OCR MEI M2 2014 June — Question 1 17 marks

Exam BoardOCR MEI
ModuleM2 (Mechanics 2)
Year2014
SessionJune
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeDirect collision with direction reversal
DifficultyModerate -0.3 This is a standard M2 momentum and collisions question with straightforward application of conservation of momentum and Newton's experimental law. Part (a) involves direct collision with clear setup and routine algebra; part (b) requires 2D momentum conservation with given trig ratios that simplify calculations. All steps are textbook-standard with no novel insight required, making it slightly easier than average A-level maths.
Spec6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact

1
  1. A particle, P , of mass 5 kg moving with speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) collides with another particle, Q , of mass 30 kg travelling with a speed of \(\frac { u } { 3 } \mathrm {~ms} ^ { - 1 }\) towards P . The particles P and Q are moving in the same horizontal straight line with negligible resistance to their motion. As a result of the collision, the speed of P is halved and its direction of travel reversed; the speed of Q is now \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
    1. Draw a diagram showing this information. Find the velocity of Q immediately after the collision in terms of \(u\). Find also the coefficient of restitution between P and Q .
    2. Find, in terms of \(u\), the impulse of P on Q in the collision.
  2. Fig. 1 shows a small object R of mass 5 kg travelling on a smooth horizontal plane at \(6 \mathrm {~ms} ^ { - 1 }\). It explodes into two parts of masses 2 kg and 3 kg . The velocities of these parts are in the plane in which R was travelling with the speeds and directions indicated. The angles \(\alpha\) and \(\beta\) are given by \(\cos \alpha = \frac { 4 } { 5 }\) and \(\cos \beta = \frac { 3 } { 5 }\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{334b2170-3708-46db-bff7-bcad7d5fab00-2_460_1450_1050_312} \captionsetup{labelformat=empty} \caption{Fig. 1}
    \end{figure}
    1. Calculate \(u\) and \(v\).
    2. Calculate the increase in kinetic energy resulting from the explosion.

Question 1:
Part (a)(i):
AnswerMarks Guidance
AnswerMarks Guidance
Diagram: before \(u\) ms\(^{-1}\) and \(\frac{u}{3}\) ms\(^{-1}\); after \(\frac{u}{2}\) ms\(^{-1}\) and \(V\) ms\(^{-1}\); P=5kg, Q=30kgB1 Accept \(V\) in either direction. Given velocities and masses must be correct
PCLM: \(5u - 30\frac{u}{3} = -5\frac{u}{2} - 30V\)M1 PCLM. Allow sign errors only
\(V = \frac{u}{12}\)A1 Award even if direction of \(V\) used in PCLM does not match diagram, so \(\frac{u}{12}\) or \(-\frac{u}{12}\) will get this A1
Direction of \(V\) correctA1 WWW. Direction of \(V\) correct (may be implied from diagram)
\(e = \dfrac{\frac{u}{2} - \frac{u}{12}}{u + \frac{u}{3}}\)M1 FT their \(V\): allow sign errors, but must be right way up
\(= \dfrac{\frac{5}{12}}{\frac{4}{3}} = \frac{5}{16}\) (= 0.3125)A1 cao
[6]
Part (a)(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(\rightarrow +ve\)
\(30\left(-\frac{u}{12} - \left(-\frac{u}{3}\right)\right)\)M1 Allow sign errors
\(7.5u\)A1
Or: find impulse on P and reverse the signM1 A1 \(5\left(-\frac{u}{2} - u\right) = -\frac{15}{2}u\) and \(7.5u\) cao. M0A0 unless sign is reversed
Direction must be given (may be implicit from diagram)
[2]
Part (b)(i):
AnswerMarks Guidance
AnswerMarks Guidance
As the parts move at 90°, PCLM in final directionsM1
For 2kg: \(5 \times 6\sin\alpha = 2u\)M1 PCLM
\(5 \times 6 \times \frac{3}{5} = 2u\)A1 Any form
\(u = 9\)A1
For 3kg: \(5 \times 6\sin\beta = 3v\)M1 PCLM
\(5 \times 6 \times \frac{4}{5} = 3v\)A1 Any form
\(v = 8\)A1
Or PCLM \(\rightarrow\) \(5 \times 6 = 2u\sin\alpha + 3v\sin\beta\)M1 PCLM. Allow cos instead of sin if error in both terms; allow sign errors; masses need to be there. Award if embedded in vector method
\(30 = 2u \times \frac{3}{5} + 3v \times \frac{4}{5}\)A1 Any form
PCLM \(\uparrow\): \(2u\cos\alpha - 3v\cos\beta = 0\)M1 PCLM. Allow sin instead of cos if error in both terms and cos used in previous PCLM eqn; allow sign errors; masses need to be there. Award if embedded in vector method
\(2u \times \frac{4}{5} = 3v \times \frac{3}{5}\)A1 Any form
SolvingM1 A complete method involving 2 equations each in \(u\) and \(v\)
\(u = 9\)A1 cao for one of \(u\) or \(v\)
\(v = 8\)F1 for the other: FT substitution into their eqn
[7] Note: Award SC5 for \(v=6\), \(u=12\) (from cos/sin reversal). Uses velocity instead of mmtum: M0M0M1A0F1 max 2/7. Uses mass in one eqn only: M1A1M0M1A0F1 max 4/7
Part (b)(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(\Delta KE = \frac{1}{2}\times 2\times 9^2 + \frac{1}{2}\times 3\times 8^2 - \frac{1}{2}\times 5\times 6^2\)M1 M1 for attempt at difference of KE (3 terms of correct form)
\(= 87\) JA1 cao
[2]
# Question 1:

## Part (a)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Diagram: before $u$ ms$^{-1}$ and $\frac{u}{3}$ ms$^{-1}$; after $\frac{u}{2}$ ms$^{-1}$ and $V$ ms$^{-1}$; P=5kg, Q=30kg | B1 | Accept $V$ in either direction. Given velocities and masses must be correct |
| PCLM: $5u - 30\frac{u}{3} = -5\frac{u}{2} - 30V$ | M1 | PCLM. Allow sign errors only |
| $V = \frac{u}{12}$ | A1 | Award even if direction of $V$ used in PCLM does not match diagram, so $\frac{u}{12}$ or $-\frac{u}{12}$ will get this A1 |
| Direction of $V$ correct | A1 | WWW. Direction of $V$ correct (may be implied from diagram) |
| $e = \dfrac{\frac{u}{2} - \frac{u}{12}}{u + \frac{u}{3}}$ | M1 | FT their $V$: allow sign errors, but must be right way up |
| $= \dfrac{\frac{5}{12}}{\frac{4}{3}} = \frac{5}{16}$ (= 0.3125) | A1 | cao |
| **[6]** | | |

## Part (a)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\rightarrow +ve$ | | |
| $30\left(-\frac{u}{12} - \left(-\frac{u}{3}\right)\right)$ | M1 | Allow sign errors |
| $7.5u$ | A1 | |
| Or: find impulse on P **and** reverse the sign | M1 A1 | $5\left(-\frac{u}{2} - u\right) = -\frac{15}{2}u$ **and** $7.5u$ cao. M0A0 unless sign is reversed |
| Direction must be given (may be implicit from diagram) | | |
| **[2]** | | |

## Part (b)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| As the parts move at 90°, PCLM in final directions | M1 | |
| For 2kg: $5 \times 6\sin\alpha = 2u$ | M1 | PCLM |
| $5 \times 6 \times \frac{3}{5} = 2u$ | A1 | Any form |
| $u = 9$ | A1 | |
| For 3kg: $5 \times 6\sin\beta = 3v$ | M1 | PCLM |
| $5 \times 6 \times \frac{4}{5} = 3v$ | A1 | Any form |
| $v = 8$ | A1 | |
| **Or** PCLM $\rightarrow$ $5 \times 6 = 2u\sin\alpha + 3v\sin\beta$ | M1 | PCLM. Allow cos instead of sin if error in both terms; allow sign errors; masses need to be there. Award if embedded in vector method |
| $30 = 2u \times \frac{3}{5} + 3v \times \frac{4}{5}$ | A1 | Any form |
| PCLM $\uparrow$: $2u\cos\alpha - 3v\cos\beta = 0$ | M1 | PCLM. Allow sin instead of cos if error in both terms and cos used in previous PCLM eqn; allow sign errors; masses need to be there. Award if embedded in vector method |
| $2u \times \frac{4}{5} = 3v \times \frac{3}{5}$ | A1 | Any form |
| Solving | M1 | A complete method involving 2 equations each in $u$ and $v$ |
| $u = 9$ | A1 | cao for one of $u$ or $v$ |
| $v = 8$ | F1 | for the other: FT substitution into their eqn |
| **[7]** | | Note: Award SC5 for $v=6$, $u=12$ (from cos/sin reversal). Uses velocity instead of mmtum: M0M0M1A0F1 max 2/7. Uses mass in one eqn only: M1A1M0M1A0F1 max 4/7 |

## Part (b)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\Delta KE = \frac{1}{2}\times 2\times 9^2 + \frac{1}{2}\times 3\times 8^2 - \frac{1}{2}\times 5\times 6^2$ | M1 | M1 for attempt at difference of KE (3 terms of correct form) |
| $= 87$ J | A1 | cao |
| **[2]** | | |

---
1
\begin{enumerate}[label=(\alph*)]
\item A particle, P , of mass 5 kg moving with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ collides with another particle, Q , of mass 30 kg travelling with a speed of $\frac { u } { 3 } \mathrm {~ms} ^ { - 1 }$ towards P . The particles P and Q are moving in the same horizontal straight line with negligible resistance to their motion. As a result of the collision, the speed of P is halved and its direction of travel reversed; the speed of Q is now $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Draw a diagram showing this information.

Find the velocity of Q immediately after the collision in terms of $u$. Find also the coefficient of restitution between P and Q .
\item Find, in terms of $u$, the impulse of P on Q in the collision.
\end{enumerate}\item Fig. 1 shows a small object R of mass 5 kg travelling on a smooth horizontal plane at $6 \mathrm {~ms} ^ { - 1 }$. It explodes into two parts of masses 2 kg and 3 kg . The velocities of these parts are in the plane in which R was travelling with the speeds and directions indicated. The angles $\alpha$ and $\beta$ are given by $\cos \alpha = \frac { 4 } { 5 }$ and $\cos \beta = \frac { 3 } { 5 }$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{334b2170-3708-46db-bff7-bcad7d5fab00-2_460_1450_1050_312}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Calculate $u$ and $v$.
\item Calculate the increase in kinetic energy resulting from the explosion.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI M2 2014 Q1 [17]}}