| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: different start times, same height |
| Difficulty | Standard +0.3 This is a standard two-particle SUVAT problem requiring careful tracking of different time references and straightforward application of kinematic equations. Part (i) involves setting up equations for both balls and solving simultaneously (routine but requires care with timing). Part (ii) requires finding when ball A hits ground, then calculating ball B's motion including its maximum height to find total distance. While multi-step, all techniques are standard M1 fare with no novel insights required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s_A = \frac{1}{2}g \times 2.5^2\ (= 31.25)\) | B1 | |
| \([s_B = 20 \times 1.5 - \frac{1}{2}g \times 1.5^2]\ (= 18.75)\) | M1 | For using \(s = ut + \frac{1}{2}at^2\) |
| \(\frac{1}{2}g \times 2.5^2 + 20 \times 1.5 - \frac{1}{2}g \times 1.5^2\), Height is \(50\text{ m}\) | AG A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(50 = 0.5g t_A^2 \quad (t_A = 3.16)\) | B1 | For using \(s = \frac{1}{2}at^2\) |
| \(t_B = \sqrt{10} - 1 = 2.16\) | B1 | |
| To top: \(0^2 = 20^2 - 2gs_B \rightarrow s_B = 20\) | B1 | |
| To top: \([0 = 20 - gt_B] \rightarrow t_B = 2\); Downwards: \([s_B = \frac{1}{2}g(0.16)^2]\ (= 0.13)\) | M1 | For using \(v = u + at\) to find time to top for B and \(s = \frac{1}{2}at^2\) to find downwards distance for B |
| Total distance is \(20.1\text{ m}\) | A1 | [5] |
# Question 4:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s_A = \frac{1}{2}g \times 2.5^2\ (= 31.25)$ | B1 | |
| $[s_B = 20 \times 1.5 - \frac{1}{2}g \times 1.5^2]\ (= 18.75)$ | M1 | For using $s = ut + \frac{1}{2}at^2$ |
| $\frac{1}{2}g \times 2.5^2 + 20 \times 1.5 - \frac{1}{2}g \times 1.5^2$, Height is $50\text{ m}$ | AG A1 | [3] |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $50 = 0.5g t_A^2 \quad (t_A = 3.16)$ | B1 | For using $s = \frac{1}{2}at^2$ |
| $t_B = \sqrt{10} - 1 = 2.16$ | B1 | |
| To top: $0^2 = 20^2 - 2gs_B \rightarrow s_B = 20$ | B1 | |
| To top: $[0 = 20 - gt_B] \rightarrow t_B = 2$; Downwards: $[s_B = \frac{1}{2}g(0.16)^2]\ (= 0.13)$ | M1 | For using $v = u + at$ to find time to top for B **and** $s = \frac{1}{2}at^2$ to find downwards distance for B |
| Total distance is $20.1\text{ m}$ | A1 | [5] |
---4 A ball $A$ is released from rest at the top of a tall tower. One second later, another ball $B$ is projected vertically upwards from ground level near the bottom of the tower with a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The two balls are at the same height 1.5 s after ball $B$ is projected.\\
(i) Show that the height of the tower is 50 m .\\
(ii) Find the length of time for which ball $B$ has been in motion when ball $A$ reaches the ground. Hence find the total distance travelled by ball $B$ up to the instant when ball $A$ reaches the ground.
\hfill \mbox{\textit{CAIE M1 2016 Q4 [8]}}