Edexcel M1 2018 Specimen — Question 3 10 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2018
SessionSpecimen
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSUVAT in 2D & Gravity
TypeVertical motion: energy loss on impact
DifficultyModerate -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.
Spec3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles

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.
  1. 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,
  2. find the value of \(R\). \includegraphics[max width=\textwidth, alt={}, center]{6ab8838f-d6f8-4761-8def-1022d97d4e82-07_2252_51_315_36}
    VIAV SIHI NI BIIIM ION OCVGHV SIHI NI GHIYM ION OCVJ4V SIHI NI JIIYM ION OC
    \includegraphics[max width=\textwidth, alt={}, center]{6ab8838f-d6f8-4761-8def-1022d97d4e82-09_2249_45_318_37}

Question 3:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark 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):
AnswerMarks Guidance
Answer/WorkingMark 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 \(a\) substitutedA1
\(R = 1680\) or \(1700\)A1
Alternative for last 6 marks:
AnswerMarks Guidance
Answer/WorkingMark 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 \\
\hline
\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]}}