OCR MEI M2 2016 June — Question 1 17 marks

Exam BoardOCR MEI
ModuleM2 (Mechanics 2)
Year2016
SessionJune
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeDirect collision with speed relationships
DifficultyModerate -0.3 This is a standard M2 momentum and collisions question covering routine applications: conservation of momentum, coefficient of restitution, and relative velocity. All parts follow textbook methods with straightforward arithmetic. Part (ii) requires showing an impossibility, but this is a common M2 exercise type. Slightly easier than average A-level due to the mechanical nature of the calculations.
Spec6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions
1
  1. Two model railway trucks are moving freely on a straight horizontal track when they are in a direct collision. The trucks are P of mass 0.5 kg and Q of mass 0.75 kg . They are initially travelling in the same direction. Just before they collide P has a speed of \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and Q has a speed of \(1 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), as shown in Fig. 1.1. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-2_263_640_484_715} \captionsetup{labelformat=empty} \caption{Fig. 1.1}
    \end{figure}
    1. Suppose that the speed of P is halved in the collision and that its direction of motion is not changed. Find the speed of Q immediately after the collision and find the coefficient of restitution.
    2. Show that it is not possible for both the speed of P to be halved in the collision and its direction of motion to be reversed. Both of the model trucks have flat horizontal tops. They are each travelling at the speeds they had immediately after the collision. Part of the mass of Q is a small object of mass 0.1 kg at rest at the edge of the top of Q nearest P . The object falls off, initially with negligible velocity relative to Q .
    3. Determine the speed of Q immediately after the object falls off it, making your reasoning clear. Part of the mass of P is an object of mass 0.05 kg that is fired horizontally from the top of P , parallel to and in the opposite direction to the motion of P . Immediately after the object is fired, it has a speed of \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) relative to P .
    4. Determine the speed of P immediately after the object has been fired from it.
  2. The velocities of a small object immediately before and after an elastic collision with a horizontal plane are shown in Fig. 1.2. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-2_172_741_1987_644} \captionsetup{labelformat=empty} \caption{Fig. 1.2}
    \end{figure} Show that the plane cannot be smooth.
Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(v = u + at\) applied correctlyM1 Allow sign errors
\(v = 15\) m/sA1 cao
Please share the pages containing the actual mark scheme questions and answers.
Question 1:
Part (a)(i)
AnswerMarks Guidance
AnswerMarks Guidance
Take \(+ve \rightarrow\), PCLM: \(0.5\times4+0.75\times1=0.5\times2+0.75\times v_Q\)M1 Application of PCLM. Allow sign errors
Correct equation. Any formA1 Correct. Any form
\(v_Q=\frac{7}{3}\)A1 Exact or anything that rounds to 2.33 or better
NEL: \(\frac{\frac{7}{3}-2}{1-4}=-e\)M1 NEL. Accept sign errors but not approach/separation
\(e=\frac{1}{9}\) (or 0.11)A1ft ft their \(v_Q\)
[5]
Part (a)(ii)
AnswerMarks Guidance
AnswerMarks Guidance
Suppose direction reversed. Given LM conserved: \(0.5\times4+0.75\times1=-0.5\times2+0.75\times v_Q\), so \(v_Q=5\)B1 Award for the correct LM equation
NEL gives \(\frac{5+2}{1-4}=-e\) so \(e=\frac{7}{3}\)M1 Using their re-calculated \(v_Q\) (not equal to 7/3 from (i))
\(e>1\) so not an elastic collisionE1 www
[3]
OR: KE after collision \(=10.375\), before collision \(=4.375\)M1
Increase in energy not possible, because no work put into systemE1
Part (a)(iii)
AnswerMarks Guidance
AnswerMarks Guidance
No (external horizontal) force acts on the truck so no change in momentum of truck (less object)B1 Force or momentum considered or correct momentum equation
So no change in velocity. Still \(\frac{7}{3}\) m s\(^{-1}\)B1 FT their value from (i): seen
[2]
Part (a)(iv)
AnswerMarks Guidance
AnswerMarks Guidance
Before 0.5 kg at 2 m s\(^{-1}\) \(\rightarrow\); After 0.05 kg at \(U\) m s\(^{-1}\) \(\leftarrow\) and 0.45 kg at \(V\) m s\(^{-1}\) \(\rightarrow\); PCLM: \(0.5\times2=-0.05\times U+0.45\times V\)M1 Allow if \((10-2)=8\) used instead of \(U\)
M1Allow only sign errors
\(U+V=10\)B1 oe: relative velocity used correctly
\(V=3\) so 3 m s\(^{-1}\)A1 cao
[4] SC1 Using \(U=10\), giving \(V=10/3\)
Part (b)
AnswerMarks Guidance
AnswerMarks Guidance
Consider the LM parallel to the planeM1 Accept considering the horizontal components of velocity before and after o.e. and arguing/stating they should be the same
Before: \(m\times10\cos60=5m\); After: \(m\times6\cos40\approx4.6m\)A1 Need not include \(m\). Using sine gets 0/3
Not the same. (LM not conserved.) Plane cannot be smooth.E1 Accept arguments from velocity
[3] 
**Question 1:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $v = u + at$ applied correctly | M1 | Allow sign errors |
| $v = 15$ m/s | A1 | cao |

Please share the pages containing the actual mark scheme questions and answers.

# Question 1:

## Part (a)(i)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Take $+ve \rightarrow$, PCLM: $0.5\times4+0.75\times1=0.5\times2+0.75\times v_Q$ | M1 | Application of PCLM. Allow sign errors |
| Correct equation. Any form | A1 | Correct. Any form |
| $v_Q=\frac{7}{3}$ | A1 | Exact or anything that rounds to 2.33 or better |
| NEL: $\frac{\frac{7}{3}-2}{1-4}=-e$ | M1 | NEL. Accept sign errors but not approach/separation |
| $e=\frac{1}{9}$ (or 0.11) | A1ft | ft their $v_Q$ |
| **[5]** | | |

## Part (a)(ii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Suppose direction reversed. Given LM conserved: $0.5\times4+0.75\times1=-0.5\times2+0.75\times v_Q$, so $v_Q=5$ | B1 | Award for the correct LM equation |
| NEL gives $\frac{5+2}{1-4}=-e$ so $e=\frac{7}{3}$ | M1 | Using their re-calculated $v_Q$ (not equal to 7/3 from (i)) |
| $e>1$ so not an elastic collision | E1 | www |
| **[3]** | | |
| OR: KE after collision $=10.375$, before collision $=4.375$ | M1 | |
| Increase in energy not possible, because no work put into system | E1 | |

## Part (a)(iii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| No (external horizontal) force acts on the truck so no change in momentum of truck (less object) | B1 | Force or momentum considered or correct momentum equation |
| So no change in velocity. Still $\frac{7}{3}$ m s$^{-1}$ | B1 | FT their value from (i): seen |
| **[2]** | | |

## Part (a)(iv)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Before 0.5 kg at 2 m s$^{-1}$ $\rightarrow$; After 0.05 kg at $U$ m s$^{-1}$ $\leftarrow$ and 0.45 kg at $V$ m s$^{-1}$ $\rightarrow$; PCLM: $0.5\times2=-0.05\times U+0.45\times V$ | M1 | Allow if $(10-2)=8$ used instead of $U$ |
| | M1 | Allow only sign errors |
| $U+V=10$ | B1 | oe: relative velocity used correctly |
| $V=3$ so 3 m s$^{-1}$ | A1 | cao |
| **[4]** | | SC1 Using $U=10$, giving $V=10/3$ |

## Part (b)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Consider the LM parallel to the plane | M1 | Accept considering the horizontal components of velocity before and after o.e. and arguing/stating they should be the same |
| Before: $m\times10\cos60=5m$; After: $m\times6\cos40\approx4.6m$ | A1 | Need not include $m$. Using sine gets 0/3 |
| Not the same. (LM not conserved.) Plane cannot be smooth. | E1 | Accept arguments from velocity |
| **[3]** | | |

---
1
\begin{enumerate}[label=(\alph*)]
\item Two model railway trucks are moving freely on a straight horizontal track when they are in a direct collision.

The trucks are P of mass 0.5 kg and Q of mass 0.75 kg . They are initially travelling in the same direction. Just before they collide P has a speed of $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and Q has a speed of $1 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, as shown in Fig. 1.1.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-2_263_640_484_715}
\captionsetup{labelformat=empty}
\caption{Fig. 1.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Suppose that the speed of P is halved in the collision and that its direction of motion is not changed. Find the speed of Q immediately after the collision and find the coefficient of restitution.
\item Show that it is not possible for both the speed of P to be halved in the collision and its direction of motion to be reversed.

Both of the model trucks have flat horizontal tops. They are each travelling at the speeds they had immediately after the collision.

Part of the mass of Q is a small object of mass 0.1 kg at rest at the edge of the top of Q nearest P . The object falls off, initially with negligible velocity relative to Q .
\item Determine the speed of Q immediately after the object falls off it, making your reasoning clear.

Part of the mass of P is an object of mass 0.05 kg that is fired horizontally from the top of P , parallel to and in the opposite direction to the motion of P . Immediately after the object is fired, it has a speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ relative to P .
\item Determine the speed of P immediately after the object has been fired from it.
\end{enumerate}\item The velocities of a small object immediately before and after an elastic collision with a horizontal plane are shown in Fig. 1.2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-2_172_741_1987_644}
\captionsetup{labelformat=empty}
\caption{Fig. 1.2}
\end{center}
\end{figure}

Show that the plane cannot be smooth.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI M2 2016 Q1 [17]}}
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