| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| 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 question requiring (a) conservation of momentum for an inelastic collision with simple arithmetic, and (b) application of work-energy principle or equations of motion with constant acceleration. Both parts are standard M1 textbook exercises with no conceptual challenges, though slightly easier than average due to the direct setup and minimal algebraic manipulation required. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt at conservation of momentum: \(1.2v = (1.2 + 0.004) \times 40\) | M1 | |
| \(v = \frac{602}{15}\) | A1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0^2 = (40)^2 + 2 \times 0.04 \times a\) giving \([a = -20000]\), or \(0.04 = \frac{0+40}{2}t\) gets \(t = 0.002\), so \(0 = 40 + 0.002a\ [a=-20000]\) | M1 | Use of 'suvat' method to get equation in \(a\). Allow sign errors. Allow \(\pm 20000\). Do not allow 4 in place of 0.04. Allow use of 40.1 or \(\frac{602}{15}\) for velocity in place of 40. |
| Attempt Newton's Second Law vertically: \([-R + (1.2+0.004)g = (1.2+0.004) \times a]\), \([-R + 12.04 = 1.204a]\) | M1 | Must have correct number of relevant terms. Allow sign errors, but terms including masses must be effectively added. Do not allow any mass other than \((1.2 + 0.004)\). |
| \(R = 24\,100\ \text{N}\ \left[24\,092.04 = \frac{602301}{25}\right]\) | A1 | WWW. Note: use of wrong sign for \(g\) leads to answers 24 067.96 (max M1M1A0). Note: Missing weight term gets 24 080 (max M1M0A0). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Change in PE \(= 1.204g \times 0.04\ [= 0.4816]\), or change in KE \(= \frac{1}{2} \times 1.204 \times (40)^2\ [= 963.2]\) | B1 | Allow use of 40.1 or \(\frac{602}{15}\) for velocity in place of 40. B0 if extra KE terms present. |
| \(1.204g \times 0.04 + \frac{1}{2} \times 1.204 \times (40)^2 = 0.04R\) | M1 | Attempt at work-energy equation. Must have correct number of relevant terms, dimensionally correct; allow sign errors. Do not allow 4 in place of 0.04. Allow use of 40.1 or \(\frac{602}{15}\) for velocity in place of 40. |
| \(R = 24\,100\ \text{N}\ \left[24\,092.04 = \frac{602301}{25}\right]\) | A1 | WWW. Note: use of wrong sign for \(g\) leads to 24 067.96 (max B1M1A0). Missing PE term gets 24 080 (max B1M0A0). |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt at conservation of momentum: $1.2v = (1.2 + 0.004) \times 40$ | M1 | |
| $v = \frac{602}{15}$ | A1 | oe |
---
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0^2 = (40)^2 + 2 \times 0.04 \times a$ giving $[a = -20000]$, or $0.04 = \frac{0+40}{2}t$ gets $t = 0.002$, so $0 = 40 + 0.002a\ [a=-20000]$ | M1 | Use of 'suvat' method to get equation in $a$. Allow sign errors. Allow $\pm 20000$. Do not allow 4 in place of 0.04. Allow use of 40.1 or $\frac{602}{15}$ for velocity in place of 40. |
| Attempt Newton's Second Law vertically: $[-R + (1.2+0.004)g = (1.2+0.004) \times a]$, $[-R + 12.04 = 1.204a]$ | M1 | Must have correct number of relevant terms. Allow sign errors, but terms including masses must be effectively added. Do not allow any mass other than $(1.2 + 0.004)$. |
| $R = 24\,100\ \text{N}\ \left[24\,092.04 = \frac{602301}{25}\right]$ | A1 | WWW. Note: use of wrong sign for $g$ leads to answers 24 067.96 (max M1M1A0). Note: Missing weight term gets 24 080 (max M1M0A0). |
**Alternative method for 2(b) using energy:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Change in PE $= 1.204g \times 0.04\ [= 0.4816]$, or change in KE $= \frac{1}{2} \times 1.204 \times (40)^2\ [= 963.2]$ | B1 | Allow use of 40.1 or $\frac{602}{15}$ for velocity in place of 40. B0 if extra KE terms present. |
| $1.204g \times 0.04 + \frac{1}{2} \times 1.204 \times (40)^2 = 0.04R$ | M1 | Attempt at work-energy equation. Must have correct number of relevant terms, dimensionally correct; allow sign errors. Do not allow 4 in place of 0.04. Allow use of 40.1 or $\frac{602}{15}$ for velocity in place of 40. |
| $R = 24\,100\ \text{N}\ \left[24\,092.04 = \frac{602301}{25}\right]$ | A1 | WWW. Note: use of wrong sign for $g$ leads to 24 067.96 (max B1M1A0). Missing PE term gets 24 080 (max B1M0A0). |
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\includegraphics[max width=\textwidth, alt={}, center]{99f20949-471d-4da3-a680-ec24abf6baa5-03_510_604_260_769}
A machine for driving a nail into a block of wood causes a hammerhead to drop vertically onto the top of a nail. The mass of the hammerhead is 1.2 kg and the mass of the nail is 0.004 kg (see diagram). The hammerhead hits the nail with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and remains in contact with the nail after the impact. The combined hammerhead and nail move immediately after the impact with speed $40 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate $v$, giving your answer as an exact fraction.
\item The nail is driven 4 cm into the wood.
Find the constant force resisting the motion.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q2 [5]}}