A2 June 2025 Q7
7.
Two uniform spheres \(P\) and \(Q\) have equal radii. The mass of \(P\) is 0.3 kg and the mass of \(Q\) is 0.4 kg. The spheres are moving on a smooth horizontal plane when they collide obliquely.
Immediately before the collision,
- \(P\) is moving with speed \(5\ \text{m s}^{-1}\) at \(30^\circ\) to the line of centres of the spheres
- \(Q\) is moving with speed \(3\ \text{m s}^{-1}\) at \(60^\circ\) to the line of centres of the spheres
Immediately after the collision,
- \(P\) is moving with speed \(v\ \text{m s}^{-1}\) at \(60^\circ\) to the line of centres of the spheres
- \(Q\) is moving with speed \(w\ \text{m s}^{-1}\) at \(\theta^\circ\) to the line of centres of the spheres
as in the plan view shown in Figure 4.
| Scheme | Marks | AO |
|---|---|---|
 | ||
| Velocity of \(P\) perpendicular to the line of centres unchanged | M1 | 3.4 |
| \(5\sin 30^\circ = v\sin 60^\circ\) | A1 | 1.1b |
| \(\dfrac{5}{2} = \dfrac{v\sqrt{3}}{2} \quad\Rightarrow v = \dfrac{5\sqrt{3}}{3}\) * | A1* | 2.2a |
| (3) |
Notes
M1: Use the model for components of \(P\) perpendicular to the line of centres. Condone sin/cos confusion if clearly working perpendicular to the line of centres.
A1: A correct unsimplified equation in \(v\). For example
- \(5\sin 30^\circ = v\sin 60^\circ\)
- \(5\sin 30^\circ = x\tan 60^\circ \;\Rightarrow\; v = \sqrt{\left(\dfrac{5}{2}\right)^2 + x^2} = \ldots\)
- \(\begin{pmatrix} 5\cos 30^\circ \\ 5\sin 30^\circ \end{pmatrix} \bullet \begin{pmatrix} -v\cos 60^\circ \\ 5\sin 30^\circ \end{pmatrix} = 0\)
A1*: Obtain given answer from complete and correct working. Must see ‘\(v = \ldots\)’ and the exact expression. There must be at least one line of working between the initial equation and the given answer.
| Scheme | Marks | AO |
|---|---|---|
| ||
| Velocity of \(Q\) perpendicular to the line of centres unchanged | M1 | 3.4 |
| Either \(w\sin\theta^\circ = 3\sin 60^\circ \left(= \dfrac{3\sqrt{3}}{2}\right)\) or \(a = 3\sin 60^\circ \left(= \dfrac{3\sqrt{3}}{2}\right)\) | A1 | 1.1b |
| CLM parallel to the line of centres | M1 | 3.4 |
| Either \(0.3 \times 5\cos 30^\circ - 0.4 \times 3\cos 60^\circ = 0.4w\cos\theta^\circ - 0.3v\cos 60^\circ\) Or \(0.3 \times 5\cos 30^\circ - 0.4 \times 3\cos 60^\circ = 0.4b - 0.3x\) | A1 | 1.1b |
| Solve for \(w\) or \(\theta\) using \(\begin{cases} w\sin\theta^\circ = \dfrac{3\sqrt{3}}{2} \\[4pt] w\cos\theta^\circ = \dfrac{5\sqrt{3} - 3}{2} \end{cases}\) or \(\begin{cases} a = \dfrac{3\sqrt{3}}{2} \\[4pt] b = \dfrac{5\sqrt{3} - 3}{2} \end{cases}\) | dM1 | 3.1a |
| \(\theta = 43 \;\; (42.552\ldots)\) or \(w = 3.8 \;\; (3.84182\ldots)\) | A1 | 2.2a |
| \(\theta = 43 \;\; (42.552\ldots)\) and \(w = 3.8 \;\; (3.84182\ldots)\) | A1 | 2.2a |
| (7) |
Notes
M1: Use the model for components of \(Q\) perpendicular to the line of centres. Condone sin/cos confusion if clearly working perpendicular to the line of centres.
M0 if speeds are used instead of components.
A1: Correct unsimplified equation.
M1: Correct use of CLM parallel to the line of centres. Dimensionally correct. Need all terms. No need to replace \(v\). Condone sign errors. Condone sin/cos confusion consistent with perpendicular components. Must have correct mass-velocity pairings.
M0 if speeds are used instead of components.
A1: Correct unsimplified equation (no need to replace \(v\)).
dM1: Dependent on both previous M marks. Complete method using two correctly formed equations to obtain \(w\) or \(\theta\). i.e. form an equation in \(w\) or \(\theta\) only and solve to reach \(w = \ldots\) or \(\theta = \ldots\)
If numerical values appear without working out, a complete method using a calculator may be implied by correct answers following correct equations.
A1: One value correct to 2 sf or better, \(\theta = 43\ (42.552\ldots)\) or \(w = 3.8\ (3.84182\ldots)\)
\(\theta\) must be in degrees.
A1: Both values correct to 2 sf or better, \(\theta = 43\ (42.552\ldots)\) and \(w = 3.8\ (3.84182\ldots)\)
\(\theta\) must be in degrees.
| Scheme | Marks | AO |
|---|---|---|
| ||
| NEL parallel to the line of centres | M1 | 3.1b |
| A1ft | 1.1b |
| \(e = 0.73 \quad (0.73300\ldots)\) | A1 | 1.1b |
| (3) | ||
| (13 marks) |
Notes
M1: Correct use of NEL parallel to the line of centres. Must form an equation in \(e\) and their components parallel to the line of centres. Condone sign errors. Condone sin/cos confusion consistent with perpendicular components. If seen in earlier work, NEL must be used in (c) to earn the marks here.
M0 if separation and approach are on the wrong side.
M0 if speeds are used instead of velocity components.
A1ft: Correct unsimplified equation (in \(e\) and their components parallel to the line of centres).
A1: Correct answer, 2 sf or better.