| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Pile-driver or hammer impact |
| Difficulty | Moderate -0.5 This is a straightforward two-part mechanics problem requiring (a) conservation of momentum for a perfectly inelastic collision (standard formula application) and (b) work-energy principle or equations of motion with constant resistance. Both parts are routine M1 techniques with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02a Work done: concept and definition6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(120 \times 8 = 120v + 40v\) | M1 | Applying conservation of momentum |
| \(v = 6 \text{ ms}^{-1}\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1600 - 4800 = 160a\) leading to \(a = -20\) | M1 | Applying Newton's 2nd law to the system |
| \(0 = 6^2 + 2 \times (-20) \times s\) | M1 | Use of constant acceleration equations such as \(v^2 = u^2 + 2as\) |
| Distance travelled by post \(= 0.9\) m | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Initial \(\text{KE} = \frac{1}{2} \times 160 \times 6^2\) | M1 | Use of \(\text{KE} = \frac{1}{2}mv^2\) for combined mass |
| \(\frac{1}{2} \times 160 \times 6^2 + 160 \times 10 \times s = 4800s\) | M1 | Forms work/energy equation |
| Distance travelled by post \(= 0.9\) m | A1 |
## Question 1:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $120 \times 8 = 120v + 40v$ | M1 | Applying conservation of momentum |
| $v = 6 \text{ ms}^{-1}$ | A1 | |
| | **Total: 2** | |
### Part (b):
**Main Method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1600 - 4800 = 160a$ leading to $a = -20$ | M1 | Applying Newton's 2nd law to the system |
| $0 = 6^2 + 2 \times (-20) \times s$ | M1 | Use of constant acceleration equations such as $v^2 = u^2 + 2as$ |
| Distance travelled by post $= 0.9$ m | A1 | |
| | **Total: 3** | |
**Alternative Method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial $\text{KE} = \frac{1}{2} \times 160 \times 6^2$ | M1 | Use of $\text{KE} = \frac{1}{2}mv^2$ for combined mass |
| $\frac{1}{2} \times 160 \times 6^2 + 160 \times 10 \times s = 4800s$ | M1 | Forms work/energy equation |
| Distance travelled by post $= 0.9$ m | A1 | |1\\
\includegraphics[max width=\textwidth, alt={}, center]{cb2cec83-6f8d-4c13-90a1-03bbf4e4452f-03_471_613_254_766}
A metal post is driven vertically into the ground by dropping a heavy object onto it from above. The mass of the object is 120 kg and the mass of the post is 40 kg (see diagram). The object hits the post with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and remains in contact with it after the impact.
\begin{enumerate}[label=(\alph*)]
\item Calculate the speed with which the combined post and object moves immediately after the impact.
\item There is a constant force resisting the motion of magnitude 4800 N .
Calculate the distance the post is driven into the ground.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q1 [5]}}