| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2002 |
| Session | November |
| 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 straightforward SUVAT problem requiring standard kinematic equations for vertical motion under gravity. Part (i) involves finding when A reaches maximum height (v=0) then calculating B's position at that time. Part (ii) requires setting up and solving a simple equation relating the two heights. Both parts use routine M1 techniques with no conceptual challenges beyond applying familiar formulas. |
| Spec | 3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For using \(v = u - gt\), with \(v = 0\), to find \(t\): \([5 - 10t = 0]\) | M1 | Using \(a = +g\) in \(v = u + at\) scores M0 at first stage |
| Time to maximum height of \(A\) is \(\frac{5}{g}\) | A1 | |
| For using \(h = ut - \frac{1}{2}gt^2\) and evaluating \(h_B(0.5) - h_A(0.5)\) | M1 | Allow error in sign of \(\frac{1}{2}gt^2\) terms for both M1 and first A1 in (ii) |
| Difference in heights is \(1.5\text{ m}\) (\(1.53\) from \(g = 9.81\) or \(g = 9.8\)) | A1 | Total: 4 |
| SR: difference in maximum heights (max 1 out of 4): \(1.95\text{ m}\) (\(1.99\) from \(g = 9.81\) or \(9.8\)) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For attempting to solve \(h_B - h_A = 0.9\) for \(t\): \([8t - 5t = 0.9]\) | M1 | |
| \(t = 0.3\) | A1 | |
| For using \(h = ut - \frac{1}{2}gt^2\) with the value of \(t\) found: \([h = 5 \times 0.3 - \frac{1}{2} \times 10 \times 0.09]\) | M1 | Using \(a = +g\) in \(s = ut + \frac{1}{2}at^2\) scores M0 at second stage |
| Height of \(A\) is \(1.05\text{ m}\) (\(1.06\) from \(g = 9.81\) or \(g = 9.8\)) | A1 | Total: 4 |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For using $v = u - gt$, with $v = 0$, to find $t$: $[5 - 10t = 0]$ | M1 | Using $a = +g$ in $v = u + at$ scores M0 at first stage |
| Time to maximum height of $A$ is $\frac{5}{g}$ | A1 | |
| For using $h = ut - \frac{1}{2}gt^2$ and evaluating $h_B(0.5) - h_A(0.5)$ | M1 | Allow error in sign of $\frac{1}{2}gt^2$ terms for both M1 and first A1 in (ii) |
| Difference in heights is $1.5\text{ m}$ ($1.53$ from $g = 9.81$ or $g = 9.8$) | A1 | **Total: 4** |
| SR: difference in maximum heights (max 1 out of 4): $1.95\text{ m}$ ($1.99$ from $g = 9.81$ or $9.8$) | B1 | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For attempting to solve $h_B - h_A = 0.9$ for $t$: $[8t - 5t = 0.9]$ | M1 | |
| $t = 0.3$ | A1 | |
| For using $h = ut - \frac{1}{2}gt^2$ with the value of $t$ found: $[h = 5 \times 0.3 - \frac{1}{2} \times 10 \times 0.09]$ | M1 | Using $a = +g$ in $s = ut + \frac{1}{2}at^2$ scores M0 at second stage |
| Height of $A$ is $1.05\text{ m}$ ($1.06$ from $g = 9.81$ or $g = 9.8$) | A1 | **Total: 4** |
**Note:** $5^2 = 8^2 - 2g(s + 0.9)$ leads fortuitously to 'correct' answer $1.05$ in (ii) but scores 0 out of 4.
---4 Two particles $A$ and $B$ are projected vertically upwards from horizontal ground at the same instant. The speeds of projection of $A$ and $B$ are $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively. Find\\
(i) the difference in the heights of $A$ and $B$ when $A$ is at its maximum height,\\
(ii) the height of $A$ above the ground when $B$ is 0.9 m above $A$.
\hfill \mbox{\textit{CAIE M1 2002 Q4 [8]}}