IGCSE Physics Answers

Page 12
1 a) Between t = 30 and t = 45 mins
b) 7.5 mins
c) i) distance travelled = area under graph between t = 0 and t = 12½ mins ii) average speed = total distance travelled total time for journey
= total area under graph
60 mins
2 a) ∆v = 32 m/s a =10 m/s² t = ∆v = 32 = 3.2 s a 10
b)

3 a) OP – constant acceleration
PQ – constant acceleration (greater than OP)
QR – constant speed
RS – constant deceleration
b) O and S
c) 6 m/s
d) 70 s
e) Total distance travelled = area under graph
4 a) u = d = 25 = 12.5 s t 2
b) speed is decreasing
c) time for tree 3 to tree 4 is greater than for time for tree 2 to tree 3
5 a) Stage 1: 20 s

Stage 2: t = 4800 = 400 s
12
Stage 3: 80 s
b)

c) Total distance travelled = (½ × 20 × 12) + (400 × 12) + (½× 80 × 12)
= 120 + 4800 + 480 = 5400m
d) average speed = total distance = 5400 = 10.8 m/s time taken
500

Page 17
1 Weigh the car using the Newton meter.
Find the mass from mass = weight
10 m/s²
(for rest of question see description on p. 16)
2 (For first part see description on p.16)
To find the density of cork the mass, M, could still be found using the balance.
But the volume must be calculated by finding its dimensions and using cross-sectional area × length to calculate the volume.
Area = area of a circle, πr²

d = 2 × radius = 2r
Area = r²
Volume = r²l
Density = r²l m 3 a) Volume = Reading Q – Reading P
b) Mass = Reading S – Reading R
c) i) P = m v ii) p = 57.5 = 2.3 g/cm³
25
Page 17
1 l = 3.6 cm
2 A = r² - (d)² = (0.6)² = 0.28 cm²
2
2
3 V = r²l = 1.0 cm³
Volume of bundle – 10 × 10 – 10.0 cm³
P = m = 59.1 – 5.91 g/cm³ v 10.0
Page 21
1 a) length = 17.5 cm
b) load = 2.25 N

c) extension = length – initial length = 15 – 11
= 4 cm
2 a) It has both size and direction
b) f = m × a
=5×3
= 15 N
3 a) 11 cm
b) 19 cm
c) 19 − 11 = 8 cm
Page 22
4 a) Q = elastic limit
b) The extension is directly proportional to the force applied
c) In the region QR,