Pre-U Pre-U 9795/2 2014 June — Question 8 9 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2014
SessionJune
Marks9
TopicImpulse and momentum (advanced)
TypeOblique collision of spheres
DifficultyChallenging +1.8 This is an oblique collision problem requiring resolution of velocities along and perpendicular to the line of centres, application of conservation of momentum and Newton's restitution law in the impact direction, then recombination of components. While mechanically involved with multiple steps and careful bookkeeping of components, it follows a standard procedure taught in Further Maths mechanics without requiring novel insight or particularly subtle reasoning.
Spec1.10d Vector operations: addition and scalar multiplication6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts

\includegraphics{figure_8} A smooth sphere with centre \(A\) and of mass 2 kg, moving at 13 m s\(^{-1}\) on a smooth horizontal plane, strikes a smooth sphere with centre \(B\) and of mass 3 kg moving at 5 m s\(^{-1}\) on the same smooth horizontal plane. The spheres have equal radii. The directions of motion immediately before impact are at angles \(\tan^{-1}\left(\frac{2}{13}\right)\) to \(\overrightarrow{AB}\) and \(\tan^{-1}\left(\frac{4}{3}\right)\) to \(\overrightarrow{BA}\) respectively (see diagram). Given that the coefficient of restitution is \(\frac{2}{3}\), find the speeds of the spheres after impact. [9]

Question 8:
AnswerMarks Guidance
8–1
Components of speed to AB are 12 and 3 ms .
CLM: 2 × 12 – 3 × 3 = 2vA + 3vB
2
NEL: − (12+3)=v −v
3 A B
vA = –3, vB = 7
Components of velocity ⊥ to AB are 5 and 4 ms –1 .
Speed A = 32 +52 = 34 or 5.83 ms –1 ,
AnswerMarks
Speed B = 72 +42 = 65 or 8.06 ms –1 (OE)Any signs on RHS for A1
Consistent signs on RHS for
A1
(both √ on x-components)
Both
either
Both answers, correct to 3SF if
AnswerMarks
necessaryB1
M1A1√
M1A1√
A1
B1
M1
A1 [9]
Question 8:
8 | –1
Components of speed || to AB are 12 and 3 ms .
CLM: 2 × 12 – 3 × 3 = 2vA + 3vB
2
NEL: − (12+3)=v −v
3 A B
vA = –3, vB = 7
Components of velocity ⊥ to AB are 5 and 4 ms –1 .
Speed A = 32 +52 = 34 or 5.83 ms –1 ,
Speed B = 72 +42 = 65 or 8.06 ms –1 (OE) | Any signs on RHS for A1
Consistent signs on RHS for
A1
(both √ on x-components)
Both
either
Both answers, correct to 3SF if
necessary | B1
M1A1√
M1A1√
A1
B1
M1
A1 [9]
\includegraphics{figure_8}

A smooth sphere with centre $A$ and of mass 2 kg, moving at 13 m s$^{-1}$ on a smooth horizontal plane, strikes a smooth sphere with centre $B$ and of mass 3 kg moving at 5 m s$^{-1}$ on the same smooth horizontal plane. The spheres have equal radii. The directions of motion immediately before impact are at angles $\tan^{-1}\left(\frac{2}{13}\right)$ to $\overrightarrow{AB}$ and $\tan^{-1}\left(\frac{4}{3}\right)$ to $\overrightarrow{BA}$ respectively (see diagram). Given that the coefficient of restitution is $\frac{2}{3}$, find the speeds of the spheres after impact. [9]

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2014 Q8 [9]}}