| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two projectiles meeting - 2D flight |
| Difficulty | Standard +0.3 This is a standard two-projectile problem with straightforward kinematics. Part (i) is routine SUVAT application, parts (ii)-(iii) involve basic range calculations, part (iv) requires comparing positions at equal times (standard technique), and part (v) is simple verification by substitution. The multi-part structure and collision analysis add some length but no conceptual challenge beyond typical M1 projectile questions. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02e Two-dimensional constant acceleration: with vectors3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical: \(s = 10t - \frac{1}{2}(10)t^2 = 10t - 5t^2\) | B1 | Using \(s=ut+\frac{1}{2}at^2\) with \(u=10\), \(a=-10\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| A hits ground when \(10t - 5t^2 = 0 \Rightarrow t(10-5t)=0 \Rightarrow t=2\) s | M1 A1 | |
| Horizontal range of A: \(20 \times 2 = 40\) m | M1 A1 | |
| Since \(40 < 70\), A lands between initial positions | A1 | Must justify |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| B starts 15 m above ground. Height above projection: \(10t-5t^2\). Height above ground: \(15 + 10t - 5t^2 = 0\) | M1 | |
| \(5t^2 - 10t - 15 = 0 \Rightarrow t^2 - 2t - 3 = 0 \Rightarrow (t-3)(t+1)=0\) | M1 A1 | |
| \(t = 3\) s, horizontal distance \(= 20 \times 3 = 60\) m | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Paths cross when horizontal positions coincide: \(20t_A = 70 - 20t_B\) with same \(t\) gives \(40t=70\), \(t=1.75\) | M1 | |
| Height of A: \(10(1.75)-5(1.75)^2 = 17.5-15.3125=2.19\) m | A1 | |
| Height of B: \(15+10(1.75)-5(1.75)^2 = 15+2.19=17.19\) m | A1 | |
| Heights differ so particles do not collide | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| A projected at \(t=0\), B projected 2 s earlier so B has been travelling for \(t_B = t_A + 2\) | M1 | |
| At \(t_A = 0.75\): A horizontal position \(= 20(0.75) = 15\) m from A's start | A1 | |
| B horizontal position from B's start \(= 20(2.75) = 55\) m, so from A's start \(= 70-55=15\) m | A1 | |
| Height of A: \(10(0.75)-5(0.75)^2 = 7.5-2.8125=4.6875\) m | M1 | |
| Height of B above ground: \(15+10(2.75)-5(2.75)^2=15+27.5-37.8125=4.6875\) m | A1 | Same height confirmed, hence collision |
# Question 8:
**(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical: $s = 10t - \frac{1}{2}(10)t^2 = 10t - 5t^2$ | B1 | Using $s=ut+\frac{1}{2}at^2$ with $u=10$, $a=-10$ |
**(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| A hits ground when $10t - 5t^2 = 0 \Rightarrow t(10-5t)=0 \Rightarrow t=2$ s | M1 A1 | |
| Horizontal range of A: $20 \times 2 = 40$ m | M1 A1 | |
| Since $40 < 70$, A lands between initial positions | A1 | Must justify |
**(iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| B starts 15 m above ground. Height above projection: $10t-5t^2$. Height above ground: $15 + 10t - 5t^2 = 0$ | M1 | |
| $5t^2 - 10t - 15 = 0 \Rightarrow t^2 - 2t - 3 = 0 \Rightarrow (t-3)(t+1)=0$ | M1 A1 | |
| $t = 3$ s, horizontal distance $= 20 \times 3 = 60$ m | A1 A1 | |
**(iv)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Paths cross when horizontal positions coincide: $20t_A = 70 - 20t_B$ with same $t$ gives $40t=70$, $t=1.75$ | M1 | |
| Height of A: $10(1.75)-5(1.75)^2 = 17.5-15.3125=2.19$ m | A1 | |
| Height of B: $15+10(1.75)-5(1.75)^2 = 15+2.19=17.19$ m | A1 | |
| Heights differ so particles do not collide | A1 | |
**(v)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| A projected at $t=0$, B projected 2 s earlier so B has been travelling for $t_B = t_A + 2$ | M1 | |
| At $t_A = 0.75$: A horizontal position $= 20(0.75) = 15$ m from A's start | A1 | |
| B horizontal position from B's start $= 20(2.75) = 55$ m, so from A's start $= 70-55=15$ m | A1 | |
| Height of A: $10(0.75)-5(0.75)^2 = 7.5-2.8125=4.6875$ m | M1 | |
| Height of B above ground: $15+10(2.75)-5(2.75)^2=15+27.5-37.8125=4.6875$ m | A1 | Same height confirmed, hence collision |8 In this question the value of $\boldsymbol { g $ should be taken as $\mathbf { 1 0 } \mathbf { m ~ s } ^ { \mathbf { - 2 } }$.}
As shown in Fig. 8, particles A and B are projected towards one another. Each particle has an initial speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ vertically and $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ horizontally. Initially A and B are 70 m apart horizontally and B is 15 m higher than A . Both particles are projected over horizontal ground.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{52d6c914-b204-4587-a82e-fbab6693fcf8-6_476_1111_518_475}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Show that, $t$ seconds after projection, the height in metres of each particle above its point of projection is $10 t - 5 t ^ { 2 }$.\\
(ii) Calculate the horizontal range of A . Deduce that A hits the horizontal ground between the initial positions of A and B .\\
(iii) Calculate the horizontal distance travelled by B before reaching the ground.\\
(iv) Show that the paths of the particles cross but that the particles do not collide if they are projected at the same time.
In fact, particle A is projected 2 seconds after particle B .\\
(v) Verify that the particles collide 0.75 seconds after A is projected.
\hfill \mbox{\textit{OCR MEI M1 2007 Q8 [18]}}