| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical motion: energy loss on impact |
| Difficulty | Moderate -0.3 Part (a) is a straightforward SUVAT calculation using v² = u² + 2as with standard values. Part (b) requires conservation of momentum for the collision, then applying Newton's second law or energy methods with resistance force, but follows a standard M1 template with clear given values and no conceptual surprises. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles |
| VIAV SIHI NI BIIIM ION OC | VGHV SIHI NI GHIYM ION OC | VJ4V SIHI NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(7^2 = 2 \times 9.8h\) | M1 | Use of \(v^2 = u^2 + 2as\) with \(u=0, v=7\), or alternative complete method to find \(h\) |
| \(h = 2.5\) | A1 | Condone \(h = -2.5\) in working but final answer must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(9 \times 7 = 10.5\, u\) | M1 | Use CLM to find speed of blocks after impact. Condone additional factor of \(g\) throughout. |
| \(u = 6\) | A1 | |
| \(0^2 = 6^2 - 2a \times 0.12\) | M1 | Use of \(v^2 = u^2 + 2as\) with \(u=6, v=0\). Allow for their \(u\) and \(v=0\). Allow for \(u=7, v=0\). Accept alternative suvat method to form equation in \(a\). Condone use of 12 for 0.12 |
| Correctly substituted equation in \(a\) with \(u=6, s=0.12\) (implied by \(a=150\)) | A1 | |
| \((\downarrow)\; 10.5g - R = 10.5 \times (-a)\) | M1 | Use of \(F=ma\) with their \(a \neq \pm g\). Must have all 3 terms and 10.5. Condone sign error(s) |
| \((\downarrow)\; 10.5g - R = 10.5 \times (-150)\) | A1 | Unsimplified equation with \(a\) substituted and at most one error (their \(a\) with wrong sign is 1 error) |
| Correct unsimplified equation with \(a\) substituted | A1 | |
| \(R = 1680\) or \(1700\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times 10.5 \times 6^2 + 10.5 \times 9.8 \times 0.12 = R \times 0.12\) | M2 | Energy equation (needs all three terms) |
| A3 | \(-1\) each error; A1A1A0 for 1 error, A1A0A0 for 2 errors | |
| \(R = 1680\) or \(1700\) | A1 |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $7^2 = 2 \times 9.8h$ | M1 | Use of $v^2 = u^2 + 2as$ with $u=0, v=7$, or alternative complete method to find $h$ |
| $h = 2.5$ | A1 | Condone $h = -2.5$ in working but final answer must be positive |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $9 \times 7 = 10.5\, u$ | M1 | Use CLM to find speed of blocks after impact. Condone additional factor of $g$ throughout. |
| $u = 6$ | A1 | |
| $0^2 = 6^2 - 2a \times 0.12$ | M1 | Use of $v^2 = u^2 + 2as$ with $u=6, v=0$. Allow for their $u$ and $v=0$. Allow for $u=7, v=0$. Accept alternative suvat method to form equation in $a$. Condone use of 12 for 0.12 |
| Correctly substituted equation in $a$ with $u=6, s=0.12$ (implied by $a=150$) | A1 | |
| $(\downarrow)\; 10.5g - R = 10.5 \times (-a)$ | M1 | Use of $F=ma$ with their $a \neq \pm g$. Must have all 3 terms and 10.5. Condone sign error(s) |
| $(\downarrow)\; 10.5g - R = 10.5 \times (-150)$ | A1 | Unsimplified equation with $a$ substituted and at most one error (their $a$ with wrong sign is 1 error) |
| Correct unsimplified equation with $a$ substituted | A1 | |
| $R = 1680$ or $1700$ | A1 | |
**Alternative for last 6 marks:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 10.5 \times 6^2 + 10.5 \times 9.8 \times 0.12 = R \times 0.12$ | M2 | Energy equation (needs all three terms) |
| | A3 | $-1$ each error; A1A1A0 for 1 error, A1A0A0 for 2 errors |
| $R = 1680$ or $1700$ | A1 | |
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3. A block $A$ of mass 9 kg is released from rest from a point $P$ which is a height $h$ metres above horizontal soft ground. The block falls and strikes another block $B$ of mass 1.5 kg which is on the ground vertically below $P$. The speed of $A$ immediately before it strikes $B$ is $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The blocks are modelled as particles.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $h$.
Immediately after the impact the blocks move downwards together with the same speed and both come to rest after sinking a vertical distance of 12 cm into the ground. Assuming that the resistance offered by the ground has constant magnitude $R$ newtons,
\item find the value of $R$.\\
\includegraphics[max width=\textwidth, alt={}, center]{6ab8838f-d6f8-4761-8def-1022d97d4e82-07_2252_51_315_36}\\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIAV SIHI NI BIIIM ION OC & VGHV SIHI NI GHIYM ION OC & VJ4V SIHI NI JIIYM ION OC \\
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\end{tabular}
\end{center}
\includegraphics[max width=\textwidth, alt={}, center]{6ab8838f-d6f8-4761-8def-1022d97d4e82-09_2249_45_318_37}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2018 Q3 [10]}}