| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Loss of energy in collision |
| Difficulty | Moderate -0.8 This is a straightforward energy conservation problem with clearly stated energy loss. Part (a) requires simple PE→KE→KE-8J→PE calculation. Part (b) needs basic SUVAT equations applied twice. Both parts follow standard textbook methods with no problem-solving insight required, making it easier than average. |
| Spec | 3.02h Motion under gravity: vector form6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{PE} = 1.6 \times 10 \times 5\ [= 80\text{ J}]\) or \(v\downarrow = \sqrt{2 \times 10 \times 5}\ [= 10]\), \(\text{KE} = \frac{1}{2} \times 1.6 \times 10^2\ [= 80\text{ J}]\) | B1 | Either finds PE loss, or uses \(v^2 = u^2 + 2as\) to find velocity and hence KE on reaching ground |
| \(1.6 \times 10 \times 5 = 1.6 \times 10 \times h + 8\) or \(\frac{1}{2} \times 1.6 \times v^2 = 80 - 8,\ v\uparrow = \sqrt{90}\), \(0 = 90 + 2\times(-10)\times h\) | M1 | Using Initial PE = Final PE + Loss in KE, or using \(\text{KE} = \frac{1}{2}mv^2\) to find initial velocity upwards and either \(v^2 = u^2 + 2as\) or KE loss = PE gain to form equation in \(h\) |
| \(h = 4.5\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(5 = 0 + \frac{1}{2} \times 10 \times t^2\) leading to \(t = 1\) | M1 | Use of \(s = ut + \frac{1}{2}gt^2\) for downward motion, or \(s = \frac{1}{2}(u+v)t\), or \(v = u + gt\) for downward motion |
| \(4.5 = 0 - \frac{1}{2} \times (-10) \times t^2\) leading to \(t = \sqrt{0.9}\) | M1 | Use of \(s = vt - \frac{1}{2}(-g)t^2\) for upward motion, or \(s = \frac{1}{2}(u+v)t\), or \(v = u - gt\) for upward motion |
| \(t = 1.95\) s | A1 |
## Question 3:
### Part (a):
| $\text{PE} = 1.6 \times 10 \times 5\ [= 80\text{ J}]$ or $v\downarrow = \sqrt{2 \times 10 \times 5}\ [= 10]$, $\text{KE} = \frac{1}{2} \times 1.6 \times 10^2\ [= 80\text{ J}]$ | B1 | Either finds PE loss, or uses $v^2 = u^2 + 2as$ to find velocity and hence KE on reaching ground |
| $1.6 \times 10 \times 5 = 1.6 \times 10 \times h + 8$ or $\frac{1}{2} \times 1.6 \times v^2 = 80 - 8,\ v\uparrow = \sqrt{90}$, $0 = 90 + 2\times(-10)\times h$ | M1 | Using Initial PE = Final PE + Loss in KE, or using $\text{KE} = \frac{1}{2}mv^2$ to find initial velocity upwards and either $v^2 = u^2 + 2as$ or KE loss = PE gain to form equation in $h$ |
| $h = 4.5$ m | A1 | |
### Part (b):
| $5 = 0 + \frac{1}{2} \times 10 \times t^2$ leading to $t = 1$ | M1 | Use of $s = ut + \frac{1}{2}gt^2$ for downward motion, or $s = \frac{1}{2}(u+v)t$, or $v = u + gt$ for downward motion |
| $4.5 = 0 - \frac{1}{2} \times (-10) \times t^2$ leading to $t = \sqrt{0.9}$ | M1 | Use of $s = vt - \frac{1}{2}(-g)t^2$ for upward motion, or $s = \frac{1}{2}(u+v)t$, or $v = u - gt$ for upward motion |
| $t = 1.95$ s | A1 | |
---3 A ball of mass 1.6 kg is released from rest at a point 5 m above horizontal ground. When the ball hits the ground it instantaneously loses 8 J of kinetic energy and starts to move upwards.
\begin{enumerate}[label=(\alph*)]
\item Use an energy method to find the greatest height that the ball reaches after hitting the ground.
\item Find the total time taken, from the initial release of the ball until it reaches this greatest height.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q3 [6]}}