| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical motion: energy loss on impact |
| Difficulty | Moderate -0.3 This is a straightforward energy conservation problem with standard SUVAT kinematics. Part (i) requires calculating initial KE (mgh = 19.6J), subtracting the energy loss (6.8J remaining), then converting back to height. Part (ii) uses basic SUVAT with g=9.8 m/s². While it involves two stages of motion, each step follows routine procedures with no conceptual challenges or problem-solving insight required—slightly easier than average due to its mechanical nature. |
| Spec | 3.02h Motion under gravity: vector form6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| PE loss \(= 0.4g \times 5 = 20\) J | B1 | |
| Initial \(KE_{up} = 0.4g \times 5 - 12.8 = 7.2\) J | B1 | |
| \([0.4gh = 2g - 12.8]\) | M1 | Uses PE gain \(=\) KE loss to form equation in \(h\) |
| Height reached is \(1.8\) m | A1 [4] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v^2 = 2 \times 10 \times 5 \rightarrow (v = 10)\) | B1 | |
| \(KE\ \text{loss} = \frac{1}{2}\ 0.4(10^2 - v_{up}^2) = 12.8\) | B1 | |
| \([v_{up} = 60,\quad 0 = 6^2 - 2gh]\) | M1 | Uses \(v^2 = u^2 - 2gs\) to form equation in \(h\) |
| Height reached is \(1.8\) m | A1 [4] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.4gh = 12.8\) | M1 | Uses PE gain \(=\) KE loss |
| \(h = 3.2\) m | A1 | |
| \([\text{Height reached} = 5 - 12.8/0.4g]\) | M1 | Uses height reached \(= 5 -\) 'height not reached' |
| Height reached is \(1.8\) m | A1 [4] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times 0.4v^2 = 12.8\ (v=8)\) and \([8^2 = 0^2 + 2gh]\) | M1 | Uses KE loss \(= 12.8\) and \(v^2 = u^2 + 2gs\) |
| \(h = 3.2\) m | A1 | |
| \([\text{Height reached} = 5 - 3.2]\) | M1 | Uses height reached \(= 5 -\) 'height not reached' |
| Height reached is \(1.8\) m | A1 [4] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5 = 0 + \frac{1}{2}g t_{down}^2\ (t_{down} = 1)\) | B1 | |
| \(0 = 6 - gt_{up}\) or \(1.8 = \frac{1}{2}g t_{up}^2\ (t_{up} = 0.6)\) | B1 | |
| Total time is \(1.6\) s | B1 [3] |
## Question 4:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| PE loss $= 0.4g \times 5 = 20$ J | B1 | |
| Initial $KE_{up} = 0.4g \times 5 - 12.8 = 7.2$ J | B1 | |
| $[0.4gh = 2g - 12.8]$ | M1 | Uses PE gain $=$ KE loss to form equation in $h$ |
| Height reached is $1.8$ m | A1 [4] | AG |
**First Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v^2 = 2 \times 10 \times 5 \rightarrow (v = 10)$ | B1 | |
| $KE\ \text{loss} = \frac{1}{2}\ 0.4(10^2 - v_{up}^2) = 12.8$ | B1 | |
| $[v_{up} = 60,\quad 0 = 6^2 - 2gh]$ | M1 | Uses $v^2 = u^2 - 2gs$ to form equation in $h$ |
| Height reached is $1.8$ m | A1 [4] | AG |
**Second Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.4gh = 12.8$ | M1 | Uses PE gain $=$ KE loss |
| $h = 3.2$ m | A1 | |
| $[\text{Height reached} = 5 - 12.8/0.4g]$ | M1 | Uses height reached $= 5 -$ 'height not reached' |
| Height reached is $1.8$ m | A1 [4] | AG |
**Third Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 0.4v^2 = 12.8\ (v=8)$ and $[8^2 = 0^2 + 2gh]$ | M1 | Uses KE loss $= 12.8$ and $v^2 = u^2 + 2gs$ |
| $h = 3.2$ m | A1 | |
| $[\text{Height reached} = 5 - 3.2]$ | M1 | Uses height reached $= 5 -$ 'height not reached' |
| Height reached is $1.8$ m | A1 [4] | AG |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5 = 0 + \frac{1}{2}g t_{down}^2\ (t_{down} = 1)$ | B1 | |
| $0 = 6 - gt_{up}$ or $1.8 = \frac{1}{2}g t_{up}^2\ (t_{up} = 0.6)$ | B1 | |
| Total time is $1.6$ s | B1 [3] | |
---4 A small ball of mass 0.4 kg is released from rest at a point 5 m above horizontal ground. At the instant the ball hits the ground it loses 12.8 J of kinetic energy and starts to move upwards.\\
(i) Show that the greatest height above the ground that the ball reaches after hitting the ground is 1.8 m .\\
(ii) Find the time taken for the ball's motion from its release until reaching this greatest height.
\hfill \mbox{\textit{CAIE M1 2014 Q4 [7]}}