| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, direction deflected given angle |
| Difficulty | Challenging +1.2 This is a standard M3 oblique collision problem requiring systematic application of momentum conservation (perpendicular and parallel to line of centres), Newton's experimental law, and impulse calculation. While it involves multiple parts and careful component resolution, it follows a well-established procedure taught in all M3 courses with no novel insight required—making it moderately above average difficulty but routine for this topic. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Component of \(A\) along line of centres before: \(3\cos60° = \frac{3}{2}\) | B1 | |
| Component of \(B\) along line of centres before: \(-5\cos60° = -\frac{5}{2}\) | B1 | Must have correct sign/direction |
| Component of \(A\) perpendicular to line of centres unchanged: \(3\sin60° = \frac{3\sqrt{3}}{2}\) | B1 | Perpendicular components unchanged |
| Conservation of momentum along line of centres: \(2(\frac{3}{2}) + 4(-\frac{5}{2}) = 2v_A + 4v_B\) | M1 | Correct momentum equation |
| \(3 - 10 = 2v_A + 4v_B \Rightarrow 2v_A + 4v_B = -7\) | A1 | |
| \(B\) moves perpendicular to line of centres, so \(v_B = 0\) along line of centres | M1 | Using given condition |
| \(v_A = -\frac{7}{4}\) along line of centres | A1 | |
| Speed of \(A = \sqrt{\left(\frac{7}{4}\right)^2 + \left(\frac{3\sqrt{3}}{2}\right)^2} = \sqrt{\frac{49}{16}+\frac{27}{4}} = \sqrt{\frac{157}{16}} \approx 3.13 \text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\alpha = \frac{3\sqrt{3}/2}{7/4} = \frac{6\sqrt{3}}{7}\) | M1 | Correct use of components |
| \(\alpha \approx 68°\) | A1 | Accept \(67°\) or \(68°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e = \frac{v_B - v_A}{u_A - u_B}\) along line of centres \(= \frac{0-(-7/4)}{3/2-(-5/2)} = \frac{7/4}{4}\) | M1 | Correct use of NEL |
| \(e = \frac{7}{16}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Impulse on \(B\) \(= 4(0-(-5/2)) \times ... \) along line of centres \(= 4 \times \frac{7}{4} = 7\) Ns | M1 | Change in momentum of \(B\) |
| Impulse \(= 7\) Ns | A1 |
# Question 6:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Component of $A$ along line of centres before: $3\cos60° = \frac{3}{2}$ | B1 | |
| Component of $B$ along line of centres before: $-5\cos60° = -\frac{5}{2}$ | B1 | Must have correct sign/direction |
| Component of $A$ perpendicular to line of centres unchanged: $3\sin60° = \frac{3\sqrt{3}}{2}$ | B1 | Perpendicular components unchanged |
| Conservation of momentum along line of centres: $2(\frac{3}{2}) + 4(-\frac{5}{2}) = 2v_A + 4v_B$ | M1 | Correct momentum equation |
| $3 - 10 = 2v_A + 4v_B \Rightarrow 2v_A + 4v_B = -7$ | A1 | |
| $B$ moves perpendicular to line of centres, so $v_B = 0$ along line of centres | M1 | Using given condition |
| $v_A = -\frac{7}{4}$ along line of centres | A1 | |
| Speed of $A = \sqrt{\left(\frac{7}{4}\right)^2 + \left(\frac{3\sqrt{3}}{2}\right)^2} = \sqrt{\frac{49}{16}+\frac{27}{4}} = \sqrt{\frac{157}{16}} \approx 3.13 \text{ ms}^{-1}$ | A1 | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\alpha = \frac{3\sqrt{3}/2}{7/4} = \frac{6\sqrt{3}}{7}$ | M1 | Correct use of components |
| $\alpha \approx 68°$ | A1 | Accept $67°$ or $68°$ |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e = \frac{v_B - v_A}{u_A - u_B}$ along line of centres $= \frac{0-(-7/4)}{3/2-(-5/2)} = \frac{7/4}{4}$ | M1 | Correct use of NEL |
| $e = \frac{7}{16}$ | A1 | |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Impulse on $B$ $= 4(0-(-5/2)) \times ... $ along line of centres $= 4 \times \frac{7}{4} = 7$ Ns | M1 | Change in momentum of $B$ |
| Impulse $= 7$ Ns | A1 | |6 Two smooth spheres, $A$ and $B$, have equal radii and masses 2 kg and 4 kg respectively.
The spheres are moving on a smooth horizontal surface and collide. As they collide, $A$ has velocity $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $60 ^ { \circ }$ to the line of centres of the spheres, and $B$ has velocity $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $60 ^ { \circ }$ to the line of centres, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{79a08adc-ba78-4afb-96ef-ed595ad373d8-16_291_844_607_468}
Just after the collision, $B$ moves in a direction perpendicular to the line of centres.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $A$ immediately after the collision.
\item Find the acute angle, correct to the nearest degree, between the velocity of $A$ and the line of centres immediately after the collision.
\item Find the coefficient of restitution between the spheres.
\item Find the magnitude of the impulse exerted on $B$ during the collision.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2014 Q6 [12]}}