Challenging +1.2 This is a standard M3 oblique collision problem requiring resolution of velocities parallel and perpendicular to the line of centres, application of conservation of momentum and Newton's restitution law. While it involves multiple steps (resolving components, applying two equations, finding the resultant angle), the method is entirely routine for M3 students and follows a well-practiced algorithm with no novel insight required. The trigonometric setup (sin α = cos β = 0.6) is straightforward. This is moderately above average difficulty due to the multi-step nature and being from Further Maths M3, but remains a textbook application.
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\includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-4_369_953_269_596}
Two smooth uniform spheres \(A\) and \(B\), of equal radius, have masses 3 kg and 4 kg respectively. They are moving on a horizontal surface, each with speed \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), when they collide. The directions of motion of \(A\) and \(B\) make angles \(\alpha\) and \(\beta\) respectively with the line of centres of the spheres, where \(\sin \alpha = \cos \beta = 0.6\) (see diagram). The coefficient of restitution between the spheres is 0.75 . Find the angle that the velocity of \(A\) makes, immediately after impact, with the line of centres of the spheres. [0pt]
[10]
Initial i components of velocity for A and B are \(4\ \text{ms}^{-1}\) and \(3\ \text{ms}^{-1}\) respectively
B1
May be implied
\(3\times4 + 4\times3 = 3a + 4b\)
M1
For using p.c.mmtm. parallel to l.o.c.
A1
\(0.75(4-3) = b - a\)
M1
For using NEL
A1
M1
For attempting to find \(a\)
\(a = 3\)
A1
Depends on all three M marks
Final j component of velocity for A is \(3\ \text{ms}^{-1}\)
B1
May be implied
Angle with l.o.c. is \(45°\) or \(135°\)
A1ft
ft incorrect value of \(a\ (\neq 0)\) only
SR notes: \(3\times3 + 4\times4 = 3a + 4b\) and \(b - a = 0.75(3-4)\); M1 M1 as scheme and A1 for *both* equations; \(a = 4\) M1 as scheme A1; j component for A is \(4\ \text{ms}^{-1}\) B1; Angle \(\tan^{-1}(4/4) = 45°\) M1 as scheme A1
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Initial **i** components of velocity for A and B are $4\ \text{ms}^{-1}$ and $3\ \text{ms}^{-1}$ respectively | B1 | May be implied |
| $3\times4 + 4\times3 = 3a + 4b$ | M1 | For using p.c.mmtm. parallel to l.o.c. |
| | A1 | |
| $0.75(4-3) = b - a$ | M1 | For using NEL |
| | A1 | |
| | M1 | For attempting to find $a$ |
| $a = 3$ | A1 | Depends on all three M marks |
| Final **j** component of velocity for A is $3\ \text{ms}^{-1}$ | B1 | May be implied |
| Angle with l.o.c. is $45°$ or $135°$ | A1ft | ft incorrect value of $a\ (\neq 0)$ only |
**SR notes:** $3\times3 + 4\times4 = 3a + 4b$ and $b - a = 0.75(3-4)$; M1 M1 as scheme and A1 for *both* equations; $a = 4$ M1 as scheme A1; **j** component for A is $4\ \text{ms}^{-1}$ B1; Angle $\tan^{-1}(4/4) = 45°$ M1 as scheme A1
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\includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-4_369_953_269_596}
Two smooth uniform spheres $A$ and $B$, of equal radius, have masses 3 kg and 4 kg respectively. They are moving on a horizontal surface, each with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, when they collide. The directions of motion of $A$ and $B$ make angles $\alpha$ and $\beta$ respectively with the line of centres of the spheres, where $\sin \alpha = \cos \beta = 0.6$ (see diagram). The coefficient of restitution between the spheres is 0.75 . Find the angle that the velocity of $A$ makes, immediately after impact, with the line of centres of the spheres.\\[0pt]
[10]
\hfill \mbox{\textit{OCR M3 2009 Q5 [10]}}