CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
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Coursebook answers
Chapter 2
d
Self-assessment questions
1
change in velocity ∆v = (18 − 0) = 18 m s−1
e
time taken ∆t = 6.0 s
so, acceleration a = ∆∆vt =
2
= 3.0 m s−2
−12
20
= −0.60 m s−2
time taken ∆t = 20 s
7
The magnitude of the deceleration is 0.60 m s−2
3
a
Rearranging the equation a = v –t u gives
v = u + at
so, after 1 s, v = 0.0 + (9.81 × 1) = 9.81 ≈
9.8 m s−1
b
4
After 3 s, v = 0.0 + (9.81 × 3) = 29.4 m s ≈
29 m s−1
Dots evenly spaced, then getting steadily
closer together
If l1 = length of first section of interrupt card,
t1 = time when first section enters light gate,
t2 = time when first section exits light gate,
l2 = length of second section of interrupt card,
t3 = time when second section enters light gate,
t4 = time when second section exits light gate,
then:
l1
= 0.05
= 0.25 m s−1
t2 − t1 0.20 − 0.0
l2
0.05
= 1.0 m s−1
=
t4 − t3 0.35 − 0.30
Δt = t3 − t1 = 0.30 − 0.0 = 0.30 s
so, acceleration a = ∆∆vt = 1.00−.300.25 = 2.5 m s−2
or first ticker-tape section, length l1 = 10 cm,
F
time taken t1 = 5 × 0.02 = 0.10 s
so, initial velocity u =
t
20
= 1.0 m s−1
0
5
l2 0.16
=
t2 0.10
= 1.6 m s−1
ections of tape are adjacent, so time between
S
start of first section and start of final section,
Δt = time taken by first section = 5 × 0.02 =
0.10 s
so, acceleration a = ∆∆vt = 1.60.−101.0 = 6.0 m s−2
10
0
l1 0.10
=
t1 0.10
or second ticker-tape section, length
F
l2 = 16 cm, time taken t2 = 5 × 0.02 = 0.10 s
v /ms–1
30
so, final velocity v =
10 15 20 25 30 t / s
9
b, c During first 10 s, acceleration a = ∆∆vt =
= 3.0 m s−2
1
= −2.0 m s−2
From area under graph: 525 m
final velocity v =
v
a
−30
15
initial velocity u =
−1
8
5
0 – 30
= ((30 – 15)) =
6
18
6.0
change in velocity ∆v = (11 − 23) = −12 m s−1
so, acceleration a = ∆∆vt =
During last 15 s, acceleration a = ∆∆vt
30
10
aWe know u, a and t and we want to know
v, so we use the equation
velocity v = u + at = 0.0 + (2.0 × 10)
= 20 m s−1
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
We know u, a and t and we want to know
s, so we use the equation
distance s = ut +
1
2
c
1
2
at2 = 0.0 +
× 2.0 × 10 × 10 = 100 m
v − u = 24 − 0
a
2.0
acceleration, a =
v − u = 20 − 4.0
t
100
v + u = 20 + 4.0
2
2
average velocity, vavg =
= 12 m s−1
We could use s = ut +
1
2
d
isplacement of car = area under graph
d
= (area of rectangle with side 8 m s−1 and
length 30 s) + (area of triangle with side
12 m s−1 and base 30 s)
at but given that
v = u + 2as
so, final velocity, v =
= (8 × 30) + ( 12 × 12 × 30) = 420 m
displacement of car, s = ut +
e
1
2
at2
= (20 × 30) + ( 12 × (−0.40) × 30 × 30)
= 600 − 180 = 420 m
16 a
Calculate distance fallen for each time
using s = ut +
1
2
at2, with u = 0
Time / s
0
1.0
2.0
Displacement / m
0
4.9
19.6 44.1 78.5
b
3.0
4.0
Graph is a parabola through the origin.
80
(8.0 )
2
70
u 2 + 2as
− 2 × 1.0 × 18 =
60
100 = 10 m s−1
v 2 − u 2 = ( 0 )2 − (30 )2 = 900
2a
2 × ( −7 )
14
distance, s =
= 64.3 m ≈ 64 m
13 We know v, a and s and we want to know u,
so we rearrange the equation v2 = u2 + 2as
into u2 = v2 − 2as, so initial speed, u =
50
40
30
20
10
v 2 − 2as = ( 0.0 )2 − 2 × ( −6.5 ) × 50 = 650
= 25.5 m s−1
his is just over the speed limit.
T
14 a
t = 7.5 s; v = 220 m s−1
raw a tangent to the curve at point
D
P. Read off two sets of values from the
tangent to find the gradient. For example:
at time t1 = 0 s, v1 ≈ 60 m s
at time t2 = 12 s, v2 ≈ 300 m s−1
so, approximately, acceleration
a = ∆∆vt =
300 − 60
12 − 0
0
c
−1
0.0
2.0
Time/ s
3.0
4.0
I n 2.5 s, stone falls 30.6 m ≈ 31 m. Check
using
s = ut +
≈ 31 m
d
1.0
1
2
at2 = 0 + ( 12 × 9.81 × 2.5 × 2.5)
time taken = 2.86 s ≈ 2.9 s
Check by rearranging, remembering that
u = 0, so that time t = 2.86 s ≈ 2.9 s
17 a
We know s and a, and that u = 0, and we
need to find t.
= 20 m s−2
15 a
The car is slowing down with constant
(uniform) deceleration.
2
= −0.40 m s−2
2
12 We know u, v and a and we want to know s, so
we rearrange the equation v2 = u2 + 2as, so that
b
8 − 20
30 − 0
2
11 We know u, a and s and we want to know v, so
we use the equation
=
acceleration a = ∆∆vt =
= 0.16 m s−2
we have worked out the average speed,
it is simpler to use distance, s = vavg × t
= 12 × 100 = 1200 m
2
c
= 12 s
10 aWe know u, v and t and we want to know
a, so we use the equation
c
initial velocity v1 = 20 m s−1; final velocity
v2 = 8 m s−1
We know u, v and a and we want to know
t, so we rearrange the equation v = u + at
so that
time t =
b
b
Displacement / m
b
Rearrange s = ut + 12 at2, remembering that
u = 0,
0.8
so that time t = 2sa = 29×.81
≈ 0.40 s
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
b
We know s and a, and that u = 0, and we
need to find v.
Use v2 = u2 + 2as so that impact velocity, v
= u 2 + 2as = ( 0 )2 + 2 × 9.81× 0.8 = 15.7
≈ 4.0 m s−1
18 a
Using the method in the worked example,
calculate the average speed of the steel
ball =
s = 2.10
t 0.67
= 3.134 m s−1
Then find the values of v and u
final speed, v = 2 × 3.134 m s−1 = 6.268 m s−1
initial speed, u = 0.0 m s−1
Substitute these values into the equation
for acceleration
v − u = 6.268
t
0.67
a =
= 9.36 m s−2 ≈ 9.4 m s−2
b
Air resistance; delay in release of ball
c
ercentage uncertainty in time = 0.02/0.67
p
× 100 = 3%
percentage uncertainty in g = 2 × 3 = 6%
or largest value of g = 9.94 m s−2 giving an
absolute uncertainty of 0.58 m s−2 and a
58
percentage uncertainty of 90..36
× 100 = 6%
2.0
1
2
at2 gives
−25 = 20t + 12 × (−9.81) × t2
so, 4.9t2 − 20t − 25 = 0 or approximately 5t2 −
20t − 25 = 0, which can be simplified to
t2 − 4t − 5 = (t − 5)(t + 1) = 0
so, time taken to reach the foot of the cliff = 5 s
(i.e. 1 s more). Accurate answer is 5.08 ≈ 5.1 s.
In solving the quadratic equation, you
will have found a second solution, t = −1 s.
Obviously, the stone could not take a negative
time to reach the foot of the cliff. However,
this solution does have a meaning: it tells us
that, if the stone had been thrown upwards
from the foot of the cliff at the correct speed, it
would have been travelling upwards at 20 m s−1
as it passed the top of the cliff at t = 0 s.
23 a
Use v = u + at to calculate v, remembering
that a = −9.81 m s−2
Velocity
30 20.19 10.38 0.57 −9.24 −19.05
/ m s−1
0
1.0
2.0
3.0
4.0
5.0
b
30
25
20
0
0.5
1.0
1.5
2.0
2.5
3.0 t2 / s2
Because s = 12 at2 the gradient is 12 g, the
acceleration of free fall, g ≈ 1.6 m s−2
his object is not falling on the Earth,
T
perhaps on the Moon
20 Drop an object towards the sensor, but
take care not to break it. A better method
is to use a sloping ramp with a trolley;
gradually increase the angle of slope.
Deduce the value of the acceleration when
the ramp is vertical.
21 a
Fx = 17.3 N ≈ 17 N; Fy ≈ 10 N
vx = 1.7 m s−1; vy = −4.7 m s−1
Velocity / m s–1
0.5
b
Fx = 77.3 N ≈ 77 N; Fy = 20.7 N ≈ 21 N
Substituting in s = ut +
1.0
c
d
22 The stone’s displacement now is s = −25 m
1.5
b
ax = −5.2 m s−2; ay = −3.0 m s−2
Time
/s
19 a
h/m
0
c
15
10
5
0
–5
2.0
1.0
3.0
4.0
5.0
Time / s
–10
–15
–20
c
3.1 s
24 a
Horizontal speed remains constant after
being thrown (ignoring air resistance), so:
horizontal velocity = st = 124..00 = 3.0 m s−1
b
or vertical distance, use s = ut + 12 at2,
F
remembering that u = 0
s = ut +
1
2
at2 = 0 +
1
2
× (−9.81) × 4.0 × 4.0
= −78.5 m, so height of cliff is 78.5 m
3
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
25 a
vertical component of velocity =
8 × sin 40° = 5.14 ≈ 5.1 m s−1
b
vertical component of velocity = 0 m s−1
c
Rearrange v = u + at, so that time t =
=
0 − 5.14
= 0.524 ≈ 0.52 s
−9.81
d
orizontal component of velocity
h
= 8 × cos 40° = 6.13 ≈ 6.1 m s−1
e
ssume horizontal component of velocity
A
is constant and use
distance s = ut +
= 3.21 ≈ 3.2 m
4
v−u
a
1
2
at2 = 6.1 × 0.52 + 0
26 F
irst, calculate the time taken for the projectile
to return to the ground.
initial vertical velocity, uver = 40 × sin 45°
= 28.3 m s−1
We know the vertical distance travelled
when the projectile hits the ground = 0 m so,
rearrange s = ut + 12 at2 to find t
0 = 28.3t + 12 × 9.81t2 = 28.3t + 4.905t2
so, t = 0 (when the projectile is launched) or
t = 5.77 s (when it returns to the ground)
Assume horizontal velocity is constant,
uhor = 40 × cos 45° = 28.3 m s−1
so, horizontal distance s = ut = 28.3 × 5.77
= 163 m ≈ 160 m
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020