Edexcel M4 2010 June — Question 2 14 marks

Exam BoardEdexcel
ModuleM4 (Mechanics 4)
Year2010
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeOblique collision, vector velocity form
DifficultyStandard +0.3 This is a standard M4 oblique collision problem with straightforward application of momentum conservation and Newton's law of restitution, followed by a routine wall collision. The line of centres being parallel to i simplifies the problem significantly (perpendicular components unchanged). All steps are algorithmic with no novel insight required, making it slightly easier than average.
Spec1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication6.02d Mechanical energy: KE and PE concepts6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03f Impulse-momentum: relation6.03k Newton's experimental law: direct impact

2. Two smooth uniform spheres \(S\) and \(T\) have equal radii. The mass of \(S\) is 0.3 kg and the mass of \(T\) is 0.6 kg . The spheres are moving on a smooth horizontal plane and collide obliquely. Immediately before the collision the velocity of \(S\) is \(\mathbf { u } _ { 1 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the velocity of \(T\) is \(\mathbf { u } _ { 2 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The coefficient of restitution between the spheres is 0.5 . Immediately after the collision the velocity of \(S\) is \(( - \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\) and the velocity of \(T\) is \(( \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). Given that when the spheres collide the line joining their centres is parallel to \(\mathbf { i }\),
  1. find
    1. \(\mathbf { u } _ { 1 }\),
    2. \(\mathbf { u } _ { 2 }\). After the collision, \(T\) goes on to collide with a smooth vertical wall which is parallel to \(\mathbf { j }\). Given that the coefficient of restitution between \(T\) and the wall is also 0.5 , find
  2. the angle through which the direction of motion of \(T\) is deflected as a result of the collision with the wall,
  3. the loss in kinetic energy of \(T\) caused by the collision with the wall.

Question 2:
Part (a):
AnswerMarks Guidance
Working/AnswerMarks Notes
\(0.3v - 0.6w = 0.3\), giving \(v - 2w = 1\)M1A1
\(\frac{1}{2}(v + w) = 2\), giving \(v + w = 4\)M1A1
\(w = 1,\ v = 3\)
(i) \(\mathbf{u}_1 = 3\mathbf{i} + 2\mathbf{j}\) (ii) \(\mathbf{u}_2 = -\mathbf{i} + \mathbf{j}\)A1A1
Total(6)
Part (b):
AnswerMarks Guidance
Working/AnswerMarks Notes
\(v = 0.5\)B1
\(\tan\theta = 0.5\)M1 \(\tan\theta =\) their \(v\)
\(\theta = 26.6°\)A1
their \(\theta + 45°\)M1
Deflection angle \(= 45 + 26.6 = 71.6°\)A1
Total(5)
Part (c):
AnswerMarks Guidance
Working/AnswerMarks Notes
\(\text{KE Loss} = \frac{1}{2} \times 0.6 \times \left\{(1^2 + 1^2) - (1^2 + v^2)\right\}\)M1A1
\(= 0.225\ \text{J}\)A1
Total(3)
Question Total14
## Question 2:

### Part (a):

| Working/Answer | Marks | Notes |
|---|---|---|
| $0.3v - 0.6w = 0.3$, giving $v - 2w = 1$ | M1A1 | |
| $\frac{1}{2}(v + w) = 2$, giving $v + w = 4$ | M1A1 | |
| $w = 1,\ v = 3$ | | |
| (i) $\mathbf{u}_1 = 3\mathbf{i} + 2\mathbf{j}$ (ii) $\mathbf{u}_2 = -\mathbf{i} + \mathbf{j}$ | A1A1 | |
| **Total** | **(6)** | |

### Part (b):

| Working/Answer | Marks | Notes |
|---|---|---|
| $v = 0.5$ | B1 | |
| $\tan\theta = 0.5$ | M1 | $\tan\theta =$ their $v$ |
| $\theta = 26.6°$ | A1 | |
| their $\theta + 45°$ | M1 | |
| Deflection angle $= 45 + 26.6 = 71.6°$ | A1 | |
| **Total** | **(5)** | |

### Part (c):

| Working/Answer | Marks | Notes |
|---|---|---|
| $\text{KE Loss} = \frac{1}{2} \times 0.6 \times \left\{(1^2 + 1^2) - (1^2 + v^2)\right\}$ | M1A1 | |
| $= 0.225\ \text{J}$ | A1 | |
| **Total** | **(3)** | |
| **Question Total** | **14** | |
2. Two smooth uniform spheres $S$ and $T$ have equal radii. The mass of $S$ is 0.3 kg and the mass of $T$ is 0.6 kg . The spheres are moving on a smooth horizontal plane and collide obliquely. Immediately before the collision the velocity of $S$ is $\mathbf { u } _ { 1 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the velocity of $T$ is $\mathbf { u } _ { 2 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The coefficient of restitution between the spheres is 0.5 . Immediately after the collision the velocity of $S$ is $( - \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and the velocity of $T$ is $( \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. Given that when the spheres collide the line joining their centres is parallel to $\mathbf { i }$,
\begin{enumerate}[label=(\alph*)]
\item find
\begin{enumerate}[label=(\roman*)]
\item $\mathbf { u } _ { 1 }$,
\item $\mathbf { u } _ { 2 }$.

After the collision, $T$ goes on to collide with a smooth vertical wall which is parallel to $\mathbf { j }$. Given that the coefficient of restitution between $T$ and the wall is also 0.5 , find
\end{enumerate}\item the angle through which the direction of motion of $T$ is deflected as a result of the collision with the wall,
\item the loss in kinetic energy of $T$ caused by the collision with the wall.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M4 2010 Q2 [14]}}