CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Coursebook answers
Chapter 6
7
Exam-style questions
1
C[1]
2
B[1]
3
D[1]
4
he law of conservation of momentum
T
applies if the Earth is considered to rise as
the ball falls. The momentum of the Earth
upwards equals the momentum of the ball
downwards.[1]
5
he weight of the ball has an equal and
T
upwards force on the Earth due to Newton’s
third law.[1]
a
momentum of ball before striking wall
= mass × velocity = 2 × 3.0 = 6.0 kg m s−1
towards the ball[1]
iiIn an inelastic collision, momentum is
conserved but not kinetic energy.[1]
8
momentum after striking the wall
= 6.0 kg m s−1 away from the wall[1]
change in momentum of ball = 12 kg m s−1
away from the wall[1]
b
6
here is no change in kinetic energy as the
T
ball’s speed and mass are unchanged.[1]
[units of mass] × [units of velocity] =
kg m s−1[1]
c
Using v2 = 2as,
v = ( 2 × 3.5 × 40 ) = 280 = 16.7 m s−1[1]
so, momentum = mass × speed
= 900 × 16.7, so momentum
= 1.5 × 104 kg m s−1[1]
d
c ombined momentum to left
= 3.0 × 4.0 − 2.0 × 4.0 = 4.0 kg m s−1[1]
combined mass = 8.0 kg[1]
so, velocity after collision
4.0
=
= 0.50 m s−1 to the left[1]
8.0
1
b
c hange in momentum =
momentum after − momentum before[1]
= 0.35 × 2.5 − 0.35 × (−2.8) =
1.855 kg m s−1 ≈ 1.9 kg m s−1[1]
c
hen the table (plus the Earth) is also
W
considered, then the initial momentum of
the ball is equal to the final momentum
of the ball added to the momentum of
the snooker table, and so momentum is
conserved.[1]
a
change in momentum =
mass × change in velocity[1]
= 1100 × (−24) = −26 400 N s ≈
−26 000 N s[1]
change in momentum
b force =
[1]
time
26 400
=
= 1320 N ≈ 1300 N[1]
20
c average speed during braking = 12 m s−1[1]
a
linear momentum = mass × velocity[1]
b
a
i In an elastic collision, both momentum
and kinetic energy are conserved.[1]
9
so, distance travelled in 20 s = 12 × 20 =
240 m[1]
a
momentum = mass × velocity[1]
= 0.10 × 0.40 = 0.040 kg m s−1[1]
b
or each marble, component of
F
momentum in x-direction = half of
original momentum = 0.020 kg m s−1[1]
0.020
so, momentum of one marble = cos
45°
−1
= 0.0283 kg m s [1]
and velocity = 00.0283
= 0.283 m s−1 ≈
.10
−1
0.28 m s [1]
c
k.e. before = 12 mv2 = 12 × 0.10 × 0.402 =
0.0080 J[1]
k.e. after = 2 × 12 × 0.10 × 0.2832 = 0.0080 J
[1]
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
10 a
initial momentum of ball = 0.16 × 25
= 4.0 kg m s−1[1]
b
change in momentum = 4.0 − (−4.0)
= 8.0 kg m s−1[1]
b
force =
change in momentum
time
=
c
The bat slows down.[1]
The law of conservation of momentum
requires that the change in momentum
of the ball and of the bat are equal but in
opposite directions.[1]
Energy is neither created nor destroyed,
but thermal energy (heat/internal energy)
and sound are created from the drop in
k.e. (of the bat).[1]
The impact is non-elastic.[1]
11 a
The total momentum before the collision
is equal to the total momentum after the
interaction.[1]
The system is closed or there are no
external forces acting.[1]
b
Momentum of alpha-particle in one
direction must equal that of uranium
nucleus in the exactly opposite
direction for the change to be zero.[1]
8
0.003 [1]
= 2667 N ≈ 2700 N
(This is the force on the ball but is equal
and opposite to the force on the bat.)[1]
ii6.65 × 10−27 × vα + 3.89 × 10−25 × vx = 0
[1]
va
iii
= −58.5 ≈ −58 or −59[1]
vx
13 a
Momentum and kinetic energy[1]
b
To stop motion/momentum of the gun,
the soldier must provide a force.[1]
∆p
iiiF =
; 140 = n × 8.96[1]
∆t
number of bullets per second = 15.6 or
15 or 16[1]
14 a
Change in
Initial
Final
momentum / kinetic
kinetic
kg m s−1
energy / J energy / J
0.35v = 0.25 × 30[1]
iichange in momentum
= 0.25 × 30 − 0.25 × 21.4[1]
change in momentum = 2.14 ≈
2.1 kg m s−1 or 2.15 ≈ 2.2 kg m s−1[1]
iiichange in total kinetic energy =
× 0.25 × 302 −
1
2
1
2
× 0.35 × 21.42[1]
change in total k.e. = 32.4 ≈ 32 J[1]
i
momentum = 0.014 × 640 = 8.96 or
momentum ≈ 9.0 kg m s−1[1]
iiBullets leave with momentum forwards
and gun has equal momentum
backwards.[1]
i
final momentum = initial momentum
v = 21.4 ≈ 21 m s−1[1]
i
Momentum is conserved, as there
are no external forces / the system
is closed.[1]
truck X
6.0 × 104
2.5 × 105
4.0 × 104
truck Y
6.0 × 104
1.5 × 104
1.35 × 105
One mark for each correct change in
momentum[2]
One mark for correct kinetic energy
values for X[2]
One mark for correct kinetic energy
values for Y.[2]
t otal initial k.e. = 2.65 × 105 J and total
final k.e. = 1.75 × 105 J[1]
ivThe arrow stops and the ball moves
off with a speed of 30 m s−1[1]
b
Relative speed remains unaltered in
an elastic collision, 30 m s−1[1]
Collision is not elastic, because the total
k.e. has decreased in the collision[1]
6.0 × 10 4
∆p
c force =
=
[1]
1.6
∆t
3.75 × 104 ≈ 3.7 or 3.8 × 104 N[1]
12 a
i The total kinetic energy before the
collision is equal to the total kinetic
energy after the collision.[1]
iiIn a completely inelastic collision, the
maximum amount of kinetic energy is
lost (subject to the law of conservation
of momentum, which must be
obeyed).[1]
2
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020