| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Collision followed by wall impact |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring conservation of momentum and Newton's restitution law, followed by a straightforward kinematics calculation. The multi-step nature and wall rebound add slight complexity, but all techniques are routine textbook applications with no novel insight required. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks |
|---|---|
| CLM: \(3m + 4m = mv + 2mw\) \((7 = v + 2w)\) | M1A1 |
| Impact: \(w - v = \frac{2}{3}(3 - 2)\) | M1A1 |
| Solve for \(w\): \(w = \frac{23}{9}\) *Given answer* | M1 A1 |
| \(v = \frac{17}{9}\) | A1 (7) |
| Notes: First M1 for momentum equn with correct terms, condone sign errors, consistent extra g's or missing m's; First A1 for a correct equation in \(v\) and \(w\) only; Second M1 for Impact Law, correct way up, condone sign errors; Second A1 for a correct equation in \(v\) and \(w\) only; Third M1 for solving for \(w\); Third A1 for \(w = \frac{23}{9}\) Given answer; Fourth A1 for \(v = \frac{17}{9}\) Accept 1.9 or better. N.B. Treat this as a B1 dependent on scoring the first 4 marks. |
| Answer | Marks |
|---|---|
| Speed of \(B\) after hitting wall \(= \frac{23}{18}\) (m s\(^{-1}\)) | B1 |
| Time for \(B\) to get to the 2nd collision \(= \frac{3}{23} + \frac{d}{23}\) | M1 A1 |
| Time for \(A\) to get to the 2nd collision \(= \frac{3-d}{17}\) | A1 |
| Equate times to give equation in \(d\) only: \(\frac{3}{23} + \frac{d}{23} = \frac{3-d}{\frac{17}{9}}\) | M1A1 |
| \(d = \frac{6}{19}\) (0.32 or better) | A1 (7) |
| Notes: First B1 for 23/18 seen. Accept 1.3 or better; First M1 for attempt at a complete expression, in \(d\) only, for time for \(B\) to arrive at 2nd collision, must be \((\frac{3}{23} + \frac{d}{23})\); First A1 for a correct expression; Second A1 for time for \(A\) to get to collision \(= \frac{3-d}{\frac{17}{9}}\); Second M1 for equating the times; must be 3 terms; Third A1 for a correct equation; Fourth A1 for \(d = \frac{6}{19}\) oe or 0.32 or better |
| Answer | Marks |
|---|---|
| Speed of \(B\) after hitting wall \(= \frac{23}{18}\) (m s\(^{-1}\)) | B1 |
| First M1 for \(\frac{3}{23}\) which is time for \(B\) to reach wall. Allow if they try to divide or the division is implied even if it is done incorrectly. | |
| First A1 for 51/23 oe and accept 2.2 or better | |
| Second A1 for \((3 - 51/23)\) oe and accept 0.8 or better. | |
| Second M1 for time to meet = (a calculated 'separation distance' / (23/18 + their v)) | |
| Third A1 for a correct time (108/437 oe) accept 0.25 or better |
| Answer | Marks |
|---|---|
| First M1 for attempt to differentiate path equation (both powers going down) | |
| First A1 for a correct expression | |
| Second DM1, dependent on first M1, for putting \(x = 20\) and using \(\tan^{-1}\) to obtain an angle | |
| Second A1 for 66⁰ or 66.2⁰ below the horizontal or 24⁰ or 23.8⁰ to the downward vertical OR the angle marked on a clear diagram with an arrow. | |
| First M1 for using \(x = 20\) in horizontal motion equn to obtain \(t = \frac{20}{7}\) and using \(v = u + at\) to obtain vertical speed or any other complete method e.g. put \(x = 20\) in equation of path to obtain \(y\), which is then used in \(v^2 = u^2 + 2as\) vertically to obtain vertical speed. | |
| First A1 for a correct expression for vertical speed | |
| Second DM1, dependent on first M1, for using ratio of components and \(\tan^{-1}\) to obtain an angle | |
| OR finding the velocity vector AND referring to its direction ('parallel to' can score M1 but not the A1). | |
| Second A1 for 66⁰ or 66.2⁰ below the horizontal or in the direction of ( parallel to is M1A0) \((7i - 16j)\) or \((7i - 15.9j)\) or a multiple. |
| Answer | Marks |
|---|---|
| First M1 for attempt to differentiate path equation (both powers going down) | |
| First A1 for a correct expression | |
| Second DM1, dependent on first M1, for putting \(x = 20\) and using \(\tan^{-1}\) to obtain an angle | |
| Second A1 for 66⁰ or 66.2⁰ below the horizontal or 24⁰ or 23.8⁰ to the downward vertical OR the angle marked on a clear diagram with an arrow. |
## 7a
| CLM: $3m + 4m = mv + 2mw$ $(7 = v + 2w)$ | M1A1 |
| Impact: $w - v = \frac{2}{3}(3 - 2)$ | M1A1 |
| Solve for $w$: $w = \frac{23}{9}$ *Given answer* | M1 A1 |
| $v = \frac{17}{9}$ | A1 (7) |
| **Notes:** First M1 for momentum equn with correct terms, condone sign errors, consistent extra g's or missing m's; First A1 for a correct equation in $v$ and $w$ only; Second M1 for Impact Law, correct way up, condone sign errors; Second A1 for a correct equation in $v$ and $w$ only; Third M1 for solving for $w$; Third A1 for $w = \frac{23}{9}$ **Given answer**; Fourth A1 for $v = \frac{17}{9}$ Accept 1.9 or better. N.B. Treat this as a B1 dependent on scoring the first 4 marks. |
## 7b
| Speed of $B$ after hitting wall $= \frac{23}{18}$ (m s$^{-1}$) | B1 |
| Time for $B$ to get to the 2nd collision $= \frac{3}{23} + \frac{d}{23}$ | M1 A1 |
| Time for $A$ to get to the 2nd collision $= \frac{3-d}{17}$ | A1 |
| Equate times to give equation in $d$ only: $\frac{3}{23} + \frac{d}{23} = \frac{3-d}{\frac{17}{9}}$ | M1A1 |
| $d = \frac{6}{19}$ (0.32 or better) | A1 (7) |
| **Notes:** First B1 for 23/18 seen. Accept 1.3 or better; First M1 for attempt at a complete expression, in $d$ only, for time for $B$ to arrive at 2nd collision, must be $(\frac{3}{23} + \frac{d}{23})$; First A1 for a correct expression; Second A1 for time for $A$ to get to collision $= \frac{3-d}{\frac{17}{9}}$; Second M1 for equating the times; must be 3 terms; Third A1 for a correct equation; Fourth A1 for $d = \frac{6}{19}$ oe or 0.32 or better |
## 7b alt
| Speed of $B$ after hitting wall $= \frac{23}{18}$ (m s$^{-1}$) | B1 |
| First M1 for $\frac{3}{23}$ which is time for $B$ to reach wall. Allow if they try to divide or the division is implied even if it is done incorrectly. | |
| First A1 for 51/23 oe and accept 2.2 or better | |
| Second A1 for $(3 - 51/23)$ oe and accept 0.8 or better. | |
| Second M1 for time to meet = (a calculated 'separation distance' / (23/18 + their v)) | |
| Third A1 for a correct time (108/437 oe) accept 0.25 or better | |
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# Question 6b (continued from earlier page):
## 6b alt (additional notes)
| First M1 for attempt to differentiate path equation (both powers going down) |
| First A1 for a correct expression |
| Second DM1, dependent on first M1, for putting $x = 20$ and using $\tan^{-1}$ to obtain an angle |
| Second A1 for 66⁰ or 66.2⁰ below the horizontal or 24⁰ or 23.8⁰ to the downward vertical OR the angle marked on a clear diagram with an arrow. |
| First M1 for using $x = 20$ in horizontal motion equn to obtain $t = \frac{20}{7}$ and using $v = u + at$ to obtain vertical speed or any other complete method e.g. put $x = 20$ in equation of path to obtain $y$, which is then used in $v^2 = u^2 + 2as$ vertically to obtain vertical speed. |
| First A1 for a correct expression for vertical speed |
| Second DM1, dependent on first M1, for using ratio of components and $\tan^{-1}$ to obtain an angle |
| OR finding the velocity vector AND referring to its direction ('parallel to' can score M1 but not the A1). |
| Second A1 for 66⁰ or 66.2⁰ below the horizontal or in the direction of ( parallel to is M1A0) $(7i - 16j)$ or $(7i - 15.9j)$ or a multiple. |
## 6b alt (final notes)
| First M1 for attempt to differentiate path equation (both powers going down) |
| First A1 for a correct expression |
| Second DM1, dependent on first M1, for putting $x = 20$ and using $\tan^{-1}$ to obtain an angle |
| Second A1 for 66⁰ or 66.2⁰ below the horizontal or 24⁰ or 23.8⁰ to the downward vertical OR the angle marked on a clear diagram with an arrow. |7. Two particles $A$ and $B$, of mass $m$ and $2 m$ respectively, are moving in the same direction along the same straight line on a smooth horizontal surface, with $B$ in front of $A$. Particle $A$ has speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and particle $B$ has speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Particle $A$ collides directly with particle $B$. The coefficient of restitution between $A$ and $B$ is $\frac { 2 } { 3 }$. The direction of motion of both particles is not changed by the collision. Immediately after the collision, $A$ has speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $B$ has speed $w \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $w = \frac { 23 } { 9 }$.
\item Find the value of $v$.
When $A$ and $B$ collide they are 3 m from a smooth vertical wall which is perpendicular to their direction of motion. After the collision with $A$, particle $B$ hits the wall and rebounds. The coefficient of restitution between $B$ and the wall is $\frac { 1 } { 2 }$.
There is a second collision between $A$ and $B$ at a point $d \mathrm {~m}$ from the wall.
\end{enumerate}\item Find the value of $d$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2016 Q7 [14]}}