CAIE M2 2006 November — Question 4 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2006
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeTime when specific condition met
DifficultyStandard +0.3 This is a standard projectile motion problem requiring students to use symmetry of parabolic flight. Given initial conditions and one time, students must find the second time at the same height using kinematic equations, then calculate horizontal distance. It involves routine application of projectile formulas with straightforward arithmetic, slightly above average due to the multi-step nature and need to recognize the symmetry property.
Spec3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model
4 A stone is projected from a point on horizontal ground with speed \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\theta\) above the horizontal, where \(\sin \theta = \frac { 4 } { 5 }\). At time 1.2 s after projection the stone passes through the point \(A\). Subsequently the stone passes through the point \(B\), which is at the same height above the ground as \(A\). Find the horizontal distance \(A B\).
Question 4:
Method 1
AnswerMarks Guidance
Answer/WorkingMark Guidance
Height at \(A = 16.8\)B1
\(25 \times 0.8t - \frac{1}{2}10t^2 = 25 \times 0.8 \times 1.2 - \frac{1}{2}10 \times 1.2^2\)M1
\(t = (1.2),\ 2.8\)A1
\(AB = 25(2.8 - 1.2)0.6\)M1 For using Distance \(= V(t_B - t_A)\cos\theta\)
Distance \(= 24\)mA1 (5)
Method 2
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(R = 2 \times 25^2 \times 0.8 \times 0.6 \div 10\ (= 60)\)B1 Using \(R = \dfrac{2V^2 \sin\theta\cos\theta}{g}\) \((\theta = 53.13)\)
M1For using Distance \(= R - 2x(1.2)\)
DM1For using \(x = Vt\cos\theta\)
\(AB = 2 \times 25^2 \times 0.8 \times 0.6 \div 10 - 2 \times 25 \times 1.2 \times 0.6\)A1ft
Distance is \(24\)mA1 (5)
Method 3
AnswerMarks Guidance
Answer/WorkingMark Guidance
Height at \(A = 16.8\)B1
\(16.8 = \frac{4x}{3} - \dfrac{x^2}{45}\) or equivalentM1 For using trajectory equation from formula booklet
\(x = 18\) or \(x = 42\)A1
Distance \(= 42 - 18\)M1
Distance is \(24\)mA1 (5) 
## Question 4:

### Method 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| Height at $A = 16.8$ | B1 | |
| $25 \times 0.8t - \frac{1}{2}10t^2 = 25 \times 0.8 \times 1.2 - \frac{1}{2}10 \times 1.2^2$ | M1 | |
| $t = (1.2),\ 2.8$ | A1 | |
| $AB = 25(2.8 - 1.2)0.6$ | M1 | For using Distance $= V(t_B - t_A)\cos\theta$ |
| Distance $= 24$m | A1 (5) | |

### Method 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = 2 \times 25^2 \times 0.8 \times 0.6 \div 10\ (= 60)$ | B1 | Using $R = \dfrac{2V^2 \sin\theta\cos\theta}{g}$ $(\theta = 53.13)$ |
| | M1 | For using Distance $= R - 2x(1.2)$ |
| | DM1 | For using $x = Vt\cos\theta$ |
| $AB = 2 \times 25^2 \times 0.8 \times 0.6 \div 10 - 2 \times 25 \times 1.2 \times 0.6$ | A1ft | |
| Distance is $24$m | A1 (5) | |

### Method 3
| Answer/Working | Mark | Guidance |
|---|---|---|
| Height at $A = 16.8$ | B1 | |
| $16.8 = \frac{4x}{3} - \dfrac{x^2}{45}$ or equivalent | M1 | For using trajectory equation from formula booklet |
| $x = 18$ or $x = 42$ | A1 | |
| Distance $= 42 - 18$ | M1 | |
| Distance is $24$m | A1 (5) | |

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4 A stone is projected from a point on horizontal ground with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\theta$ above the horizontal, where $\sin \theta = \frac { 4 } { 5 }$. At time 1.2 s after projection the stone passes through the point $A$. Subsequently the stone passes through the point $B$, which is at the same height above the ground as $A$. Find the horizontal distance $A B$.

\hfill \mbox{\textit{CAIE M2 2006 Q4 [5]}}
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