M2 COLLISIONS PAST QUESTIONS
1.
A small ball A of mass 3m is moving with speed u in a straight line on a smooth horizontal
table. The ball collides directly with another small ball B of mass m moving with speed u
1
towards A along the same straight line. The coefficient of restitution between A and B is .
2
The balls have the same radius and can be modelled as particles.
(a)
Find
(i)
the speed of A immediately after the collision,
(ii)
the speed of B immediately after the collision.
M
After the collision B hits a smooth vertical wall which is perpendicular to the direction of
2
motion of B. The coefficient of restitution between B and the wall is .
5
(b)
Find the speed of B immediately after hitting the wall.
(2)
The first collision between A and B occurred at a distance 4a from the wall. The balls collide
again T seconds after the first collision.
(c)
Show that T =
112a
.
15u
(6)
(Total 15 marks)
2.
Two particles, P, of mass 2m, and Q, of mass m, are moving along the same straight line on a
smooth horizontal plane. They are moving in opposite directions towards each other and collide.
Immediately before the collision the speed of P is 2u and the speed of Q is u. The coefficient of
restitution between the particles is e, where e < 1. Find, in terms of u and e,
(i)
the speed of P immediately after the collision,
(ii)
the speed of Q immediately after the collision.
(Total 7 marks)
3.
Particles A, B and C of masses 4m, 3m and m respectively, lie at rest in a straight line on a
smooth horizontal plane with B between A and C. Particles A and B are projected towards each
other with speeds u m s–1 and v m s–1 respectively, and collide directly.
As a result of the collision, A is brought to rest and B rebounds with speed kv m s–1. The
3
coefficient of restitution between A and B is .
4
City of London Academy
1
(a)
Show that u = 3v.
(6)
(b)
Find the value of k.
(2)
Immediately after the collision between A and B, particle C is projected with speed 2v m s–1
towards B so that B and C collide directly.
(c)
Show that there is no further collision between A and B.
(4)
(Total 12 marks)
4.
A particle P of mass 3m is moving in a straight line with speed 2u on a smooth horizontal table.
It collides directly with another particle Q of mass 2m which is moving with speed u in the
opposite direction to P. The coefficient of restitution between P and Q is e.
(a)
Show that the speed of Q immediately after the collision is
1
5
(9e + 4)u.
(5)
The speed of P immediately after the collision is
(b)
Show that e =
1
4
1
2
u.
.
(4)
The collision between P and Q takes place at the point A. After the collision Q hits a smooth
fixed vertical wall which is at right-angles to the direction of motion of Q. The distance from A
to the wall is d.
(c)
Show that P is a distance
3
d from the wall at the instant when Q hits the wall.
5
(4)
Particle Q rebounds from the wall and moves so as to collide directly with particle P at the point
1
B. Given that the coefficient of restitution between Q and the wall is ,
5
(d)
find, in terms of d, the distance of the point B from the wall.
(4)
(Total 17 marks)
5.
A particle A of mass 4m is moving with speed 3u in a straight line on a smooth horizontal table.
The particle A collides directly with a particle B of mass 3m moving with speed 2u in the same
direction as A. The coefficient of restitution between A and B is e. Immediately after the
collision the speed of B is 4eu.
City of London Academy
2
(a)
Show that e 
3
.
4
(5)
(b)
Find the total kinetic energy lost in the collision.
(4)
(Total 9 marks)
6.
Two small smooth spheres A and B have equal radii. The mass of A is 2m kg and the mass of B
is m kg. The spheres are moving on a smooth horizontal plane and they collide. Immediately
before the collision the velocity of A is (2i – 2j) m s–1 and the velocity of B is (–3i – j) m s–1.
Immediately after the collision the velocity of A is (i – 3j) m s–1. Find the speed of B
immediately after the collision.
(Total 5 marks)
7.
2
A small smooth ball B, moving on a horizontal plane, collides with a fixed vertical wall.
Immediately before the collision the angle between the direction of motion of B and the wall is
2θ where 0o < θ < 45°. Immediately after the collision the angle between the direction of motion
of B and the wall is θ as shown in the diagram above. Given that the coefficient of restitution
3
between B and the wall is , find the value of tanθ.
8
(Total 8 marks)
8.
A particle P of mass 2m is moving with speed 2u in a straight line on a smooth horizontal plane.
A particle Q of mass 3m is moving with speed u in the same direction as P. The particles collide
1
directly. The coefficient of restitution between P and Q is .
2
(a)
Show that the speed of Q immediately after the collision is
8
u.
3
(5)
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3
(b)
Find the total kinetic energy lost in the collision.
(5)
After the collision between P and Q, the particle Q collides directly with a particle R of mass m
which is at rest on the plane. The coefficient of restitution between Q and R is e.
(c)
Calculate the range of values of e for which there will be a second collision between P
and Q.
(7)
(Total 17 marks)
9.
Two small spheres P and Q of equal radius have masses m and 5m respectively. They lie on a
smooth horizontal table. Sphere P is moving with speed u when it collides directly with sphere
1
Q which is at rest. The coefficient of restitution between the spheres is e, where e > .
5
(a)
u
5e  1.
6
(i)
Show that the speed of P immediately after the collision is
(ii)
Find an expression for the speed of Q immediately after the collision, giving your
answer in the form λu, where λ is in terms of e.
(6)
Three small spheres A, B and C of equal radius lie at rest in a straight line on a smooth
horizontal table, with B between A and C. The spheres A and C each have mass 5m, and the
mass of B is m. Sphere B is projected towards C with speed u. The coefficient of restitution
4
between each pair of spheres is .
5
(b)
Show that, after B and C have collided, there is a collision between B and A.
(3)
(c)
Determine whether, after B and A have collided, there is a further collision between B and
C.
(4)
(Total 13 marks)
10.
A particle P of mass m is moving in a straight line on a smooth horizontal table. Another
particle Q of mass km is at rest on the table. The particle P collides directly with Q. The
direction of motion of P is reversed by the collision. After the collision, the speed of P is v and
1
the speed of Q is 3v. The coefficient of restitution between P and Q is .
2
(a)
Find, in terms of v only, the speed of P before the collision.
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4
(3)
(b)
Find the value of k.
(3)
After being struck by P, the particle Q collides directly with a particle R of mass 11m which is
at rest on the table. After this second collision, Q and R have the same speed and are moving in
opposite directions. Show that
(c)
the coefficient of restitution between Q and R is
3
,
4
(4)
(d)
there will be a further collision between P and Q.
(2)
(Total 12 marks)
11.
Two particles A and B move on a smooth horizontal table. The mass of A is m, and the mass of
B is 4m. Initially A is moving with speed u when it collides directly with B, which is at rest on
the table. As a result of the collision, the direction of motion of A is reversed. The coefficient of
restitution between the particles is e.
(a)
Find expressions for the speed of A and the speed of B immediately after the collision.
(7)
In the subsequent motion, B strikes a smooth vertical wall and rebounds. The wall is
perpendicular to the direction of motion of B. The coefficient of restitution between B and the
wall is 54 . Given that there is a second collision between A and B,
(b)
show that
1
9
e
.
4
16
(5)
Given that e 
(c)
1
,
2
find the total kinetic energy lost in the first collision between A and B.
(3)
(Total 15 marks)
12.
A particle A of mass 2m is moving with speed 3u in a straight line on a smooth horizontal table.
The particle collides directly with a particle B of mass m moving with speed 2u in the opposite
direction to A. Immediately after the collision the speed of B is 83 u and the direction of motion
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5
of B is reversed.
(a)
Calculate the coefficient of restitution between A and B.
(6)
(b)
Show that the kinetic energy lost in the collision is 7mu2.
(3)
After the collision B strikes a fixed vertical wall that is perpendicular to the direction of motion
14
of B. The magnitude of the impulse of the wall on B is 3 mu.
(c)
Calculate the coefficient of restitution between B and the wall.
(4)
(Total 13 marks)
13.
Two small spheres A and B have mass 3m and 2m respectively. They are moving towards each
other in opposite directions on a smooth horizontal plane, both with speed 2u, when they collide
directly. As a result of the collision, the direction of motion of B is reversed and its speed is
unchanged.
(a)
Find the coefficient of restitution between the spheres.
(7)
Subsequently, B collides directly with another small sphere C of mass 5m which is at rest. The
coefficient of restitution between B and C is 53 .
(b)
Show that, after B collides with C, there will be no further collisions between the spheres.
(7)
(Total 14 marks)
14.
A particle P of mass 3m is moving with speed 2u in a straight line on a smooth horizontal table.
The particle P collides with a particle Q of mass 2m moving with speed u in the opposite
direction to P. The coefficient of restitution between P and Q is e.
(a)
Show that the speed of Q after the collision is
1
5
u(9e + 4).
(5)
As a result of the collision, the direction of motion of P is reversed.
(b)
Find the range of possible values of e.
(5)
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6
Given that the magnitude of the impulse of P on Q is
(c)
32
5
mu,
find the value of e.
(4)
(Total 14 marks)
15.
[In this question i and j are perpendicular unit vectors in a horizontal plane.]
A ball has mass 0.2 kg. It is moving with velocity (30i) m s–1 when it is struck by a bat. The bat
exerts an impulse of (–4i + 4j) Ns on the ball.
Find
(a)
the velocity of the ball immediately after the impact,
(3)
(b)
the angle through which the ball is deflected as a result of the impact,
(2)
(c)
the kinetic energy lost by the ball in the impact.
(4)
(Total 9 marks)
16.
Two small smooth spheres, P and Q, of equal radius, have masses 2m and 3m respectively. The
sphere P is moving with speed 5u on a smooth horizontal table when it collides directly with Q,
which is at rest on the table. The coefficient of restitution between P and Q is e.
(a)
Show that the speed of Q immediately after the collision is 2(1 + e)u.
(5)
After the collision, Q hits a smooth vertical wall which is at the edge of the table and
perpendicular to the direction of motion of Q. The coefficient of restitution between Q and the
wall is f, 0 < f  1.
(b)
Show that, when e = 0.4, there is a second collision between P and Q.
(3)
Given that e = 0.8 and there is a second collision between P and Q,
(c)
find the range of possible values of f.
(3)
(Total 11 marks)
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7
17.
A smooth sphere A of mass m is moving with speed u on a smooth horizontal table when it
collides directly with another smooth sphere B of mass 3m, which is at rest on the table. The
coefficient of restitution between A and B is e. The spheres have the same radius and are
modelled as particles.
(a)
Show that the speed of B immediately after the collision is
1
4
(1 + e)u.
(5)
(b)
Find the speed of A immediately after the collision.
(2)
Immediately after the collision the total kinetic energy of the spheres is
(c)
1
6
mu2.
Find the value of e.
(6)
(d)
Hence show that A is at rest after the collision.
(1)
(Total 14 marks)
18.
A uniform sphere A of mass m is moving with speed u on a smooth horizontal table when it
collides directly with another uniform sphere B of mass 2m which is at rest on the table. The
spheres are of equal radius and the coefficient of restitution between them is e. The direction of
motion of A is unchanged by the collision.
(a)
Find the speeds of A and B immediately after the collision.
(7)
(b)
Find the range of possible values of e.
(2)
After being struck by A, the sphere B collides directly with another sphere C, of mass 4m and
of the same size as B. The sphere C is at rest on the table immediately before being struck by B.
The coefficient of restitution between B and C is also e.
(c)
Show that, after B has struck C, there will be a further collision between A and B.
(6)
(Total 15 marks)
19.
A smooth sphere P of mass 2m is moving in a straight line with speed u on a smooth horizontal
table. Another smooth sphere Q of mass m is at rest on the table. The sphere P collides directly
with Q. The coefficient of restitution between P and Q is 13 . The spheres are modelled as
particles.
(a)
Show that, immediately after the collision, the speeds of P and Q are
City of London Academy
5
9
u and
8
9
u
8
respectively.
(7)
After the collision, Q strikes a fixed vertical wall which is perpendicular to the direction of
motion of P and Q. The coefficient of restitution between Q and the wall is e. When P and Q
collide again, P is brought to rest.
(b)
Find the value of e.
(7)
(c)
Explain why there must be a third collision between P and Q.
(1)
(Total 15 marks)
20.
A smooth sphere is moving with speed U in a straight line on a smooth horizontal plane. It
strikes a fixed smooth vertical wall at right angles. The coefficient of restitution between the
sphere and the wall is 12 .
Find the fraction of the kinetic energy of the sphere that is lost as a result of the impact.
(Total 5 marks)
21.
A smooth sphere S of mass m is moving with speed u on a smooth horizontal plane. The sphere
S collides with another smooth sphere T, of equal radius to S but of mass km, moving in the
same straight line and in the same direction with speed u, 0 <  < 12 . The coefficient of
restitution between S and T is e.
Given that S is brought to rest by the impact,
(a)
show that e 
1  k
.
k (1   )
(6)
(b)
Deduce that k > 1.
(3)
(Total 9 marks)
1.
(a)
(i)
Con. of Mom: 3mu  mu  3mv  mw
2u  3v  w
N.L.R:
1
2
u  u   w  v
u  wv
City of London Academy
(1)
M1# A1
M1# A1
(2)
9
(ii)
(1) – (2)
u  4v
v  14 u
In (2)
u  w  14 u
w  54 u
(b)
B to wall: N.L.R:
5
4
u  52  V
V  12 u
DM1#
A1
A17
M1
A1ft2
(c)
5
16a
time  4a  u 
4
5u
1 16a 4
Dist. Travelled by A  u 
 a
4
5u 5
1
1
In t secs, A travels ut , B travels ut
4
2
B to wall:
Collide when speed of approach 
4a 
B1ft
B1ft
1
1
ut  ut , distance to cover = M1$
2
4
4
a
5
4a  54 a 16a 4 64a

 
3
u
5
3u 15u
4
16a 64a 112a
Total time 


5u 15u 15u
*
t
DM1$ A1
A16
[15]
2u
u



2.
2m
m


v1
v2
CLM: 4mu – mu = 2mv1 + mv2
M1 A1
i.e. 3u = 2v1 + v2
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10
NIL:
3eu = –v1 + v2
M1 A1
v1 = u(1 – e)
DM1 A1
v2 = u(1 + 2e)
A1
[7]
3.
(a)
Conservation of momentum:
4mu – 3mv = 3mkv
M1A1
Impact law:
kv 
3
u  v 
4
M1A1
Eliminate k:
4mu – 3mv = 3m ×
3
u  v 
4
DM1
u = 3v (Answer given)
A16
3
3v  v , k  3
4
(b)
kv 
(c)
Impact law: (kv + 2v) e = vC – vB (5ve = vC – vB)
B1
Conservation of momentum : 3× kv –1 × 2v =
3vB + vC (7v = 3vB + vc)
B1
M1,A12
v
7 – 5e  0 hence no
4
further collision with A.
Eliminate vC : v B 
M1 A14
[12]
4.
(a)
Correct use of NEL
y – x = e(2u + u) o.e.
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M1 *
A1
11
CLM (→): 3m(2u) + 2m(–u) = 3m(x) + 2m(y) (  4u = 3x + 2y)
Hence x = y – 3eu, 4u = 3(y – 3eu) + 2y, (u(9e + 4) = 5y)
Hence, speed of Q = 15 (9e  4)u
AG
(b)
1
x = y – 3eu  (9e  4)u – 3eu
5
M1 *
1
2u
Hence, speed P  (4 – 6e)u  (2 – 3e) o.e.
5
5
A1
1
2u
x  u  (2 – 3e)  5u  8u – 12eu, 12e  3
2
5
gives, e =
3

12
e  41
B1
d * M1
A1 cso5
& solve for e
AG
d * M1
A1
Or
Using NEL correctly with given speeds of P and Q
3eu = 15 (9e  4)u – 12 u
3eu =
9
5
eu  54 u – 12 u ,
6
5
(c)
3e  95 e  54 – 12
M1 *
A1
& solve for e
d * M1
3
e  10
 e  15
 e  14 .
60
A14
Time taken by Q from A to the wall =
Distance moved by P in this time =
d  4d 
 
y  5u 
M1+
u d u  4d  2
 (    d )
2 y 2  5u  5
d 
Distance of P from wall = d –   ;  d – 52 d  53 d
 y
A1
d+M1;
AG
A1 cso
or
Ratio speed P:speed Q = x:y =
1 1 9
1 5
u : (  4)u  u : u  2 : 5
2 5 4
2 4
So if Q moves a distance d, P will move a distance
Distance of P from wall = d –
(d)
2
5
d ;  53 d
2
5
AG
After collision with wall, speed Q  15 y  15 54u   14 u
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d
M1+
A1
d+M1;A14
cso
their y
B1ft
12
Time for P, T AB 
3d
–x
5
1
u
2
Hence T AB  TWB 
, Time for Q, TWB 
3d
–x
5
1
u
2

x
u
from their y
1
4
B1ft
x
u
M1
1
4
gives, 2( 35d – x)  4 x  35d – x  2 x, 3x  3d
 x  15 d
5
A1 cao
or
After collision with wall, speed Q  15 y  15 54u   14 u
their y
B1ft
speed P  x  12 u, speed P: new speed Q  12 u : 14 u  2 : 1
from their y
Distance of B from wall =
1 3d d
 ;
3 5
5
their
B1 ft
1
2 1
M1; A14
2nd or
After collision with wall, speed Q  15 y  15 54u   14 u
their y
B1ft
from their y
B1ft
Combined speed of P and Q  12 u  14 u  34 u
Time from wall to 2nd collision =
3d
5
3u
4
 35d  34u  45ud
Distance of B from
wall = (their speed)x(their time) =
u 4d

;  15 d
4 5u
M1; A14
[17]
5.
(a)
3u
2u
4m
3m
x
y = 4eu
LM
12mu + 6mu = 4mx + 12meu
NEL
4eu – x = eu
Eliminating x to obtain equation in e
3
Leading to e = (*)
4
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B1
M1A1
DM1
cso
A1
5
13
(b)
x = 3eu or
9
u or 4.5u – 3eu
4
seen or implied in (b)
B1
2
1
1
1
1
9 
4m(3u ) 2  3m(2u ) 2  4m u   3m(3u ) 2
2
2
2 4 
2
ft their x
5
3
 24mu 2  23 mu 2 mu 2 = 0.375mu2
8
8
Loss in KE =
M1A1ft
A1
4
[9]
6.
2m(2i – 2j) + m(–3i – j) = 2m(i – 3j) + mv
(i – 5j) = (2i – 6j) + v
(–i + j) = v
v  (1) 2  12  2 m s–1
M1A1
M1A1
A1
cwo
DM1A1
[5]
7.
u cos 2 = v cos 
3
u sin 2 = v sin 
8
3 tan 2 = 8 tan 
6 tan
 8 tan
1  tan 2 
1
tan2  =
(tan   0)
4
1
tan  =
2
M1 A1
M1 A1
M1
M1
M1 A1 8
[8]
8.
(a)
2u
u
2m
3m
x
LM
y
4mu + 3mu = 2mx + 3my
1
NEL y  x  u
2
8
Solving to y = u *
5
(b)
x=
11
u or equivalent
10
City of London Academy
M1 A1
B1
cso
M1 A1
5
B1
14
Energy loss

2
2


1
 11   1
8  
 2m (2u ) 2   u     3m u 2   u  


2
 10   2
 5  


M1 A(2,1,0)
9
mu 2
20
A1
5
(c)
8
5u
3m
m
s
t
24
LM
mu = 3ms + mt
5
8
NEL t  s  eu
5
2
Solving to s = u (3  e)
3
11
2
u  u (3  e)
For a further collision
10
5
1
e>
4
M1 A1
B1
M1 A1
M1
ignore e ≤ 1
[17]
9.
u
5m
m
V
(a)
(b)
W
CLM: mv + 5mw = mu
NLI: w – v = eu
1
1
Solve v: v = (1 – 5e)u, so speed = (5e – 1)u
6
6
(NB – answer given on paper)
1
Solve w: w = (1 + e)u
6
* The M’s are dependent on having equations (not necessarily
correct) for CLM and NLI
1
4
(1  5. )u = –½u
6
3
velocity < 0  change of direction  B hits A
After B hits C, velocity of B = “v” =
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B1
B1
M1*A1
M1*A16
M1A1
A1CSO3
15
(c)
velocity of C after =
3
u
10
B1
When B hits A, “u” = ½u, so velocity of B after = –½(–½u) =
Travelling in the same direction but
1
u
4
1 3
  no second collision
4 10
B1
Conservation of momentum – signs consistent with their
diagram/between the two equations
B1
Impact equation
M1
Attempt to eliminate w
A1
correct expression for v. Q asks for speed so final answer must be
verified positive with reference to e > 1/5.
B1
M1A1cso4
Answer given so watch out for fudges.
M1
Attempt to eliminate v
A1
correct expression for w
M1
Substitute for e in speed or velocity of P to obtain v in terms of u.
Alternatively, can obtain v in terms of w.
A1
5w 

(+/–)u/2  v  

3 

A1
CSO Justify direction (and correct conclusion)
B1
speed of C = value of w = ()
3u
(Must be referred to in (c)
10
to score the B1.)
B1
1
5
speed of B after second collision () u or () w
4
6
M1
Comparing their speed of B after 2nd collision with their speed of C
after first collision.
A1
CSO. Correct conclusion.
[13]
10.
(a)
City of London Academy
16
u
km
m
v
3v
NEL 3v – (–V) = eu
u = 8v
(b)
M1A1
A1
LM 8mv = –mv + 3kmv
(m × (u) = –mv + 3kmv)
k=3
ft their u
3
M1A1ft
A13
(c)
3y
km
11m
LM 9mv = –3my + 11my
NEL 2y = e × 3v
9
3
y= ve=
*
8
4
(d)
y=
ft their k
cso
9
v > v  further collision between P and Q
8
M1A1ft
M1
A14
M1A12
A1 is cso – watch out for incorrect statements re. velocity
[12]
11.
(a)
u
m
4m
v
(b)
w
mu = 4mw – mv
M1 A1
eu = w + v
M1 A1
1 e
4e 1
w(
)u, v  (
)u
5
5
Indep M1 A1 A1
w  (
4  4e
)u
25
B1 f.t.
Second collision  w > v
City of London Academy
17

4  4e 4e  1

25
5
M1
Also v > 0  e > 1 / 4
(c)
Hence result
B1
5
KE lost = ½ mu2  [½ 4m{(u / 5)(1 + e)}2 + ½ m{(u/5)(4e  1)}2]

M1A1f.t.
3
mu2
10
A1 cao 3
[15]
12.
(a)
3u
2u
2m
m
x
LM
8u/3
6mu – 2mu = 2mx + 8 mu
3
M1 A1
 x  2 u 
3 

NEL
(b)
8 u – x = 5ue
3
M1 A1
Solving to e = 2
5
M1 A1
Initial K.E. = 1 × 2m(3u)2 + 1 × m (2u)2 = 11 mu2
2
2
2
2
Final K.E. = 1 × 2m  2 u  + 1 × m  8 u  = 4mu2 both
2
2
3 
3 
Change in K.E. = 7mu2
(c)
6
m  8 u  v   14 mu
3
 3
M1
M1 Subtracting and simplifying
M1 A1
2
to kmu
A1cso
3
M1 A1
(v = 2u)
e= 2 3
8
4
3
M1 A1
4
[13]
City of London Academy
18
13.
2u
(a)
2u
3m
2m
v
2u
CLM: 6mu – 4mu = 3mv + 4mu
2
v=– u
3
NLI: 2u – v = e.4u
8
2
 4eu = u  e = .
3
3
2u
(b)
M1 A1
A1
M1 A1
M1 A1 7
0
2m
5m
x
y
5my + 2mx = 4mu
3
6
y – x = .2u = u
5
5
2
Solve: x = – u
7
2
2
u < u so B does not overtake A
7
3
So no more collisions
M1 A1
A1
M1 A1
M1
A1cso 7
[14]
2u
u
3m
2m
14.
x
(a)
y
LM 6mu – 2mu = 3mx + 2my
y – x = 3eu
NEL
(b)
M1 A1
B1
Solving to y =
1
u(9e + 4) (*) cso
5
M1 A1 5
Solving to x =
2
u(2 – 3e)
5
M1 A1
oe
x<0e>
City of London Academy
2
3
M1 A1
19
2
<e1
3
A1ft
5
ft their e for glb
(c)
2m[
1
32
u(9e + 4) + u] =
mu
5
5
M1 A1
7
9
awrt 0.78
Solving to e =
M1 A1 4
[14]
15.
I = mv – mu
–4i + 4j = 0.2v – 0.2 × 30i
v = 10i + 20j (m s–1)
(a)
20
10
θ = 63.4°
tan θ =
(b)
M1 A1
A1
3
M1
A1
2
accept awrt 63° ot 1.1°
1
× 0.2 × (102 + 202) (= 50) ft their v
2
1
1
K.E. lost =
× 0.2 × 302 –
× 0.2 × (102 + 202)
2
2
= 40 (J) cao
(c)
Final K.E. =
M1 A1ft
M1
A1
4
[9]
5u
16.
(a)
(b)
2m
3m
x
y
LM 10mu = 2mx + 3my
NEL y – x = 5eu
M1 A1
B1
Solving to y = 2(1 + e)u (*)
cso M1 A1
x = 2u – 3eu finding x, with or without e = 0.4
M1
City of London Academy
20
(c)
x = 0.8u
A1
x > 0  P moves towards wall and Q rebounds from wall
 second collision
ft any positive x
A1 ft
x = – 0.4u
Speed of Q on rebound is 3.6fu
For second collision 3.6 fu > 0.4u
1
f>
ignore f  1
9
3
B1
M1
A1
3
[11]
17.
(a)
u
m
v1
0
3m
v2
v2 =
(b)
(c)
(d)
CLM: mu = mv1 + 3 mv2
NIL: eu =  v1 + v2
solving,
u
(1 + e)*
4
B1
M1 A1
dep. M1
A1
5
Solving for v1; u (1  3e)
4
M1 A1 2
1
1
1
u2
u2
m
(1  3e)2 +
3m
(1 + e)2 =
mu2
6
2
2
16
16
1
e2 =
9
1
e=
3
v1 =
M1 A1 f.t. A1
dep. M1 A1
A1
1
u
(1  3  ) = 0  at rest.
3
4
A1 c.s.o.
[14]
18.
(a)
u
o
A m e 2m B
v1
v2
mu = mv1 + 2mv2
eu = v1 + v2
v1 =
u
u
(1 – 2e); v2 = (1 + e)
3
3
City of London Academy
6
M1 A1
M1 A1
M1 A1 A1
7
21
u
(1 – 2e) > 0  e <
3
(b)
v1 > 0 
(c)
o
v2
B 2m 4m e
1
2
M1 A1 c.s.o.
v3
v4
2mv2 = 2mv3 + 4mv4
ev2 = v3 + v4
v3 =
M1
v2
u
(1 – 2e) = (1 – 2e)(1 + e)
3
9
M1 A1
Further collision if
v1 > v3
u
u
i.e. if
(1 – 2e) > (1 – 2e)(1 + e)
3
9
i.e. if 3 > 1 + e (as (1 – 2e) > 0)
i.e. if 2 > e
which is always true, so further collision occurs
M1
M1
A1 cso 6
[15]
19.
(a)
L.M. 2u = 2x + y
1
NEL y  x = u
3
5
Solving to x = u (*)
9
8
y = u (*)
9
(b)
()
L.M
M1 A1
M1 A1
M1 A1
A1
7
8
eu
9
B1
8
10
u  eu = w
9
9
M1 A1
8 
1 5
 u  eu 
3 9
9 
25
Solving to e =
32
accept 0.7812s
NEL w =
(c)
M1 A1
M1 A1 7
Q still has velocity and will bounce back from wall
colliding with stationary P.
B1
[15]
20.
v=
1
2
u
City of London Academy
B1
22
1
1
2
KE loss =
=
m (u 2 – ( 12 u )2)
M1
3mu2
8
A1
fraction of KE lost =
1
3
3mu2
 mu 2 =
2
4
8
M1 A15
[5]
21.
(a)
S
(b)
u
 u
mu + kmu = kmv
m
km T
u
(1 + k) = v
k
M1 A1

0

v
v = e(u – u) = eu(1 – )
M1 A1

u
(1 + k) = eu(1 – )
k
M1

(1  k )
= e(T)
k (1   )
A16
1  k
 1  1 + k  k(1 – )
k (1   )

1
k
1  2
since 0 <  <
1
2
M1
A1
, k > 1 (T)
A13
[9]
1.
There were also a significant number of fully correct answers to this question. Most candidates completed parts (a) and (b) well but
part (c) proved to be rather more demanding.
For part (a) the majority of candidates understood and applied the conservation of linear momentum and the law of restitution
correctly. The equations were usually consistent despite the occasional lack of a clear diagram, and only a very small minority got
the restitution equation the wrong way round.
However, subsequent errors in the speed of A and B were common – these arose from simple sign errors in the initial equations or
more commonly from minor processing errors. These basic errors can be costly – unexpected outcomes should be checked carefully
to avoid continuing to work with unrealistic situations. A few candidates surprisingly failed to substitute ½ in for e and then
struggled through with their answers for the rest of the question.
Full marks were usually scored for part (b) with only a minority of candidates with method errors in the use of the restitution
equation. The question asked for the speed of B after the collision, so the final answer should have been positive, which was not
always the case.
There were a wide variety of approaches to part (c). It was pleasing to see that some candidates could produce the given expression
fluently with their methods clearly laid out. The most successful approach involved finding the separation of the two particles as B
impacted with the wall and then to use the relative velocity to find the time taken to cover this separation. Some used the ratio of
distances travelled or set up an equation in T.
City of London Academy
23
Many could find the time to B hitting the wall and the distance travelled by A in this time, but got no further, having run out of time
or having no idea how to proceed further.
A few solutions went off in entirely the wrong direction, either by thinking that another collision was needed (considering CLM and
NEL again) or by attempting to use methods which implied non-zero acceleration. A worrying handful did not use the correct
relationship between distance speed and time (e.g time = distance × speed was seen).
Expressions involving a and u were often badly written and these symbols would become interchanged during the rearrangement
and simplification of terms. For example, a fractional term such as
5
u
4
was written without due care so that the u migrated to the
bottom of the fraction during manipulation. Clear layout and a description of the symbols are vital in this kind of question to avoid
these careless errors and to help examiners navigate through a candidates’ work.
The given answer was helpful to some candidates who made a false start - realising their error they would often produce a better or
correct solution. However, candidates working with incorrect values from part (a) were often misled into altering work displaying
correct method in an attempt to derive the given answer. There were also some who tried to fudge incorrect processes to achieve the
given answer.
2.
This question was very well answered by the majority of candidates. The momentum and impact equations were often correct - the
most common error was lack of consistency in signs between their equations. A surprising number of candidates did not draw a
diagram which possibly made it more difficult to avoid these sign errors. Only a small number of candidates quoted the impact law
the wrong way round.
A significant number of candidates had P going to the left after the collision, obtaining a velocity of u(e – 1). However, they almost
always failed to realise that this was a negative answer, ignoring the fact that the question had asked for speed.
A
number of candidates, who started with two correct equations, went on to lose marks at the end due to algebraic errors in solving the
simultaneous equations.
3.
Parts (a) and (b) were tackled with confidence by most candidates although a few were not sufficiently careful with signs and/or had
long-winded algebraic manipulation to achieve the given result. CLM and the impact law were generally used correctly.
Part (c) proved to be more challenging and differentiated between the stronger candidates and those with confused concepts. No
mention of e in the question caused some to ignore it or, more commonly, to assume the previous value. There were various
arguments used to justify their final statement but, encouragingly, the range for e seemed to be understood.
4.
Candidates made a confident start to this question, but in parts (a) and (b), poor algebraic skills and the lack of a clear diagram with
the directions marked on it hampered weaker candidates’ attempts to set up correct and consistent (or even physically possible)
equations. The direction of P after impact was not given and those candidates who took its direction as reversed ran into problems
when finding the value of e. Many realised that they had chosen the wrong direction and went on to answer part (b) correctly but
some did not give an adequate explanation for a change of sign for their velocity of P. Algebraic and sign errors were common, and
not helped by candidates’ determination to reach the given answers.
Parts (c) and (d) caused the most problems. They could be answered using a wide variety of methods, some more formal than others.
Many good solutions were seen but unclear reasoning and methods marred several attempts. Too many solutions were sloppy, with
u or d appearing and disappearing through the working. A few words describing what was being calculated or expressed at each
stage would have helped the clarity of solutions greatly. Students need to be reminded yet again that all necessary steps need to be
shown when reaching a given answer. Too many simply stated the answer
5.
3d
5
without the explanation to support it.
(a)
Very few candidates had a problem with the two equations required. The most common errors were still inconsistency in
signs and getting the ratio the wrong way round for the impact law. This is a very costly error leading to the loss of most
marks for this part of the question.
(b)
Some candidates struggled with the algebra here. The correct values for the speeds of A and B immediately after the
collision were usually used, but, especially when candidates found the change in kinetic energy of the particles separately
and then combined their answers, they failed to subtract energies correctly. A significant number managed to make mass
and velocity disappear from their final answer having successful worked through the algebra to find the kinetic energy lost.
City of London Academy
24
6.
Many candidates failed to recognise this as one of the most straightforward questions they are ever likely to meet at this level. Those
candidates who did recognise the need for a simple application of conservation of momentum, using the vectors given, were few and
far between. It was more common for candidates to apply conservation of momentum in the i and j directions separately, and such
candidates generally did so successfully. Many candidates wanted this to be the more usual style of question, where they could work
parallel and perpendicular to the line of centres of the two spheres. Candidates who tried to find the line of centres were generally
unsuccessful. Some assumed it was in either the i or the j direction and made no headway as a result. A large number of candidates
were content to regard the velocity as the speed and did not attempt to find the magnitude of their velocity.
7.
This was another standard problem, which most candidates managed easily. The majority of candidates recognised the need to apply
conservation of momentum parallel to the wall and Newton’s experimental law perpendicular to it. Those that did so often went on
to score full marks on this question.
There were many ways through the trigonometric manipulation including working with tan throughout, or solving for cos first and
then finding tan. Errors involving the incorrect use of e were rare but sign errors in the use of Newton’s Experimental Law were
more common.
8.
This question was generally well understood and answered. Most errors were caused by poor presentation leading to carelessness.
Candidates who kept all the velocities in the direction of the original velocities usually fared better than those who reversed one or
more velocity. The clearest solutions included clearly annotated diagrams which made the relative directions of motion very clear.
In the weaker solutions it was sometimes difficult to work out the candidate’s thoughts about what happened in each collision -the
question did not give them names for the speeds after the initial collision and this gave rise to problems for some candidates who
often gave the same name to more than one variable. Candidates with an incorrect or inconsistent application of Newton’s
Experimental Law lost a lot of time trying to obtain the given answer for the speed of Q after the first collision. In part (b) although
2
most candidates attempted to form a valid expression for the change in kinetic energy, the m and u were too often discarded along
the way.
In part (c), and to a lesser extent in part (a), the tendency to want to solve simultaneous equations by substitution, rather than by
elimination, produced untidy and unwieldy expressions which often led to arithmetical errors; a shame when the original equations
were correct. Most candidates interpreted the final part correctly, although too many wanted to substitute 11/10u rather than tackle
an inequality -it was clear that many candidates were not confident in setting up an inequality. Some problems did occur where
students assumed the reversal of the direction of motion of Q following the collision but failed to take account of this in setting up
their inequality.
9.
Very few candidates noticed the link between part (a) of this questions and parts (b) and (c).
This resulted in a considerable quantity of valid but unnecessary work. The marks allocated to the three parts of the question should
give candidates an indication that each of the later parts is not expected to involve as much work as the first part.
(a)
There were many substantially correct answers to this part. Most candidates formed correct equations using restitution and
conservation of momentum. The difficulties started with the speed of P – many candidates whose answer was the negative
of the printed answer did not justify the change of sign using the information about the value of e, and those whose answer
agreed did not appreciate the need to verify that their value for velocity was in fact positive.
Candidates who changed the sign of their answer for the speed of P often went on to substitute incorrectly to find the speed
of Q. There was also evidence of some confusion over the exact meaning of the question in part (aii), with several
candidates starting by substituting  in place of e.
(b)
Most candidates elected to start the question afresh rather than use the results from part (a). Examiners were presented with
confusing diagrams which were often contradicted by the working which followed. Alternatively there was no diagram and
we had to decide for ourselves which direction the candidate assumed sphere B would move in after the collision with C.
(c)
By this stage in the working many candidates were working with an incorrect initial speed of B, and were then further
confused about the possible directions of motion of A and B after their collision. This often resulted in a page or more of
working to deduce a velocity for B after the collision with A, all for a potential score of one mark. Most candidates did
demonstrate an understanding that they needed to compare the speeds of B and C to determine whether or not there would
be a further collision.
City of London Academy
25
10.
Many scored full marks in (a) and (b). It was pleasing to see that most used correct methods for momentum and impact law
equations but disappointing to see a number of sign and arithmetic errors. Parts (c) and (d) proved to be more challenging. It did not
help that some candidates confused themselves by giving every unknown speed the same name. Part (c) required the use of two
correct equations solved simultaneously and many successfully showed that e = ¾. In (d) marks were often lost through poor
explanation. For the final mark a clear statement backed up by a comparison of speeds was required.
11.
In part (a) candidates generally understood the methods involved and were able to produce momentum and restitution equations.
Inconsistencies between directions in diagrams and equations were common and many were unable to obtain a correct expression
for the speed of A after the collision. In the second part, the rebound again caused problems with signs but most were able to set up
the first inequality. The given double inequality was sometimes fudged or just stated without any attempt to justify e > ¼. In part (c),
the correct method was usually adopted but accuracy errors were common.
12.
The first part was very well answered with the odd error of incorrect signs. It was pleasing to note that only a few gave the
restitution equation the wrong way round. Most of the errors here occurred in parts (b) and (c). There appeared to be some confusion
over the fact that the KE of every particle before and after the impact was required and that these needed to be added and subtracted
correctly in order to get a positive loss of KE. Some candidates missed out the KE of one of the particles and many subtracted to get
a negative KE. In part (c) the impulse equation was often incorrect, with an incorrect sign in the velocity term. This led to a value of
e >1. Some candidates realised that this was wrong and corrected their equation but others changed the final equation but failed to
change the signs in their original impulse equation and were unable to score full marks.
13.
The method in part (a) was generally well known and there were many good solutions. However, the second part proved to be more
challenging with only the better candidates able to set up and solve their equations correctly and even then not all were able to pick
up the final two marks - many only considered a further collision between B and C, which was impossible, instead of between A and
B.
14.
In part (a), nearly all candidates could obtain a pair of equations using conservation of linear momentum and Newton’s law of
restitution. The printed answer usually helped those who had made sign errors to correct them. Part (b) proved more difficult. The
majority of candidates first found the velocity of P. A few, on recognising that the direction of P differed from the one they had
used in part (a), reversed the direction of P and started all over again rather than correctly interpreting the results they already had.
In doing so they sometimes confused their two sets of working. Those who had found the velocity of P often gave the inequality the
wrong way round or produced fallacious work when solving the inequality. Another error, frequently seen, was to produce an
inequality from the incorrect reasoning that the speed of one particle had to be greater than the other. The fully correct range
2
3
 e 1
was rarely given. One neat, although equivalent, method of solution seen was to say that the velocity of separation had
to be greater than the velocity of Q. This gives
3eu  15 u  9e  4 
without the necessity of finding the velocity of P. Part (c)
is demanding in its algebraic requirements and the signs were difficult for many to sort out. However many completely correct
solutions were seen and the general standard of algebraic manipulation was good.
15.
Part (a) was well done, although a minority changed the sign of the velocity from
 30i  ms1
to
  30i  ms 1 , thinking
that they were taking a “rebound” into account, not understanding that the direction was implied by the vector itself. In part (b) there
was some confusion between the angle of deflection and the angle the impulse made with i. Part (c) was quickly done by those who
knew how to find the kinetic energy associated with a velocity given in vector form but many could not do this.
16.
Part (a) was well done but parts (b) and (c) proved testing. Many produced irrelevant equations for a second collision between the
spheres and wasted much time not realising that the question was whether or not a second collision occurred and not what the results
of such a collision might be. The essential distinction between (b) and (c) is that in the former it is the direction of P after the first
collision which is important whereas in (c) it is the magnitude and this was not recognised by many. It was not unusual to see a
City of London Academy
26
comparison of speeds in (b). In (c) there were many errors in manipulating the inequality and it was common to see the incorrect
inequality
17.
3.6 fu  0.4u leading, through an incorrect sequence of steps, to the “correct” f 
1
.
9
Part (a) was successfully completed by most of the candidates who attempted the question, although some thought that the two
particles had the same final speed. In part (b), most were able to obtain an expression for a velocity but the significance of the word
‘speed’ was not understood and there were perhaps only one or two correct answers from the entire candidature. In the third part
most candidates were unsuccessful in obtaining e=1/3 due to either forgetting that the mass of B was 3m or their inability to
accurately expand and collect terms from two squared brackets.
Others added the velocities of A and B together and squared the total and a significant number just “cancelled” the squares off the
three expressions in the energy equation reducing it to u = v1 + 3v2. Some may have misread the question and assumed that the
total KE before and after the collision was 1/6m u2 gaining an extra KE term. Some candidates realised from their answer to (b) that
e should come out to be 1/3 and simply fiddled their working. Even some of those successful in (c) did not justify or show by
substitution that the velocity of A was zero, merely stating the fact without any reference even to e=1/3.
18.
19.
(a)
The method here was generally well known and the momentum equation was usually correct although the NIL equation
sometimes had sign errors. Manipulation of the equations was generally poor.
(b)
This was rarely correct – a common error was to compare the two velocities. Many answers were unsupported and got no
credit.
(c)
A significant number were able to write down attempts at the two equations but only a few were able to progress further as
the algebra became more complex. A tiny number realised that the second impact was essentially the same as the first and
were able to write down the final velocities without any further working. A full and convincing argument for this part was
very rare.
The most notable characteristic about the majority of candidates’ responses to this question was the number of completely correct
solutions seen. The algebra in part (a) was quite straightforward but the same was definitely not true of part (b) and yet many could
negotiate the minimum of four variables needed in this question and obtain the correct exact answer
25
. The principles
e  32
needed to solve part (a) were well understood. When errors were made in part (b), these usually arose from not realising that there
were two separate impacts to consider, one between Q and the wall, with an unknown e and a second impact between P and Q, in
which
e  13 . Errors of sign were also seen, often resulting in a pair of incompatible equations for linear momentum and
Newton’s Experimental Law. In part (c), candidates were expected to explain that there would be a second impact between Q and
the wall, which would result in a further impact between Q and the stationary P.
20. No Report available for this question
21. No Report available for this question.
City of London Academy
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