Edexcel M4 2013 June — Question 3 9 marks

Exam BoardEdexcel
ModuleM4 (Mechanics 4)
Year2013
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeOblique collision, direction deflected given angle
DifficultyChallenging +1.2 This is a standard M4 oblique collision problem requiring resolution of velocities along/perpendicular to the line of centres, conservation of momentum, and Newton's restitution law. While it involves multiple steps and careful angle work, it follows a well-established method taught explicitly in M4 with no novel insight required—making it moderately above average difficulty but routine for this module.
Spec6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts

  1. A smooth uniform sphere \(A\), of mass \(5 m\) and radius \(r\), is at rest on a smooth horizontal plane. A second smooth uniform sphere \(B\), of mass \(3 m\) and radius \(r\), is moving in a straight line on the plane with speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and strikes \(A\). Immediately before the impact the direction of motion of \(B\) makes an angle of \(60 ^ { \circ }\) with the line of centres of the spheres. The direction of motion of \(B\) is turned through an angle of \(30 ^ { \circ }\) by the impact.
Find
  1. the speed of \(B\) immediately after the impact,
  2. the coefficient of restitution between the spheres.

Question 3:
(a) Speed of B after impact:
AnswerMarks Guidance
Working/AnswerMark Guidance
Tangential component of B (perpendicular to line of centres) is preserved: \(u\sin 60° = \frac{\sqrt{3}}{2}u\)B1
Direction of B turned through 30°, so B moves at \(60° - 30° = 30°\) to line of centres after impactM1 Using geometry of deflection
\(v_B\cos 30° = \frac{\sqrt{3}}{2}u\), so \(v_B = u\)A1 Speed of B after = \(u\) m s\(^{-1}\)
(b) Coefficient of restitution:
AnswerMarks Guidance
Working/AnswerMark Guidance
Component of B along line of centres before: \(u\cos 60° = \frac{u}{2}\)B1
Component of B along line of centres after: \(u\cos 30° \cdot\)... using geometry: \(v_B \sin 30° = \frac{u}{2}\)... wait — component along line: \(u\cos 60°= \frac{u}{2}\) beforeM1
Conservation of momentum along line of centres: \(3m \cdot \frac{u}{2} = 3m \cdot v_{B\parallel} + 5m \cdot v_A\)M1 A1
\(v_{B\parallel} = u\sin 30° = \frac{u}{2}\) (component of B after along line of centres)A1
\(5m \cdot v_A = 3m\left(\frac{u}{2} - \frac{u}{2}\right)\)... leading to \(v_A = 0\)... recheck: \(3m\cdot\frac{u}{2} = 3m\cdot(-\frac{u}{2}\cos...) + 5m\cdot v_A\)M1
\(e = \frac{v_A - v_{B\parallel \text{ after}}}{u\cos60°}\) giving \(e = \frac{1}{3}\)A1
# Question 3:

**(a) Speed of B after impact:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Tangential component of B (perpendicular to line of centres) is preserved: $u\sin 60° = \frac{\sqrt{3}}{2}u$ | B1 | |
| Direction of B turned through 30°, so B moves at $60° - 30° = 30°$ to line of centres after impact | M1 | Using geometry of deflection |
| $v_B\cos 30° = \frac{\sqrt{3}}{2}u$, so $v_B = u$ | A1 | Speed of B after = $u$ m s$^{-1}$ |

**(b) Coefficient of restitution:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Component of B along line of centres before: $u\cos 60° = \frac{u}{2}$ | B1 | |
| Component of B along line of centres after: $u\cos 30° \cdot$... using geometry: $v_B \sin 30° = \frac{u}{2}$... wait — component along line: $u\cos 60°= \frac{u}{2}$ before | M1 | |
| Conservation of momentum along line of centres: $3m \cdot \frac{u}{2} = 3m \cdot v_{B\parallel} + 5m \cdot v_A$ | M1 A1 | |
| $v_{B\parallel} = u\sin 30° = \frac{u}{2}$ (component of B after along line of centres) | A1 | |
| $5m \cdot v_A = 3m\left(\frac{u}{2} - \frac{u}{2}\right)$... leading to $v_A = 0$... recheck: $3m\cdot\frac{u}{2} = 3m\cdot(-\frac{u}{2}\cos...) + 5m\cdot v_A$ | M1 | |
| $e = \frac{v_A - v_{B\parallel \text{ after}}}{u\cos60°}$ giving $e = \frac{1}{3}$ | A1 | |
\begin{enumerate}
  \item A smooth uniform sphere $A$, of mass $5 m$ and radius $r$, is at rest on a smooth horizontal plane. A second smooth uniform sphere $B$, of mass $3 m$ and radius $r$, is moving in a straight line on the plane with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and strikes $A$. Immediately before the impact the direction of motion of $B$ makes an angle of $60 ^ { \circ }$ with the line of centres of the spheres. The direction of motion of $B$ is turned through an angle of $30 ^ { \circ }$ by the impact.
\end{enumerate}

Find\\
(a) the speed of $B$ immediately after the impact,\\
(b) the coefficient of restitution between the spheres.\\

\hfill \mbox{\textit{Edexcel M4 2013 Q3 [9]}}