IGCSE Physics 1.3-1.4 — Mass, Weight, Density Numericals (Solutions)

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IGCSE Physics 0625 · Sections 1.3-1.4 · Mass, Weight and Density — Numerical Solutions

Mass, Weight and Density — Numerical Practice: Solutions

A. Weight and Mass

A1

Weight of a 12 kg object (g = 9.8 N/kg).

1
W = mg
12 × 9.8
✓117.6 N
A2

Weight 245 N. Find mass (g = 9.8 N/kg).

1
m = W/g
245 ÷ 9.8
✓25 kg
A3

Weight of 50 kg object on the Moon (g = 1.6 N/kg).

1
W = mg
50 × 1.6
✓80 N
A4

Weight 20 N, g = 4 N/kg. Find mass.

1
m = W/g
20 ÷ 4
✓5 kg
A5

Why does mass stay the same but weight change?

✓Mass measures the quantity of matter in an object, which does not depend on location. Weight is the gravitational force on that mass, which depends on the local gravitational field strength — since the Moon’s gravitational field is weaker than Earth’s, the same mass experiences a smaller weight there.
A6

70 kg astronaut, weight 686 N. Find g, suggest location.

1
g = W/m
686 ÷ 70
✓g = 9.8 N/kg — this matches Earth’s gravitational field strength, so the astronaut is likely on Earth’s surface

B. Density

B1

Mass 250 g, volume 50 cm³. Find density.

1
ρ = m/V
250 ÷ 50
✓5 g/cm³
B2

Cube side 4 cm, mass 320 g. Find density.

1
Volume of cube
4³ = 64 cm³
2
ρ = m/V
320 ÷ 64
✓5 g/cm³
B3

Mass 45 g, density 9 g/cm³. Find volume.

1
V = m/ρ
45 ÷ 9
✓5 cm³
B4

Density 0.8 g/cm³, volume 250 cm³. Find mass.

1
m = ρV
0.8 × 250
✓200 g
B5

Empty cylinder 30 g; with 40 cm³ liquid, total 62 g. Find liquid density.

1
Mass of liquid alone
62 − 30 = 32 g
2
ρ = m/V
32 ÷ 40
✓0.8 g/cm³
B6

Water rises 60→85 cm³, stone mass 100 g. Find density.

1
Volume of stone = rise in level
85 − 60 = 25 cm³
2
ρ = m/V
100 ÷ 25
✓4 g/cm³
B7

Density 2.7 g/cm³, volume 12 cm³. Find mass.

1
m = ρV
2.7 × 12
✓32.4 g

C. Floating and Comparing Densities

C1

Density 0.95 g/cm³ — float in water (1.0 g/cm³)?

✓Yes, it will float, because its density (0.95 g/cm³) is less than the density of water (1.0 g/cm³).
C2

Densities 0.88 and 1.05 g/cm³ — which floats on top?

✓The 0.88 g/cm³ liquid floats on top, since it has the lower density.
C3

Iceberg 920 kg/m³ in seawater 1025 kg/m³. Find submerged fraction.

1
The submerged fraction equals the ratio of the object’s density to the liquid’s density
fraction = 920 ÷ 1025
✓≈ 0.90, so about 90% of the iceberg is submerged (only ~10% shows above the surface)

Mastery Test — Answer Key

  1. 245 N (25 × 9.8)
  2. 10 kg (98 ÷ 9.8)
  3. 148 N (40 × 3.7)
  4. A balance compares the weights on each side, but since both sides experience the same gravitational field strength g, this effectively cancels out and gives a direct comparison of mass.
  5. 3 g/cm³ (120 ÷ 40)
  6. 30 cm³ (60 ÷ 2)
  7. 180 g (1.2 × 150)
  8. 1 g/cm³ — mass of liquid = 85−25 = 60 g; ρ = 60 ÷ 60
  9. The 0.92 g/cm³ liquid floats on top, since it has the lower density.
  10. It will sink, because its density (1.1 g/cm³) is greater than the density of water (1.0 g/cm³).

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