Question 12 - A-Level Maths

OCR MEI Further Mechanics Major 2023 June — Question 12 13 marks

Exam Board	OCR MEI
Module	Further Mechanics Major (Further Mechanics Major)
Year	2023
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Impulse and momentum (advanced)
Type	Oblique collision of spheres
Difficulty	Challenging +1.2 This is a standard oblique collision problem requiring impulse-momentum principles and Newton's experimental law. Part (a) tests conceptual understanding and routine application of collision formulas. Part (b) involves algebraic manipulation of the energy loss equation to find constraints on λ, which requires more careful work but follows established methods. The problem is moderately challenging for Further Maths students but doesn't require novel insights—it's a well-structured textbook-style question testing mastery of collision mechanics.
Spec	6.02i Conservation of energy: mechanical energy principle 6.03b Conservation of momentum: 1D two particles 6.03d Conservation in 2D: vector momentum 6.03i Coefficient of restitution: e 6.03k Newton's experimental law: direct impact

Two small uniform smooth spheres A and B are of equal radius and have masses \(m\) and \(\lambda m\) respectively. The spheres are on a smooth horizontal surface. Sphere A is moving on the surface with velocity \(u_1 \mathbf{i} + u_2 \mathbf{j}\) towards B, which is at rest. The spheres collide obliquely. When the spheres collide, the line joining their centres is parallel to \(\mathbf{i}\). The coefficient of restitution between A and B is \(e\).

1. Explain why, when the spheres collide, the impulse of A on B is in the direction of \(\mathbf{i}\). [1]
2. Determine this impulse in terms of \(\lambda\), \(m\), \(e\) and \(u_1\). [6]

The loss in kinetic energy due to the collision between A and B is \(\frac{1}{8}mu_1^2\).

Determine the range of possible values of \(\lambda\). [6]

Show mark scheme Show mark scheme source

Question 12:

Answer	Marks	Guidance
12	(a)	(i)
parallel to the line of centres (which is in the i direction)	B1	2.4

[1]

Answer	Marks	Guidance
(a)	(ii)	M1*

the line of centres) – correct number of terms

(condone no masses present for this mark) but

condone sign errors

Answer	Marks	Guidance
M1*	3.3	Use of Newton’s experimental law (parallel to the

line of centres) – correct number of terms

m u m v m v  = +

1 A B

v − v = − e u

Answer	Marks	Guidance
A B 1	A1	1.1

e.g. m u m v m v  = − + and v +v =eu

1 A B A B 1

(v and v are the horizontal components of the

A B

velocities of A and B after impact parallel to the line

of centres)

Solving simultaneously to find either v or v in terms u , e

A B 1

Answer	Marks	Guidance
and 	M1dep*	1.1

For reference: v 11 e u 1 = − =

A 1 1   + +

u (1 e ) +

v 11 =

B  +

Answer	Marks	Guidance
M1	3.3	Using either I=mwi or I=m(v −u ) i

A 1

(condone lack of i) to get I in terms of ,m,e and u

1 only. Dependent on all previous M marks. Must be

using the correct mass

1 e  + 

Impulse = m u  i

Answer	Marks	Guidance
1 1  +	A1	2.5

Allow unsimplified e.g. m 1 u − − i -

1 1  +

condone lack of i

[6]

Answer	Marks	Guidance
(b)	12 m u 2 ( + 12 m u 2 )
1 2	B1	1.1

i direction (so B1 for 12 m u 2 )

1 u 2 ( 1 e ) 2 1 e 2   −   + 

12 m 1 12 ( m ) u 2  + ( + 12 m u 2 )

Answer	Marks	Guidance
( 1 ) 2 1 1   + + 2	B1
B1	1.1	KE after collision – B1 for each correct term (need

only consider the energy in the i direction)

u (1 e ) u (1 e )  + −

For reference: v 11 e u 1 = − =

A 1 1   + +

u (1 e ) +

v 11 =

B  +

1 1 1 e 2 1 1 e 2 1   −   + 

 − − =

Answer	Marks	Guidance
2 2 1 2 1 8   + +	M1*	1.1

Using KE before – KE after =  m u 2 to obtain an

1

8 equation in and e only

3 2 2 1   + −

For reference only : e 2 = and

4 ( 1 )  +

2 ( 4 e 2 3 ) ( 4 e 2 2 ) 1 0   − + − + =

1−e2  1 3 1  − 3 1  −

 =  e 2 = 0 1 

 

Answer	Marks	Guidance
1+  4 4  4 	M1dep*	3.1b

equation containing terms in  (possibly 2) and e 2

(not e) only and considering either e 2 0 or e 2 1

(condone ‘equals’ or ‘strict’ comparison). Or

obtaining an equation in (possibly 2  ) and e 2 (not

e) only and setting e = 1 or e = 0

1 

Answer	Marks	Guidance
3	A1	2.2a

of  stated

[6]

Question 12:
12 | (a) | (i) | Spheres are smooth and so the impulse acts in a direction
parallel to the line of centres (which is in the i direction) | B1 | 2.4
[1]
(a) | (ii) | M1* | 3.3 | Use of conservation of linear momentum (parallel to
the line of centres) – correct number of terms
(condone no masses present for this mark) but
condone sign errors
M1* | 3.3 | Use of Newton’s experimental law (parallel to the
line of centres) – correct number of terms
m u m v m v  = +
1 A B
v − v = − e u
A B 1 | A1 | 1.1 | Use of NEL must be consistent with CLM
e.g. m u m v m v  = − + and v +v =eu
1 A B A B 1
(v and v are the horizontal components of the
A B
velocities of A and B after impact parallel to the line
of centres)
Solving simultaneously to find either v or v in terms u , e
A B 1
and  | M1dep* | 1.1 | u (1 e ) u (1 e )  + −
For reference: v 11 e u 1 = − =
A 1 1   + +
u (1 e ) +
v 11 =
B  +
M1 | 3.3 | Using either I=mwi or I=m(v −u ) i
A 1
(condone lack of i) to get I in terms of ,m,e and u
1
only. Dependent on all previous M marks. Must be
using the correct mass
1 e  + 
Impulse = m u  i
1 1  + | A1 | 2.5 | u (1 e )   − 
Allow unsimplified e.g. m 1 u − − i -
1 1  +
condone lack of i
[6]
(b) | 12 m u 2 ( + 12 m u 2 )
1 2 | B1 | 1.1 | KE before collision - or just considering energy in the
i direction (so B1 for 12 m u 2 )
1
u 2 ( 1 e ) 2 1 e 2   −   + 
12 m 1 12 ( m ) u 2  + ( + 12 m u 2 )
( 1 ) 2 1 1   + + 2 | B1
B1 | 1.1 | KE after collision – B1 for each correct term (need
only consider the energy in the i direction)
u (1 e ) u (1 e )  + −
For reference: v 11 e u 1 = − =
A 1 1   + +
u (1 e ) +
v 11 =
B  +
1 1 1 e 2 1 1 e 2 1   −   + 
 − − =
2 2 1 2 1 8   + + | M1* | 1.1 | 1
Using KE before – KE after =  m u 2 to obtain an
1
8
equation in and e only
3 2 2 1   + −
For reference only : e 2 = and
4 ( 1 )  +
2 ( 4 e 2 3 ) ( 4 e 2 2 ) 1 0   − + − + =
1−e2  1 3 1  − 3 1  −
 =  e 2 = 0 1 
 
1+  4 4  4  | M1dep* | 3.1b | Re-arranging to make e2(or e) the subject from an
equation containing terms in  (possibly 2) and e 2
(not e) only and considering either e 2 0 or e 2 1
(condone ‘equals’ or ‘strict’ comparison). Or
obtaining an equation in (possibly 2  ) and e 2 (not
e) only and setting e = 1 or e = 0
1

3 | A1 | 2.2a | A0 if strict inequality or an upper limit for the value
of  stated
[6]

Show LaTeX source

Two small uniform smooth spheres A and B are of equal radius and have masses $m$ and $\lambda m$ respectively. The spheres are on a smooth horizontal surface.

Sphere A is moving on the surface with velocity $u_1 \mathbf{i} + u_2 \mathbf{j}$ towards B, which is at rest. The spheres collide obliquely. When the spheres collide, the line joining their centres is parallel to $\mathbf{i}$.

The coefficient of restitution between A and B is $e$.

\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Explain why, when the spheres collide, the impulse of A on B is in the direction of $\mathbf{i}$. [1]
\item Determine this impulse in terms of $\lambda$, $m$, $e$ and $u_1$. [6]
\end{enumerate}
\end{enumerate}

The loss in kinetic energy due to the collision between A and B is $\frac{1}{8}mu_1^2$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Determine the range of possible values of $\lambda$. [6]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2023 Q12 [13]}}

This paper (13 questions)

View full paper

Q1 4 Q2 4 Q3 5 Q4 6 Q5 7 Q6 9 Q7 9 Q8 8 Q9 12 Q10 16 Q11 12 Q12 13 Q13 15