t - 5t^2$ | M1 | For using $s = ut - \frac{1}{2}gt^2$ for $P_1$ and $P_2$ |
| $30t - 10t = 25$ | M1 | For using $s_1 = s_2 + 25$ and attempting to solve for $t$ |
| $t = 1.25$ | A1 | |
| $v_1 = 30 - 10\times1.25$ or $v_2 = 10 - 10\times1.25$ | M1 | For using $v = u - gt$ (either case) or for calculating $s_1$ and substituting into $v_1^2 = 30^2 - 2\times10s_1$ or calculating $s_2$ and substituting into $v_2^2 = 10^2 - 2\times10s_2$ |
| Velocities 17.5 ms⁻¹ and $-2.5$ ms⁻¹ | A1 (×5) | |
## Part (ii) OR method:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v_1 = 30 - 10t$, $v_2 = 10 - 10t$ $\Rightarrow v_1 - v_2 = 20$ | M1 | For using $v = u - gt$ for $P_1$ and $P_2$ and eliminating $t$ |
| $(30^2 - v_1^2)\div20 = (10^2 - v_2^2)\div20 + 25$ | M1, A1 | For using $v^2 = u^2 - 2gs$ for $P_1$ and $P_2$ and then $s_1 = s_2 + 25$ |
| $v_1 - v_2 = 20$, $v_1^2 - v_2^2 = 300$ | M1 | For solving simultaneous equations in $v_1$ and $v_2$ |
| Velocities 17.5 ms⁻¹ and $-2.5$ ms⁻¹ | A1 (×5) | |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $t_{up} = 3$ | B1 | |
| $3 - 1.25$ | M1 | For using $t_{\text{up and above}} = t_{up} - t_{equal}$ |
| Time is 1.75 s or $1.25 < t < 3$ | A1 (×3) | |
## Part (iii) OR method:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0 = 17.5 - 10t$ | M2 | For using $0 = u - gt$ with $u$ equal to the answer found for $v_1$ in **(ii)** |
| Time is 1.75 s or $1.25 < t < 3$ | A1 | SR (max 1 out of 3): $0 = 17.5 + 10t$ B1 ft |7 A particle $P _ { 1 }$ is projected vertically upwards, from horizontal ground, with a speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At the same instant another particle $P _ { 2 }$ is projected vertically upwards from the top of a tower of height 25 m , with a speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find\\
(i) the time for which $P _ { 1 }$ is higher than the top of the tower,\\
(ii) the velocities of the particles at the instant when the particles are at the same height,\\
(iii) the time for which $P _ { 1 }$ is higher than $P _ { 2 }$ and is moving upwards.
\hfill \mbox{\textit{CAIE M1 2004 Q7 [11]}}