OCR Further Mechanics 2023 June — Question 3 7 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2023
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeOblique collision, find velocities/angles
DifficultyChallenging +1.2 This is a 2D collision problem requiring decomposition of velocities, conservation of momentum along the line of centres, and Newton's restitution equation. While it involves multiple steps (decomposing initial velocity, applying two conservation principles, finding final velocity components, calculating angle), these are standard Further Mechanics techniques applied in a straightforward manner. The 7 marks reflect the working required rather than conceptual difficulty. Slightly above average due to being Further Maths content and requiring careful coordinate work, but follows a well-established procedure.
Spec6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts

Two smooth circular discs \(A\) and \(B\) are moving on a smooth horizontal plane when they collide. The mass of \(A\) is \(5\) kg and the mass of \(B\) is \(3\) kg. At the instant before they collide, • the velocity of \(A\) is \(4\) m s\(^{-1}\) at an angle of \(60°\) to the line of centres, • the velocity of \(B\) is \(6\) m s\(^{-1}\) along the line of centres (see diagram). \includegraphics{figure_3} The coefficient of restitution for collisions between the two discs is \(\frac{3}{4}\). Determine the angle that the velocity of \(A\) makes with the line of centres after the collision. [7]

Question 3:
AnswerMarks Guidance
3v = u = 4sin60
Ay AyB1 1.1
component of velocity of A
before collision, which is
AnswerMarks
unchanged by the collision= 23 = 3.46... may be seen on the
diagram
54cos60 + 3–6 = 5v + 3v
AnswerMarks Guidance
Ax BxM1 3.4
condone sin/cos need to be mv
for all terms
5v + 3v = –8
AnswerMarks Guidance
Ax BxA1 1.1
term equationCheck diagram for directions
3 v − v
= B x A− x
AnswerMarks Guidance
4 4 c o s 6 0  − 6M1 3.4
consistent
v −v =6 oe
AnswerMarks Guidance
Bx AxA1 1.1
term equation
v = –13/4 (= –3.25)
AnswerMarks Guidance
AxA1 1.1
Bx
2 3
t a n 1 4 6 .8  = − =  (1 dp)
AnswerMarks Guidance
3 .2 5A1 2.2a
+ 0.817 rad or -0.817 rad or 2.32
rad
[7]
Question 3:
3 | v = u = 4sin60
Ay Ay | B1 | 1.1 | Correct perpendicular
component of velocity of A
before collision, which is
unchanged by the collision | = 23 = 3.46... may be seen on the
diagram
54cos60 + 3–6 = 5v + 3v
Ax Bx | M1 | 3.4 | Conservation of momentum
condone sin/cos need to be mv
for all terms
5v + 3v = –8
Ax Bx | A1 | 1.1 | Correct rearrangement to a 3
term equation | Check diagram for directions
3 v − v
= B x A− x
4 4 c o s 6 0  − 6 | M1 | 3.4 | Restitution – needs to be
consistent
v −v =6 oe
Bx Ax | A1 | 1.1 | Correct rearrangement to a 3
term equation
v = –13/4 (= –3.25)
Ax | A1 | 1.1 | v = 11/4 (= 2.75)
Bx
2 3
t a n 1 4 6 .8  = − =  (1 dp)
3 .2 5 | A1 | 2.2a | Could be 133.2 or –46.8 or
+ 0.817 rad or -0.817 rad or 2.32
rad
[7]
Two smooth circular discs $A$ and $B$ are moving on a smooth horizontal plane when they collide. The mass of $A$ is $5$ kg and the mass of $B$ is $3$ kg.

At the instant before they collide,
• the velocity of $A$ is $4$ m s$^{-1}$ at an angle of $60°$ to the line of centres,
• the velocity of $B$ is $6$ m s$^{-1}$ along the line of centres

(see diagram).

\includegraphics{figure_3}

The coefficient of restitution for collisions between the two discs is $\frac{3}{4}$.

Determine the angle that the velocity of $A$ makes with the line of centres after the collision. [7]

\hfill \mbox{\textit{OCR Further Mechanics 2023 Q3 [7]}}