| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: different start times and heights |
| Difficulty | Standard +0.3 This is a standard two-particle SUVAT problem requiring systematic application of kinematic equations. Part (a) involves finding when a projectile is above a certain height for a given duration (straightforward quadratic). Part (b) requires setting up equations for two particles with different start times and positions, then solving simultaneously. While multi-step, it follows predictable patterns with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(40 - gt = 0\ [t=4]\) | M1 | Using \(v = u + at\) with \(u=40\), \(v=0\) and \(a=-g\) to find the time taken to reach the highest point |
| Time to top of building \(= 4 - \frac{1}{2}(4) = 2\) | A1 | May see \(t = 4+2=6\) for A1 |
| \(h = 40\times 2 - \frac{1}{2}\times 10\times 2^2\) or \(h = 40\times 6 - \frac{1}{2}\times 10\times 6^2\) | M1 | Using \(s = ut + \frac{1}{2}at^2\) with \(u=40\), \(a=-g\) and \(t=2\) or \(t=6\) to set up an equation which enables the value of \(h\) to be found |
| \(h = 60\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0 = 40^2 + 2\times(-10)\times H\) | M1 | For using \(v^2 = u^2 + 2as\) with \(u=40\), \(v=0\) and \(a=-g\) to find \(H\), the greatest height achieved |
| \(H = 80\) | A1 | |
| \(s = \frac{1}{2}\times 10\times 2^2\) | M1 | Use either \(s = vt - \frac{1}{2}at^2\) with \(v=0\), \(a=-g\), \(t=2\) or use \(s = ut + \frac{1}{2}at^2\) with \(u=0\), \(a=g\), \(t=2\) to find the distance travelled either in the final 2 seconds going up or the first 2 seconds going down |
| \(s = 20\) and so \(h = 80 - 20 = 60\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Height of first particle above ground \(= 40t - \frac{1}{2}\times 10t^2\) | B1 | |
| Height of second particle above top of building \(= 20(t-1) - \frac{1}{2}\times 10\times(t-1)^2\) | B1 | |
| \(60 + 20(t-1) - \frac{1}{2}\times 10\times(t-1)^2 = 40t - \frac{1}{2}\times 10t^2\) | M1 | Set up an equation involving expressions for displacement to enable the time at which the particles reach the same height to be found |
| \(t = 3.5\) seconds | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(h_1 = 40\times 1 - 5\times 1^2\ [=35]\) and \(v_1 = 40 - 10\times 1\ [=30]\) | B1 | Distance travelled and speed of first particle after 1 second |
| \(H_1 = 30T - 5\times T^2,\ H_2 = 20T - 5\times T^2\) | B1 | Distance travelled by both particles, \(T\) seconds after the second particle is projected |
| \(30T - 5\times T^2 = 20T - 5\times T^2 + (60-35)\) | M1 | Set up an equation in \(T\) involving expressions for displacement to enable the time at which the particles are at the same height to be found |
| \(T = 2.5\) and so time to meet \(= 2.5 + 1 = 3.5\) seconds | A1 | |
| 4 |
## Question 5:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $40 - gt = 0\ [t=4]$ | M1 | Using $v = u + at$ with $u=40$, $v=0$ and $a=-g$ to find the time taken to reach the highest point |
| Time to top of building $= 4 - \frac{1}{2}(4) = 2$ | A1 | May see $t = 4+2=6$ for A1 |
| $h = 40\times 2 - \frac{1}{2}\times 10\times 2^2$ or $h = 40\times 6 - \frac{1}{2}\times 10\times 6^2$ | M1 | Using $s = ut + \frac{1}{2}at^2$ with $u=40$, $a=-g$ and $t=2$ or $t=6$ to set up an equation which enables the value of $h$ to be found |
| $h = 60$ | A1 | |
**Alternative Method for 5(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $0 = 40^2 + 2\times(-10)\times H$ | M1 | For using $v^2 = u^2 + 2as$ with $u=40$, $v=0$ and $a=-g$ to find $H$, the greatest height achieved |
| $H = 80$ | A1 | |
| $s = \frac{1}{2}\times 10\times 2^2$ | M1 | Use either $s = vt - \frac{1}{2}at^2$ with $v=0$, $a=-g$, $t=2$ or use $s = ut + \frac{1}{2}at^2$ with $u=0$, $a=g$, $t=2$ to find the distance travelled either in the final 2 seconds going up or the first 2 seconds going down |
| $s = 20$ and so $h = 80 - 20 = 60$ | A1 | |
| | **4** | |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Height of first particle above ground $= 40t - \frac{1}{2}\times 10t^2$ | B1 | |
| Height of second particle above top of building $= 20(t-1) - \frac{1}{2}\times 10\times(t-1)^2$ | B1 | |
| $60 + 20(t-1) - \frac{1}{2}\times 10\times(t-1)^2 = 40t - \frac{1}{2}\times 10t^2$ | M1 | Set up an equation involving expressions for displacement to enable the time at which the particles reach the same height to be found |
| $t = 3.5$ seconds | A1 | |
**Alternative Method for 5(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $h_1 = 40\times 1 - 5\times 1^2\ [=35]$ and $v_1 = 40 - 10\times 1\ [=30]$ | B1 | Distance travelled **and** speed of first particle after 1 second |
| $H_1 = 30T - 5\times T^2,\ H_2 = 20T - 5\times T^2$ | B1 | Distance travelled by both particles, $T$ seconds after the second particle is projected |
| $30T - 5\times T^2 = 20T - 5\times T^2 + (60-35)$ | M1 | Set up an equation in $T$ involving expressions for displacement to enable the time at which the particles are at the same height to be found |
| $T = 2.5$ and so time to meet $= 2.5 + 1 = 3.5$ seconds | A1 | |
| | **4** | |5 A particle is projected vertically upwards with speed $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ alongside a building of height $h \mathrm {~m}$.
\begin{enumerate}[label=(\alph*)]
\item Given that the particle is above the level of the top of the building for 4 s , find $h$.
\item One second after the first particle is projected, a second particle is projected vertically upwards from the top of the building with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Denoting the time after projection of the first particle by $t \mathrm {~s}$, find the value of $t$ for which the two particles are at the same height above the ground.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q5 [8]}}