OCR MEI M2 2010 June — Question 1 17 marks

Exam BoardOCR MEI
ModuleM2 (Mechanics 2)
Year2010
SessionJune
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeDirect collision with given impulse
DifficultyModerate -0.3 This is a standard M2 mechanics question testing impulse-momentum, conservation of momentum, coefficient of restitution, and variable mass problems. All parts follow routine procedures with clearly stated values and 'show that' scaffolding. The multi-part structure and context make it slightly more involved than basic exercises, but no novel problem-solving or geometric insight is required—just systematic application of standard formulae.
Spec6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form6.03k Newton's experimental law: direct impact

1 Two sledges P and Q, with their loads, have masses of 200 kg and 250 kg respectively and are sliding on a horizontal surface against negligible resistance. There is an inextensible light rope connecting the sledges that is horizontal and parallel to the direction of motion. Fig. 1 shows the initial situation with both sledges travelling with a velocity of \(5 \mathbf { i m ~ } \mathbf { m } ^ { - 1 }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{1a605f0b-f595-4bb9-9624-f816c789ad86-2_397_1379_520_383} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} A mechanism on Q jerks the rope so that there is an impulse of \(250 \mathbf { i N s }\) on P .
  1. Show that the new velocity of \(P\) is \(6.25 \mathrm { i } \mathrm { m } \mathrm { s } ^ { - 1 }\) and find the new velocity of \(Q\). There is now a direct collision between the sledges and after the impact P has velocity \(4.5 \mathrm { i } \mathrm { m } \mathrm { s } ^ { - 1 }\).
  2. Show that the velocity of Q becomes \(5.4 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Calculate the coefficient of restitution in the collision. Before the rope becomes taut again, the velocity of P is increased so that it catches up with Q . This is done by throwing part of the load from sledge P in the \(- \mathbf { i }\) direction so that P 's velocity increases to \(5.5 \mathrm { i } \mathrm { m } \mathrm { s } ^ { - 1 }\). The part of the load thrown out is an object of mass 20 kg .
  3. Calculate the speed of separation of the object from P . When the sledges directly collide again they are held together and move as a single object.
  4. Calculate the common velocity of the pair of sledges, giving your answer correct to 3 significant figures. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{1a605f0b-f595-4bb9-9624-f816c789ad86-3_987_524_258_769} \captionsetup{labelformat=empty} \caption{not to scale he lengths are}
    \end{figure} Fig. 2 Fig. 2 shows a stand on a horizontal floor and horizontal and vertical coordinate axes \(\mathrm { O } x\) and \(\mathrm { O } y\). The stand is modelled as
    • a thin uniform rectangular base PQRS, 30 cm by 40 cm with mass 15 kg ; the sides QR and PS are parallel to \(\mathrm { O } x\),
    • a thin uniform vertical rod of length 200 cm and mass 3 kg that is fixed to the base at O , the mid-point of PQ and the origin of coordinates,
    • a thin uniform top rod AB of length 50 cm and mass \(2 \mathrm {~kg} ; \mathrm { AB }\) is parallel to \(\mathrm { O } x\).
    Coordinates are referred to the axes shown in the figure.

Question 1:
Part (i)
AnswerMarks Guidance
Answer/WorkingMark Guidance
For P: \(200 \times 5 + 250 = 200v_P\)M1 Award for Impulse-Momentum
\(v_P = 6.25\) so \(6.25\mathbf{i}\) m s\(^{-1}\)E1 Accept no \(\mathbf{i}\) and no units
For Q: \(250 \times 5 - 250 = 250v_Q\)M1 Must have impulse in opposite sense
\(v_Q = 4\) so \(4\mathbf{i}\) m s\(^{-1}\)A1 Must indicate direction. Accept no units.
Part (ii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{i}\) direction positive; PCLM: \(2250 = 200 \times 4.5 + 250w_Q\)M1 PCLM used. Allow error in LHS FT from (i)
F1Any form. FT only from (i)
\(w_Q = 5.4\) so \(5.4\mathbf{i}\) m s\(^{-1}\)E1
NEL: \(\frac{w_Q - 4.5}{4 - 6.25} = -e\)M1 NEL. Allow sign errors
A1Signs correct. FT only from (i)
\(e = 0.4\)A1 cao
Part (iii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{i}\) direction positive; suppose absolute vel of object is \(-V\mathbf{i}\)
\(200 \times 4.5 = -20V + 180 \times 5.5\)M1 Applying PCLM. All terms present. Allow sign errors.
B1Correct masses
\(V = 4.5\)A1 All correct (including signs)
speed of separation is \(5.5 + 4.5 = 10\) m s\(^{-1}\)A1
F1FT their \(V\)
Part (iv)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(180 \times 5.5 + 250 \times 5.4 = 430W\)M1 Using correct masses and velocities
\(W = 5.4418\ldots\) so \(5.44\mathbf{i}\) m s\(^{-1}\) (3 s.f.)A1 cao
# Question 1:

## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| For P: $200 \times 5 + 250 = 200v_P$ | M1 | Award for Impulse-Momentum |
| $v_P = 6.25$ so $6.25\mathbf{i}$ m s$^{-1}$ | E1 | Accept no $\mathbf{i}$ and no units |
| For Q: $250 \times 5 - 250 = 250v_Q$ | M1 | Must have impulse in opposite sense |
| $v_Q = 4$ so $4\mathbf{i}$ m s$^{-1}$ | A1 | Must indicate direction. Accept no units. |

## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{i}$ direction positive; PCLM: $2250 = 200 \times 4.5 + 250w_Q$ | M1 | PCLM used. Allow error in LHS FT from (i) |
| | F1 | Any form. FT only from (i) |
| $w_Q = 5.4$ so $5.4\mathbf{i}$ m s$^{-1}$ | E1 | |
| NEL: $\frac{w_Q - 4.5}{4 - 6.25} = -e$ | M1 | NEL. Allow sign errors |
| | A1 | Signs correct. FT only from (i) |
| $e = 0.4$ | A1 | cao |

## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{i}$ direction positive; suppose absolute vel of object is $-V\mathbf{i}$ | | |
| $200 \times 4.5 = -20V + 180 \times 5.5$ | M1 | Applying PCLM. All terms present. Allow sign errors. |
| | B1 | Correct masses |
| $V = 4.5$ | A1 | All correct (including signs) |
| speed of separation is $5.5 + 4.5 = 10$ m s$^{-1}$ | A1 | |
| | F1 | FT their $V$ |

## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $180 \times 5.5 + 250 \times 5.4 = 430W$ | M1 | Using correct masses and velocities |
| $W = 5.4418\ldots$ so $5.44\mathbf{i}$ m s$^{-1}$ (3 s.f.) | A1 | cao |

---
1 Two sledges P and Q, with their loads, have masses of 200 kg and 250 kg respectively and are sliding on a horizontal surface against negligible resistance. There is an inextensible light rope connecting the sledges that is horizontal and parallel to the direction of motion.

Fig. 1 shows the initial situation with both sledges travelling with a velocity of $5 \mathbf { i m ~ } \mathbf { m } ^ { - 1 }$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{1a605f0b-f595-4bb9-9624-f816c789ad86-2_397_1379_520_383}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}

A mechanism on Q jerks the rope so that there is an impulse of $250 \mathbf { i N s }$ on P .\\
(i) Show that the new velocity of $P$ is $6.25 \mathrm { i } \mathrm { m } \mathrm { s } ^ { - 1 }$ and find the new velocity of $Q$.

There is now a direct collision between the sledges and after the impact P has velocity $4.5 \mathrm { i } \mathrm { m } \mathrm { s } ^ { - 1 }$.\\
(ii) Show that the velocity of Q becomes $5.4 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

Calculate the coefficient of restitution in the collision.

Before the rope becomes taut again, the velocity of P is increased so that it catches up with Q . This is done by throwing part of the load from sledge P in the $- \mathbf { i }$ direction so that P 's velocity increases to $5.5 \mathrm { i } \mathrm { m } \mathrm { s } ^ { - 1 }$. The part of the load thrown out is an object of mass 20 kg .\\
(iii) Calculate the speed of separation of the object from P .

When the sledges directly collide again they are held together and move as a single object.\\
(iv) Calculate the common velocity of the pair of sledges, giving your answer correct to 3 significant figures.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{1a605f0b-f595-4bb9-9624-f816c789ad86-3_987_524_258_769}
\captionsetup{labelformat=empty}
\caption{not to scale he lengths are}
\end{center}
\end{figure}

Fig. 2

Fig. 2 shows a stand on a horizontal floor and horizontal and vertical coordinate axes $\mathrm { O } x$ and $\mathrm { O } y$. The stand is modelled as

\begin{itemize}
  \item a thin uniform rectangular base PQRS, 30 cm by 40 cm with mass 15 kg ; the sides QR and PS are parallel to $\mathrm { O } x$,
  \item a thin uniform vertical rod of length 200 cm and mass 3 kg that is fixed to the base at O , the mid-point of PQ and the origin of coordinates,
  \item a thin uniform top rod AB of length 50 cm and mass $2 \mathrm {~kg} ; \mathrm { AB }$ is parallel to $\mathrm { O } x$.
\end{itemize}

Coordinates are referred to the axes shown in the figure.\\

\hfill \mbox{\textit{OCR MEI M2 2010 Q1 [17]}}