| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.3 This is a straightforward three-part projectiles question using standard SUVAT equations. Part (i) uses time of flight formula, (ii) applies maximum height formula, and (iii) requires finding when tan(angle) = 1 using velocity components. All parts follow routine procedures with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using \(y = \dot{y}_0 t - \frac{1}{2}gt^2\) with \(y=0\), \(t=10\) or \(\dot{y} = \dot{y}_0 - gt\) with \(\dot{y}=0\), \(t=5\) | M1 | |
| \(0 = 60\sin\alpha \times 10 - \frac{1}{2} \times 10 \times 10^2\) or \(0 = 60\sin\alpha - 10 \times 5\) | A1 | |
| \(\alpha = 56.4°\) | A1 | NB: Use of \(\dot{y}_0 = 60\) in (i) and (ii) is M0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substituting \(t=5\) into \(y = \dot{y}_0 t - \frac{1}{2}gt^2\) or \(\dot{y}=0\) into \(\dot{y}^2 = \dot{y}_0^2 - 2gy\) or \(\dot{y}=0\), \(t=5\) into \(y = \dfrac{\dot{y}_0 + \dot{y}}{2}t\) | M1 | |
| Greatest height is 125 m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dot{y} = 60\sin\alpha - gT\) | B1 | |
| \(\dot{x} = 60\cos\alpha\) | B1 | |
| Attempting to solve \(\dot{x} = \dot{y}\), or complete method for equation in \(T\) using \(\dot{x} = \dot{y}\) | M1 | |
| \(T = 1.68\) | A1 |
# Question 6:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Using $y = \dot{y}_0 t - \frac{1}{2}gt^2$ with $y=0$, $t=10$ or $\dot{y} = \dot{y}_0 - gt$ with $\dot{y}=0$, $t=5$ | M1 | |
| $0 = 60\sin\alpha \times 10 - \frac{1}{2} \times 10 \times 10^2$ or $0 = 60\sin\alpha - 10 \times 5$ | A1 | |
| $\alpha = 56.4°$ | A1 | NB: Use of $\dot{y}_0 = 60$ in (i) and (ii) is M0 |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substituting $t=5$ into $y = \dot{y}_0 t - \frac{1}{2}gt^2$ or $\dot{y}=0$ into $\dot{y}^2 = \dot{y}_0^2 - 2gy$ or $\dot{y}=0$, $t=5$ into $y = \dfrac{\dot{y}_0 + \dot{y}}{2}t$ | M1 | |
| Greatest height is 125 m | A1 | |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dot{y} = 60\sin\alpha - gT$ | B1 | |
| $\dot{x} = 60\cos\alpha$ | B1 | |
| Attempting to solve $\dot{x} = \dot{y}$, or complete method for equation in $T$ using $\dot{x} = \dot{y}$ | M1 | |
| $T = 1.68$ | A1 | |
---6 A particle is projected with speed $60 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point on horizontal ground. The angle of projection is $\alpha ^ { \circ }$ above the horizontal. The particle reaches the ground again after 10 s .\\
(i) Find the value of $\alpha$.\\
(ii) Find the greatest height reached by the particle.\\
(iii) At time $T$ s after the instant of projection the direction of motion of the particle is at an angle of $45 ^ { \circ }$ above the horizontal. Find the value of $T$.
\hfill \mbox{\textit{CAIE M2 2003 Q6 [9]}}