OCR MEI M2 — Question 1

Exam BoardOCR MEI
ModuleM2 (Mechanics 2)
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeExplosion or separation of particles
DifficultyModerate -0.5 This is a straightforward application of conservation of momentum to standard scenarios (impulse calculation, separation problems, and a direct collision with given coefficient of restitution). All parts use routine methods with clear setups and no novel problem-solving required, making it slightly easier than average for A-level mechanics.
Spec6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum

1
  1. Roger of mass 70 kg and Sheuli of mass 50 kg are skating on a horizontal plane containing the standard unit vectors \(\mathbf { i }\) and \(\mathbf { j }\). The resistances to the motion of the skaters are negligible. The two skaters are locked in a close embrace and accelerate from rest until they reach a velocity of \(2 \mathrm { ims } ^ { - 1 }\), as shown in Fig. 1.1. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{5dd6ba0d-e516-4b9e-ba19-6e90520b171b-002_191_181_543_740} \captionsetup{labelformat=empty} \caption{Fig. 1.1}
    \end{figure} \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{5dd6ba0d-e516-4b9e-ba19-6e90520b171b-002_177_359_589_1051} \captionsetup{labelformat=empty} \caption{Fig. 1.1}
    \end{figure}
    1. What impulse has acted on them? During a dance routine, the skaters separate on three occasions from their close embrace when travelling at a constant velocity of \(2 \mathrm { i } \mathrm { ms } ^ { - 1 }\).
    2. Calculate the velocity of Sheuli after the separation in the following cases.
      (A) Roger has velocity \(\mathrm { ims } ^ { - 1 }\) after the separation.
      (B) Roger and Sheuli have equal speeds in opposite senses after the separation, with Roger moving in the \(\mathbf { i }\) direction.
      (C) Roger has velocity \(4 ( \mathbf { i } + \mathbf { j } ) \mathrm { ms } ^ { - 1 }\) after the separation.
  2. Two discs with masses 2 kg and 3 kg collide directly in a horizontal plane. Their velocities just before the collision are shown in Fig. 1.2. The coefficient of restitution in the collision is 0.5. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{5dd6ba0d-e516-4b9e-ba19-6e90520b171b-002_278_970_1759_594} \captionsetup{labelformat=empty} \caption{Fig. 1.2}
    \end{figure}
    1. Calculate the velocity of each disc after the collision. The disc of mass 3 kg moves freely after the collision and makes a perfectly elastic collision with a smooth wall inclined at \(60 ^ { \circ }\) to its direction of motion, as shown in Fig. 1.2.
    2. State with reasons the speed of the disc and the angle between its direction of motion and the wall after the collision.

Question 1:
Part (a)(i)
AnswerMarks Guidance
\(21 \times 2 = 4(v - 3)\) using impulse-momentumM1 Apply impulse = change in momentum
\(v = \frac{42}{4} + 3 = 13.5 \text{ ms}^{-1}\)A1 Accept \(\frac{27}{2}\)
Part (a)(ii)
AnswerMarks Guidance
\(4 \times 13.5 + 2 \times (-3) = 6v_S\)M1 Conservation of momentum, correct masses
\(v_S = \frac{54 - 6}{6} = 8 \text{ ms}^{-1}\) towards RA1 Must state direction
Part (a)(iii)
AnswerMarks Guidance
NEL: \(v_R - v_S = \frac{1}{4}(8 - v_R^{\text{before}})\) ... (1)B1 Correct NEL equation
CLM: \(6 \times 8 + 4v_R^{\text{before}} = 6 \times 5 + 4v_R^{\text{after}}\) ... (2)M1 A1 CLM with correct signs/values
Solving: \(v_R^{\text{before}} = 2 \text{ ms}^{-1}\) towards SA1
\(v_R^{\text{after}} = \frac{25}{4} = 6.25 \text{ ms}^{-1}\) away from SM1 A1 Method for solving simultaneous equations
Part (b)(i)
AnswerMarks Guidance
Speed before collision: \(v^2 = 5^2 + 2(10)(10)\), \(v = 15 \text{ ms}^{-1}\)M1 A1
Component along plane: \(15\sin\alpha = 15 \times \frac{3}{5} = 9 \text{ ms}^{-1}\)M1 Resolving along plane (unchanged)
Component perpendicular to plane (before): \(15\cos\alpha = 15 \times \frac{4}{5} = 12 \text{ ms}^{-1}\)A1
After collision speed = \(13 \text{ ms}^{-1}\), component along plane = \(9 \text{ ms}^{-1}\)
AnswerMarks
Perpendicular component after: \(\sqrt{13^2 - 9^2} = \sqrt{88} = 2\sqrt{22}\) ms\(^{-1}\)M1
Angle between velocity and plane: \(\sin\theta = \frac{\sqrt{88}}{13}\), \(\theta \approx 32.3°\)A1
\(e = \frac{2\sqrt{22}}{12} \approx 0.392\)M1 A1
Part (b)(ii)
AnswerMarks Guidance
Impulse \(= 0.2(2\sqrt{22} + 12) = 0.2 \times 16.690... \approx 3.34\) N sM1 A1 Change in perpendicular component only
# Question 1:

## Part (a)(i)
$21 \times 2 = 4(v - 3)$ using impulse-momentum | M1 | Apply impulse = change in momentum
$v = \frac{42}{4} + 3 = 13.5 \text{ ms}^{-1}$ | A1 | Accept $\frac{27}{2}$

## Part (a)(ii)
$4 \times 13.5 + 2 \times (-3) = 6v_S$ | M1 | Conservation of momentum, correct masses
$v_S = \frac{54 - 6}{6} = 8 \text{ ms}^{-1}$ towards R | A1 | Must state direction

## Part (a)(iii)
NEL: $v_R - v_S = \frac{1}{4}(8 - v_R^{\text{before}})$ ... (1) | B1 | Correct NEL equation
CLM: $6 \times 8 + 4v_R^{\text{before}} = 6 \times 5 + 4v_R^{\text{after}}$ ... (2) | M1 A1 | CLM with correct signs/values
Solving: $v_R^{\text{before}} = 2 \text{ ms}^{-1}$ towards S | A1 |
$v_R^{\text{after}} = \frac{25}{4} = 6.25 \text{ ms}^{-1}$ away from S | M1 A1 | Method for solving simultaneous equations

## Part (b)(i)
Speed before collision: $v^2 = 5^2 + 2(10)(10)$, $v = 15 \text{ ms}^{-1}$ | M1 A1 |
Component along plane: $15\sin\alpha = 15 \times \frac{3}{5} = 9 \text{ ms}^{-1}$ | M1 | Resolving along plane (unchanged)
Component perpendicular to plane (before): $15\cos\alpha = 15 \times \frac{4}{5} = 12 \text{ ms}^{-1}$ | A1 |
After collision speed = $13 \text{ ms}^{-1}$, component along plane = $9 \text{ ms}^{-1}$
Perpendicular component after: $\sqrt{13^2 - 9^2} = \sqrt{88} = 2\sqrt{22}$ ms$^{-1}$ | M1 |
Angle between velocity and plane: $\sin\theta = \frac{\sqrt{88}}{13}$, $\theta \approx 32.3°$ | A1 |
$e = \frac{2\sqrt{22}}{12} \approx 0.392$ | M1 A1 |

## Part (b)(ii)
Impulse $= 0.2(2\sqrt{22} + 12) = 0.2 \times 16.690... \approx 3.34$ N s | M1 A1 | Change in perpendicular component only

---
1
\begin{enumerate}[label=(\alph*)]
\item Roger of mass 70 kg and Sheuli of mass 50 kg are skating on a horizontal plane containing the standard unit vectors $\mathbf { i }$ and $\mathbf { j }$. The resistances to the motion of the skaters are negligible. The two skaters are locked in a close embrace and accelerate from rest until they reach a velocity of $2 \mathrm { ims } ^ { - 1 }$, as shown in Fig. 1.1.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{5dd6ba0d-e516-4b9e-ba19-6e90520b171b-002_191_181_543_740}
\captionsetup{labelformat=empty}
\caption{Fig. 1.1}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{5dd6ba0d-e516-4b9e-ba19-6e90520b171b-002_177_359_589_1051}
\captionsetup{labelformat=empty}
\caption{Fig. 1.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item What impulse has acted on them?

During a dance routine, the skaters separate on three occasions from their close embrace when travelling at a constant velocity of $2 \mathrm { i } \mathrm { ms } ^ { - 1 }$.
\item Calculate the velocity of Sheuli after the separation in the following cases.\\
(A) Roger has velocity $\mathrm { ims } ^ { - 1 }$ after the separation.\\
(B) Roger and Sheuli have equal speeds in opposite senses after the separation, with Roger moving in the $\mathbf { i }$ direction.\\
(C) Roger has velocity $4 ( \mathbf { i } + \mathbf { j } ) \mathrm { ms } ^ { - 1 }$ after the separation.
\end{enumerate}\item Two discs with masses 2 kg and 3 kg collide directly in a horizontal plane. Their velocities just before the collision are shown in Fig. 1.2. The coefficient of restitution in the collision is 0.5.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{5dd6ba0d-e516-4b9e-ba19-6e90520b171b-002_278_970_1759_594}
\captionsetup{labelformat=empty}
\caption{Fig. 1.2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Calculate the velocity of each disc after the collision.

The disc of mass 3 kg moves freely after the collision and makes a perfectly elastic collision with a smooth wall inclined at $60 ^ { \circ }$ to its direction of motion, as shown in Fig. 1.2.
\item State with reasons the speed of the disc and the angle between its direction of motion and the wall after the collision.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI M2  Q1}}
This paper (2 questions)
View full paper