Answers · Mass, weight and density

Mass, weight and density — final check key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. DK1 · Final check

Model answer. (a) The force per unit mass in a gravitational field: g = W / m, in N/kg. (b) g = 670 N ÷ 180 kg = 3.7 N/kg. (c) 180 kg, the same as on Earth. (d) Weight is the pull of a gravitational field on a mass. The field of Mars is weaker: it pulls each kilogram with 3.7 N instead of 9.8 N, so the same 180 kg has a smaller weight. (e) Compare them on an equal-arm balance; a level beam means equal masses, on any world. (Or hang each in turn on the same spring balance: equal weights in the same field mean equal masses.) (f) 3.7 m/s², because the acceleration of free fall is equal to g, here 3.7 N/kg.

A full answer includes

(a) Gravitational field strength is the gravitational force per unit mass (g = W / m), in N/kg.
(b) g = 670 N ÷ 180 kg = 3.72, so 3.7 N/kg.
(c) 180 kg.
(d) Weight is the effect of the planet’s gravitational field on the mass; Mars’s field is weaker (smaller force per kilogram), so the same mass has a smaller weight.
(e) Put the sample and the 1 kg mass on the two pans of an equal-arm balance: if the beam is level, the masses are equal. Hanging each in turn on the same spring balance also works: equal readings in the same field mean equal masses.
(f) 3.7 m/s²: the acceleration of free fall equals the gravitational field strength on Mars.

Also accept

(a) force per unit mass; (b) 3.7 N/kg; (c) 180 kg; (d) weaker field, same mass; (e) equal-arm balance level, or equal readings on one spring balance; (f) 3.7 m/s².

Watch for

“The rover’s mass on Mars is 670 ÷ 9.8 = 68 kg.” — The mass does not change. Dividing by Earth’s g mixes up two different places.
“A kitchen scale brought from Earth would show 1 kg for the sample.” — An electronic scale works out grams from the push on its pan, using Earth’s g. On Mars a 1 kg mass pushes less hard, so it would show about 0.38 kg. Compare the sample with the 1 kg mass in the same field instead.

2. DK2 · Final check

Model answer. (a) Measure the sides with a rule and multiply: V = 2.0 × 2.0 × 2.0 = 8.0 cm³; ρ = 63 g ÷ 8.0 cm³ = 7.9 g/cm³ (about the density of iron). (b) Weigh the dry pebble on a balance; put water in a measuring cylinder and read the bottom of the meniscus; lower the pebble on a thread until it is completely under water; read the new level; subtract to get the volume; density = mass ÷ volume. (c) ρ = 51.5 g ÷ 50 cm³ = 1.03 g/cm³. (d) Weighing the pebble wet, with its volume measured correctly (mass too big), or splashing water out (volume too small). (e) Empty, dry cylinder on the balance, press zero (tare), pour to the 50 cm³ mark reading the bottom of the meniscus, read the mass; or weigh it empty and full and subtract.

A full answer includes

(a) Measure each side with a rule (2.0 cm) and multiply them: V = 2.0 × 2.0 × 2.0 = 8.0 cm³; ρ = 63 g ÷ 8.0 cm³ = 7.9 g/cm³.
(b) Weigh the dry pebble; read the water level in a measuring cylinder; lower the pebble in completely; read the new level; volume = the rise; density = mass ÷ volume.
(c) 51.5 g ÷ 50 cm³ = 1.03 g/cm³.
(d) For example: weighing it wet while its volume is measured correctly (mass too big), or water splashing out so the rise is too small (volume too small).
(e) Put the empty, dry measuring cylinder on the balance and press zero (tare), pour salt water to the 50 cm³ mark (bottom of the meniscus, eye level), then read the mass; or weigh the cylinder empty and full and subtract.

Also accept

(a) 7.9 g/cm³; (b) mass on a balance, volume by displacement, divide; (c) 1.03 g/cm³; (d) wet weighing or splashing; (e) tared cylinder (or full − empty) and the 50 cm³ mark.

Watch for

“The cube’s volume is 2.0 × 3 = 6.0 cm³.” — A cube’s volume is side × side × side: 2.0 × 2.0 × 2.0 = 8.0 cm³.
“An air bubble on the pebble makes the density too big.” — A bubble adds volume, so the density comes out too small.

3. DK3 · Final check

Model answer. (a) ρ = 72 g ÷ 80 cm³ = 0.90 g/cm³; it floats in water (0.90 < 1.00). (b) It sinks in paraffin oil, because 0.90 g/cm³ is greater than 0.80 g/cm³. (c) The paraffin oil floats on the water: 0.80 g/cm³ is less than 1.00 g/cm³ and the two do not mix. (d) Whether something floats depends on its density compared with the liquid’s, not on its size: a small cork floats and a big boulder sinks, but a big log floats and a small pebble sinks.

A full answer includes

(a) 72 g ÷ 80 cm³ = 0.90 g/cm³; it floats in water because 0.90 < 1.00.
(b) No: 0.90 g/cm³ is more than paraffin oil’s 0.80 g/cm³, so it sinks in the oil.
(c) Paraffin oil is on top: it is less dense than water and they do not mix.
(d) Floating depends on density, not size: a large object of a dense material sinks, a small object of a material less dense than the liquid floats.

Also accept

(a) 0.90 g/cm³, floats; (b) sinks in paraffin oil; (c) paraffin oil on top; (d) density, not size.

Watch for

“A candle floats in any oil because oil is lighter than water.” — Compare the candle with the liquid it is in: 0.90 is more than paraffin oil’s 0.80, so it sinks there.
“Pour in more paraffin oil than water and the oil ends up underneath.” — The amount does not matter. Each cubic centimetre of paraffin oil (0.80 g) has less mass than one of water (1.00 g), so the oil floats on top however much you pour.