| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with given impulse |
| Difficulty | Moderate -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. |
| Spec | 6.03b 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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. |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 \(V\) |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
# 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]}}