| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Standard +0.3 This is a standard projectiles question requiring application of SUVAT equations and trajectory formulas. Parts (i) and (ii) are routine calculations using time of flight and maximum height formulas. Parts (iii) and (iv) add mild complexity by requiring students to find when the velocity vector makes specific angles, but this follows directly from tan(θ) = v_y/v_x using standard methods. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 65\sin\alpha\, t - \frac{1}{2}gt^2\) | B1 | May be implied |
| \(0 = 65 \times 12\sin\alpha - 5 \times 12^2\) | M1 | For using \(y(12) = 0\) and solving for \(\alpha\) |
| \(\alpha = 67.4\) | A1 | 3 |
| OR \(\dot{y} = 65\sin\alpha - gt\) | B1 | |
| \(0 = 65\sin\alpha - 10 \times 6\) | M1 | \(\dot{y}(6) = 0\) and solving for \(\alpha\) |
| \(\alpha = 67.4\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 65 \times 6 \times (60/65) - 5 \times 6^2\) | M1 | For using the value of \(\alpha\) and finding \(y(6)\) |
| Greatest height is 180 m | A1 | 2 |
| OR \(0 = 65^2(60/65)^2 - 2 \times 10 \times y\) | M1 | For solving \(0 = (65\sin\alpha)^2 - 2gy\) for \(y\) |
| Greatest height is 180 m | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dot{x} = 65\cos\alpha\) | B1 | |
| \(\dot{y} = 65\sin\alpha - 10t\) | B1 | |
| \((60 - 10t)/25 = \tan 20°\) \([T = 5.09\text{s}]\) | M1 | For using \(\dot{y}/\dot{x} = \tan 20°\) and attempting to solve |
| M1 | For using time interval \(= 12 - 2T\) | |
| Length of time is 1.82 s | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance is 45.5 m | A1ft | 1 |
## Question 7:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 65\sin\alpha\, t - \frac{1}{2}gt^2$ | B1 | May be implied |
| $0 = 65 \times 12\sin\alpha - 5 \times 12^2$ | M1 | For using $y(12) = 0$ and solving for $\alpha$ |
| $\alpha = 67.4$ | A1 | 3 | |
| **OR** $\dot{y} = 65\sin\alpha - gt$ | B1 | |
| $0 = 65\sin\alpha - 10 \times 6$ | M1 | $\dot{y}(6) = 0$ and solving for $\alpha$ |
| $\alpha = 67.4$ | A1 | (3) | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 65 \times 6 \times (60/65) - 5 \times 6^2$ | M1 | For using the value of $\alpha$ and finding $y(6)$ |
| Greatest height is 180 m | A1 | 2 | |
| **OR** $0 = 65^2(60/65)^2 - 2 \times 10 \times y$ | M1 | For solving $0 = (65\sin\alpha)^2 - 2gy$ for $y$ |
| Greatest height is 180 m | A1 | (2) | |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dot{x} = 65\cos\alpha$ | B1 | |
| $\dot{y} = 65\sin\alpha - 10t$ | B1 | |
| $(60 - 10t)/25 = \tan 20°$ $[T = 5.09\text{s}]$ | M1 | For using $\dot{y}/\dot{x} = \tan 20°$ and attempting to solve |
| | M1 | For using time interval $= 12 - 2T$ |
| Length of time is 1.82 s | A1 | 5 | |
### Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance is 45.5 m | A1ft | 1 | ft $25 \times$ candidate's ans (iii) |
**Total: 11 marks**7 A particle is projected with speed $65 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point on horizontal ground, in a direction making an angle of $\alpha ^ { \circ }$ above the horizontal. The particle reaches the ground again after 12 s . Find\\
(i) the value of $\alpha$,\\
(ii) the greatest height reached by the particle,\\
(iii) the length of time for which the direction of motion of the particle is between $20 ^ { \circ }$ above the horizontal and $20 ^ { \circ }$ below the horizontal,\\
(iv) the horizontal distance travelled by the particle in the time found in part (iii).
\hfill \mbox{\textit{CAIE M2 2007 Q7 [11]}}