OCR MEI M1 — Question 2 18 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeTwo projectiles meeting - 2D flight
DifficultyStandard +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 time delay element add length but not conceptual difficulty—all steps follow textbook methods with no novel insight required.
Spec3.02d Constant acceleration: SUVAT formulae3.02e Two-dimensional constant acceleration: with vectors3.02i Projectile motion: constant acceleration model
2 In this question the value of \(g\) should be taken as \(10 \mathrm {~m \mathrm {~s} ^ { 2 }\).} 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]
\includegraphics[alt={},max width=\textwidth]{362d5995-bd39-4b07-b6a4-63eb1dd3e69d-2_461_1114_464_505} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Show that, \(t\) seconds after projection, the height in metres of each particle above its point of projection is \(10 t - 5 t ^ { 2 }\).
  2. Calculate the horizontal range of A . Deduce that A hits the horizontal ground between the initial positions of A and B .
  3. Calculate the horizontal distance travelled by B before reaching the ground.
  4. 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 .
  5. Verify that the particles collide 0.75 seconds after A is projected.
Question 2:
Part (i)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Using \(s = ut + 0.5at^2\) with \(u = 10\) and \(a = -10\)E1 Must be clear evidence of derivation of \(-5\). Accept one calculation and no statement about the other.
1
Part (ii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Either: \(s = 0\) gives \(10t - 5t^2 = 0\)B1
so \(5t(2-t) = 0\)M1 Factorising
so \(t = 0\) or \(2\). Clearly need \(t = 2\)A1 Award 3 marks for \(t = 2\) seen WWW
Or: Time to highest point given by \(0 = 10 - 10t\)M1
Time of flight is \(2 \times 1 = 2\) sM1, A1 Dep on 1st M1. Doubling their \(t\). Properly obtained
Horizontal range is \(40\) m, as \(40 < 70\), hits the groundB1, E1 FT \(20 \times\) their \(t\). Must be clear. FT their range.
5
Part (iii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Need \(10t - 5t^2 = -15\)M1 Equate \(s = -15\) or equivalent. Allow use of \(\pm 15\)
Solving \(t^2 - 2t - 3 = 0\)M1 Method leading to solution of quadratic. Equivalent form will do.
so \((t-3)(t+1) = 0\) and \(t = 3\)A1 Obtaining \(t = 3\). Allow no reference to other root. [Award SC3 if \(t = 3\) seen WWW]
Range is \(60\) mM1, A1 Range is \(20 \times\) their \(t\) (provided \(t > 0\)). cao. CWO.
5
Part (iv)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Using (ii) & (iii), since \(40 + 60 > 70\), paths crossE1 Must be convincing. Accept sketches.
For \(0 < t \leq 2\) both have same vertical motion so B is always \(15\) m above AE1 Do not accept evaluation at one or more points alone. That B is *always* above A must be clear.
2
Part (v)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Need \(x\) components summing to \(70\): \(20 \times 0.75 + 20 \times 2.75 = 15 + 55 = 70\) so trueM1, E1 May be implied. Or correct derivation of \(0.75\) s or \(2.75\) s
Need \(y\) components the sameM1 Attempt to use \(0.75\) and \(2.75\) in two vertical height equations (accept same one or wrong one)
\(10 \times 2.75 - 5 \times 2.75^2 + 15 = 4.6875\)B1 \(0.75\) and \(2.75\) each substituted in appropriate equation. Both values correct.
\(10 \times 0.75 - 5 \times 0.75^2 = 4.6875\)E1 [Using Cartesian equation: B1, B1 each equation; M1 solving; A1 correct point; E1 Verify times]
5
Total18 
# Question 2:

## Part (i)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $s = ut + 0.5at^2$ with $u = 10$ and $a = -10$ | E1 | Must be clear evidence of derivation of $-5$. Accept one calculation and no statement about the other. |
| | **1** | |

## Part (ii)

| Answer/Working | Mark | Guidance |
|---|---|---|
| **Either:** $s = 0$ gives $10t - 5t^2 = 0$ | B1 | |
| so $5t(2-t) = 0$ | M1 | Factorising |
| so $t = 0$ or $2$. Clearly need $t = 2$ | A1 | Award 3 marks for $t = 2$ seen WWW |
| **Or:** Time to highest point given by $0 = 10 - 10t$ | M1 | |
| Time of flight is $2 \times 1 = 2$ s | M1, A1 | Dep on 1st M1. Doubling **their** $t$. Properly obtained |
| Horizontal range is $40$ m, as $40 < 70$, hits the ground | B1, E1 | FT $20 \times$ **their** $t$. Must be clear. FT **their** range. |
| | **5** | |

## Part (iii)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Need $10t - 5t^2 = -15$ | M1 | Equate $s = -15$ or equivalent. Allow use of $\pm 15$ |
| Solving $t^2 - 2t - 3 = 0$ | M1 | Method leading to solution of quadratic. Equivalent form will do. |
| so $(t-3)(t+1) = 0$ and $t = 3$ | A1 | Obtaining $t = 3$. Allow no reference to other root. [Award SC3 if $t = 3$ seen WWW] |
| Range is $60$ m | M1, A1 | Range is $20 \times$ **their** $t$ (provided $t > 0$). cao. CWO. |
| | **5** | |

## Part (iv)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Using (ii) & (iii), since $40 + 60 > 70$, paths cross | E1 | Must be convincing. Accept sketches. |
| For $0 < t \leq 2$ both have same vertical motion so B is always $15$ m above A | E1 | Do not accept evaluation at one or more points alone. That B is *always* above A must be clear. |
| | **2** | |

## Part (v)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Need $x$ components summing to $70$: $20 \times 0.75 + 20 \times 2.75 = 15 + 55 = 70$ so true | M1, E1 | May be implied. Or correct derivation of $0.75$ s or $2.75$ s |
| Need $y$ components the same | M1 | Attempt to use $0.75$ and $2.75$ in two vertical height equations (accept same one or wrong one) |
| $10 \times 2.75 - 5 \times 2.75^2 + 15 = 4.6875$ | B1 | $0.75$ and $2.75$ each substituted in appropriate equation. Both values correct. |
| $10 \times 0.75 - 5 \times 0.75^2 = 4.6875$ | E1 | [Using Cartesian equation: B1, B1 each equation; M1 solving; A1 correct point; E1 Verify times] |
| | **5** | |
| **Total** | **18** | |
2 In this question the value of $g$ should be taken as $10 \mathrm {~m \mathrm {~s} ^ { 2 }$.}
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]{362d5995-bd39-4b07-b6a4-63eb1dd3e69d-2_461_1114_464_505}
\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  Q2 [18]}}
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