| Exam Board | SPS |
|---|---|
| Module | SPS FM Mechanics (SPS FM Mechanics) |
| Year | 2026 |
| Session | January |
| Marks | 9 |
| Topic | Impulse and momentum (advanced) |
| Type | Oblique collision of spheres |
| Difficulty | Challenging +1.2 This is a two-part oblique collision problem requiring conservation of momentum along the line of centres, Newton's restitution law, and geometric reasoning with the constraint that final and initial directions are parallel. Part (a) is standard mechanics bookwork (4 marks), while part (b) requires setting up perpendicular component equations and solving a trigonometric relationship (5 marks). The constraint α + β = 90° simplifies the algebra. This is moderately above average for FM mechanics but follows established collision problem patterns without requiring exceptional insight. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
\includegraphics{figure_7}
Two uniform smooth spheres $A$ and $B$ of equal radii have masses $m$ and $\frac{1}{2}m$ respectively. The two spheres are moving on a horizontal surface when they collide. Immediately before the collision, sphere $A$ is travelling with speed $u$ and its direction of motion makes an angle $\alpha$ with the line of centres. Sphere $B$ is travelling with speed $2u$ and its direction of motion makes an angle $\beta$ with the line of centres (see diagram). The coefficient of restitution between the spheres is $\frac{2}{3}$ and $\alpha + \beta = 90°$.
\begin{enumerate}[label=(\alph*)]
\item Find the component of the velocity of $B$ parallel to the line of centres after the collision, giving your answer in terms of $u$ and $\alpha$. [4]
\end{enumerate}
The direction of motion of $B$ after the collision is parallel to the direction of motion of $A$ before the collision.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $\tan \alpha$. [5]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Mechanics 2026 Q7 [9]}}