Edexcel M2 2009 January — Question 7 17 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2009
SessionJanuary
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeCollision followed by wall impact
DifficultyStandard +0.3 This is a standard M2 momentum/collisions question with multiple parts building on each other. Parts (a)-(c) are routine 'show that' exercises applying conservation of momentum and Newton's restitution formula with given answers to verify. Part (d) requires tracking two particles after a wall collision, but follows a predictable pattern. The algebra is straightforward and the problem-solving is mechanical rather than requiring insight, making it slightly easier than average.
Spec6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact
  1. A particle \(P\) of mass \(3 m\) is moving in a straight line with speed \(2 u\) on a smooth horizontal table. It collides directly with another particle \(Q\) of mass \(2 m\) which is moving with speed \(u\) in the opposite direction to \(P\). The coefficient of restitution between \(P\) and \(Q\) is \(e\).
    1. Show that the speed of \(Q\) immediately after the collision is \(\frac { 1 } { 5 } ( 9 e + 4 ) u\).
    The speed of \(P\) immediately after the collision is \(\frac { 1 } { 2 } u\).
  2. Show that \(e = \frac { 1 } { 4 }\). The collision between \(P\) and \(Q\) takes place at the point \(A\). After the collision \(Q\) hits a smooth fixed vertical wall which is at right-angles to the direction of motion of \(Q\). The distance from \(A\) to the wall is \(d\).
  3. Show that \(P\) is a distance \(\frac { 3 } { 5 } d\) from the wall at the instant when \(Q\) hits the wall. Particle \(Q\) rebounds from the wall and moves so as to collide directly with particle \(P\) at the point \(B\). Given that the coefficient of restitution between \(Q\) and the wall is \(\frac { 1 } { 5 }\),
  4. find, in terms of \(d\), the distance of the point \(B\) from the wall.
Question 7:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Before: \(P(3m)\) moves at \(2u\), \(Q(2m)\) moves at \(-u\) (towards each other). After: \(P\) at \(x\), \(Q\) at \(y\). Correct use of NEL.M1*
\(y - x = e(2u + u)\) o.e.A1
CLM\((\rightarrow): 3m(2u) + 2m(-u) = 3m(x) + 2m(y)\ (\Rightarrow 4u = 3x + 2y)\)B1
Hence \(x = y - 3eu,\ 4u = 3(y - 3eu) + 2y,\ (u(9e+4) = 5y)\)d*M1
Hence speed of \(Q = \frac{1}{5}(9e+4)u\)A1 cso AG
Total: (5)
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x = y - 3eu = \frac{1}{5}(9e+4)u - 3eu\)M1\(^{\#}\)
Hence speed of \(P = \frac{1}{5}(4-6e)u = \frac{2u}{5}(2-3e)\) o.e.A1
\(x = \frac{1}{2}u \Rightarrow \frac{2u}{5}(2-3e) = \frac{u}{2}\ \Rightarrow 5u = 8u - 12eu \Rightarrow 12e = 3\) & solve for \(e\)d\(^{\#}\)M1
\(e = \frac{3}{12} \Rightarrow \underline{e = \frac{1}{4}}\)A1 AG
Total: (4)
Part (b) – Alternative Method:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Using NEL correctly with given speeds of \(P\) and \(Q\)M1\(^{\#}\), A1
\(3eu = \frac{1}{5}(9e+4)u - \frac{1}{2}u\)
\(3eu = \frac{9}{5}eu + \frac{4}{5}u - \frac{1}{2}u,\quad 3e - \frac{9}{5}e = \frac{4}{5} - \frac{1}{2}\) & solve for \(e\)d\(^{\#}\)M1
\(\frac{6}{5}e = \frac{3}{10} \Rightarrow e = \frac{15}{60} \Rightarrow e = \frac{1}{4}\)A1
Total: (4)
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Time taken by \(Q\) from \(A\) to wall \(= \dfrac{d}{y} = \left\{\dfrac{4d}{5u}\right\}\)M1\(^{\dagger}\)
Distance moved by \(P\) in this time \(= \dfrac{u}{2} \times \dfrac{d}{y}\ \left(= \dfrac{u}{2}\left(\dfrac{4d}{5u}\right) = \dfrac{2}{5}d\right)\)A1
Distance of \(P\) from wall \(= d - x\!\left(\dfrac{d}{y}\right) := d - \dfrac{2}{5}d = \underline{\dfrac{3}{5}d}\)d\(^{\dagger}\)M1, A1 cso AG
Total: (4)
Part (c) – Alternative Method:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Ratio speed \(P\) : speed \(Q = x : y = \frac{1}{2}u : \frac{1}{5}\!\left(\frac{9}{4}+4\right)u = \frac{1}{2}u : \frac{5}{4}u = 2:5\)M1\(^{\dagger}\), A1
So if \(Q\) moves distance \(d\), \(P\) moves distance \(\frac{2}{5}d\)
Distance of \(P\) from wall \(= d - \frac{2}{5}d := \underline{\frac{3}{5}d}\)d\(^{\dagger}\)M1, A1 cso
Total: (4)
Question (d):
Method 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
After collision with wall, speed \(Q = \frac{1}{5}y = \frac{1}{5}\left(\frac{5u}{4}\right) = \frac{1}{4}u\)B1ft follow their \(y\)
Time for \(P\): \(T_{AB} = \dfrac{\frac{3d}{5} - x}{\frac{1}{2}u}\), Time for \(Q\): \(T_{WB} = \dfrac{x}{\frac{1}{4}u}\)B1ft from their \(y\)
Hence \(T_{AB} = T_{WB} \Rightarrow \dfrac{\frac{3d}{5} - x}{\frac{1}{2}u} = \dfrac{x}{\frac{1}{4}u}\)M1
gives \(2\left(\frac{3d}{5} - x\right) = 4x \Rightarrow \frac{3d}{5} - x = 2x\), \(3x = \frac{3d}{5} \Rightarrow x = \frac{1}{5}d\)A1 cao
(4)
Method 2 (or):
AnswerMarks Guidance
Answer/WorkingMark Guidance
After collision with wall, speed \(Q = \frac{1}{5}y = \frac{1}{5}\left(\frac{5u}{4}\right) = \frac{1}{4}u\)B1ft follow their \(y\)
speed \(P = x = \frac{1}{2}u\), speed \(P\): new speed \(Q = \frac{1}{2}u : \frac{1}{4}u = 2:1\)B1ft from their \(y\)
Distance of \(B\) from wall \(= \dfrac{1}{3} \times \dfrac{3d}{5} = \dfrac{d}{5}\)M1; A1 their \(\dfrac{1}{2+1}\)
(4)
2nd Method (or):
AnswerMarks Guidance
Answer/WorkingMark Guidance
After collision with wall, speed \(Q = \frac{1}{5}y = \frac{1}{5}\left(\frac{5u}{4}\right) = \frac{1}{4}u\)B1ft follow their \(y\)
Combined speed of \(P\) and \(Q = \frac{1}{2}u + \frac{1}{4}u = \frac{3}{4}u\)
Time from wall to \(2^{\text{nd}}\) collision \(= \dfrac{\frac{3d}{5}}{\frac{3u}{4}} = \dfrac{3d}{5} \times \dfrac{4}{3u} = \dfrac{4d}{5u}\)B1ft from their \(y\)
Distance of \(B\) from wall \(= (\text{their speed}) \times (\text{their time}) = \dfrac{u}{4} \times \dfrac{4d}{5u} = \dfrac{1}{5}d\)M1; A1
(4) [17] 
## Question 7:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Before: $P(3m)$ moves at $2u$, $Q(2m)$ moves at $-u$ (towards each other). After: $P$ at $x$, $Q$ at $y$. Correct use of NEL. | M1* | |
| $y - x = e(2u + u)$ o.e. | A1 | |
| CLM$(\rightarrow): 3m(2u) + 2m(-u) = 3m(x) + 2m(y)\ (\Rightarrow 4u = 3x + 2y)$ | B1 | |
| Hence $x = y - 3eu,\ 4u = 3(y - 3eu) + 2y,\ (u(9e+4) = 5y)$ | d*M1 | |
| Hence speed of $Q = \frac{1}{5}(9e+4)u$ | A1 cso | **AG** |
| **Total: (5)** | | |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = y - 3eu = \frac{1}{5}(9e+4)u - 3eu$ | M1$^{\#}$ | |
| Hence speed of $P = \frac{1}{5}(4-6e)u = \frac{2u}{5}(2-3e)$ o.e. | A1 | |
| $x = \frac{1}{2}u \Rightarrow \frac{2u}{5}(2-3e) = \frac{u}{2}\ \Rightarrow 5u = 8u - 12eu \Rightarrow 12e = 3$ & solve for $e$ | d$^{\#}$M1 | |
| $e = \frac{3}{12} \Rightarrow \underline{e = \frac{1}{4}}$ | A1 | **AG** |
| **Total: (4)** | | |

### Part (b) – Alternative Method:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Using NEL correctly with given speeds of $P$ and $Q$ | M1$^{\#}$, A1 | |
| $3eu = \frac{1}{5}(9e+4)u - \frac{1}{2}u$ | | |
| $3eu = \frac{9}{5}eu + \frac{4}{5}u - \frac{1}{2}u,\quad 3e - \frac{9}{5}e = \frac{4}{5} - \frac{1}{2}$ & solve for $e$ | d$^{\#}$M1 | |
| $\frac{6}{5}e = \frac{3}{10} \Rightarrow e = \frac{15}{60} \Rightarrow e = \frac{1}{4}$ | A1 | |
| **Total: (4)** | | |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Time taken by $Q$ from $A$ to wall $= \dfrac{d}{y} = \left\{\dfrac{4d}{5u}\right\}$ | M1$^{\dagger}$ | |
| Distance moved by $P$ in this time $= \dfrac{u}{2} \times \dfrac{d}{y}\ \left(= \dfrac{u}{2}\left(\dfrac{4d}{5u}\right) = \dfrac{2}{5}d\right)$ | A1 | |
| Distance of $P$ from wall $= d - x\!\left(\dfrac{d}{y}\right) := d - \dfrac{2}{5}d = \underline{\dfrac{3}{5}d}$ | d$^{\dagger}$M1, A1 cso | **AG** |
| **Total: (4)** | | |

### Part (c) – Alternative Method:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Ratio speed $P$ : speed $Q = x : y = \frac{1}{2}u : \frac{1}{5}\!\left(\frac{9}{4}+4\right)u = \frac{1}{2}u : \frac{5}{4}u = 2:5$ | M1$^{\dagger}$, A1 | |
| So if $Q$ moves distance $d$, $P$ moves distance $\frac{2}{5}d$ | | |
| Distance of $P$ from wall $= d - \frac{2}{5}d := \underline{\frac{3}{5}d}$ | d$^{\dagger}$M1, A1 cso | |
| **Total: (4)** | | |

## Question (d):

**Method 1:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| After collision with wall, speed $Q = \frac{1}{5}y = \frac{1}{5}\left(\frac{5u}{4}\right) = \frac{1}{4}u$ | B1ft | follow their $y$ |
| Time for $P$: $T_{AB} = \dfrac{\frac{3d}{5} - x}{\frac{1}{2}u}$, Time for $Q$: $T_{WB} = \dfrac{x}{\frac{1}{4}u}$ | B1ft | from their $y$ |
| Hence $T_{AB} = T_{WB} \Rightarrow \dfrac{\frac{3d}{5} - x}{\frac{1}{2}u} = \dfrac{x}{\frac{1}{4}u}$ | M1 | |
| gives $2\left(\frac{3d}{5} - x\right) = 4x \Rightarrow \frac{3d}{5} - x = 2x$, $3x = \frac{3d}{5} \Rightarrow x = \frac{1}{5}d$ | A1 **cao** | |
| | **(4)** | |

**Method 2 (or):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| After collision with wall, speed $Q = \frac{1}{5}y = \frac{1}{5}\left(\frac{5u}{4}\right) = \frac{1}{4}u$ | B1ft | follow their $y$ |
| speed $P = x = \frac{1}{2}u$, speed $P$: new speed $Q = \frac{1}{2}u : \frac{1}{4}u = 2:1$ | B1ft | from their $y$ |
| Distance of $B$ from wall $= \dfrac{1}{3} \times \dfrac{3d}{5} = \dfrac{d}{5}$ | M1; A1 | their $\dfrac{1}{2+1}$ |
| | **(4)** | |

**2nd Method (or):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| After collision with wall, speed $Q = \frac{1}{5}y = \frac{1}{5}\left(\frac{5u}{4}\right) = \frac{1}{4}u$ | B1ft | follow their $y$ |
| Combined speed of $P$ and $Q = \frac{1}{2}u + \frac{1}{4}u = \frac{3}{4}u$ | | |
| Time from wall to $2^{\text{nd}}$ collision $= \dfrac{\frac{3d}{5}}{\frac{3u}{4}} = \dfrac{3d}{5} \times \dfrac{4}{3u} = \dfrac{4d}{5u}$ | B1ft | from their $y$ |
| Distance of $B$ from wall $= (\text{their speed}) \times (\text{their time}) = \dfrac{u}{4} \times \dfrac{4d}{5u} = \dfrac{1}{5}d$ | M1; A1 | |
| | **(4) [17]** | |
\begin{enumerate}
  \item A particle $P$ of mass $3 m$ is moving in a straight line with speed $2 u$ on a smooth horizontal table. It collides directly with another particle $Q$ of mass $2 m$ which is moving with speed $u$ in the opposite direction to $P$. The coefficient of restitution between $P$ and $Q$ is $e$.\\
(a) Show that the speed of $Q$ immediately after the collision is $\frac { 1 } { 5 } ( 9 e + 4 ) u$.
\end{enumerate}

The speed of $P$ immediately after the collision is $\frac { 1 } { 2 } u$.\\
(b) Show that $e = \frac { 1 } { 4 }$.

The collision between $P$ and $Q$ takes place at the point $A$. After the collision $Q$ hits a smooth fixed vertical wall which is at right-angles to the direction of motion of $Q$. The distance from $A$ to the wall is $d$.\\
(c) Show that $P$ is a distance $\frac { 3 } { 5 } d$ from the wall at the instant when $Q$ hits the wall.

Particle $Q$ rebounds from the wall and moves so as to collide directly with particle $P$ at the point $B$. Given that the coefficient of restitution between $Q$ and the wall is $\frac { 1 } { 5 }$,\\
(d) find, in terms of $d$, the distance of the point $B$ from the wall.\\

\hfill \mbox{\textit{Edexcel M2 2009 Q7 [17]}}
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