| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Free fall: time or distance |
| Difficulty | Moderate -0.8 This is a straightforward two-part SUVAT question requiring standard application of kinematic equations (v² = u² + 2as) twice with clearly defined conditions. The algebra is simple and the problem-solving approach is routine for mechanics students, making it easier than average but not trivial. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([V^2 = (V-10)^2 + 2g \times 35]\) | M1 | For using \(v^2 = u^2 + 2gs\) to obtain an equation in \(V\) only or to obtain two equations in \(V\) and \(H\) and attempting to eliminate \(H\) |
| \(20V = 100 + 70g\) | A1 | |
| \(V = 40\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = V - 10 + 10t \rightarrow t = 1\) and \(35 = (V-10) \times 1 + \frac{1}{2} \times 10 \times 1^2\) or \(35 = (V - 10 + V)/2 \times 1\) | M1, A1 | A complete method to find \(V\) by considering the final 35 m using \(v = u + at\) and either \(s = ut + \frac{1}{2}at^2\) or \(s = (u+v)/2 \times t\) |
| \(V = 40\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([40^2 = 0^2 + 20H]\) | M1 | For using \(v^2 = u^2 + 2gs\) |
| \(H = 80\) | A1 | |
| Total: 2 |
# Question 2:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[V^2 = (V-10)^2 + 2g \times 35]$ | M1 | For using $v^2 = u^2 + 2gs$ to obtain an equation in $V$ only **or** to obtain two equations in $V$ and $H$ and attempting to eliminate $H$ |
| $20V = 100 + 70g$ | A1 | |
| $V = 40$ | A1 | |
| | **Total: 3** | |
### Alternative for 2(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = V - 10 + 10t \rightarrow t = 1$ and $35 = (V-10) \times 1 + \frac{1}{2} \times 10 \times 1^2$ or $35 = (V - 10 + V)/2 \times 1$ | M1, A1 | A complete method to find $V$ by considering the final 35 m using $v = u + at$ and either $s = ut + \frac{1}{2}at^2$ or $s = (u+v)/2 \times t$ |
| $V = 40$ | A1 | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[40^2 = 0^2 + 20H]$ | M1 | For using $v^2 = u^2 + 2gs$ |
| $H = 80$ | A1 | |
| | **Total: 2** | |
---2 A particle is released from rest at a point $H \mathrm {~m}$ above horizontal ground and falls vertically. The particle passes through a point 35 m above the ground with a speed of $( V - 10 ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and reaches the ground with a speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find\\
(i) the value of $V$,\\
(ii) the value of $H$.
\hfill \mbox{\textit{CAIE M1 2015 Q2 [5]}}