February/March 2026 Paper 42

2026 · 11 questions · 49 parts · 80 marks

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1(a)1(b)1(c)

–/6 mks

2(a)(i)2(a)(ii)2(a)(iii)2(a)(iv)2(b)2(b)(i)2(b)(ii)

–/9 mks

3(a)(i)3(a)(ii)3(a)(iii)3(b)

–/7 mks

4(a)4(b)(i)4(b)(ii)4(c)

–/7 mks

5(a)5(b)(i)5(b)(ii)

–/8 mks

6(a)(i)6(a)(ii)6(a)(iii)6(b)6(b)(i)6(b)(ii)

–/7 mks

7(a)7(b)7(c)7(d)

–/7 mks

8(a)(i)8(a)(ii)8(b)8(b)(i)8(b)(ii)

–/8 mks

9(a)9(b)9(c)

–/8 mks

Q10

10(a)10(a)(i)10(a)(ii)10(b)10(b)(i)10(b)(ii)

–/5 mks

Q11

11(a)11(b)11(b)(i)11(b)(ii)

–/8 mks

Context for question 3

Figure 3.1 shows a polar bear and a person standing on thick ice. The figures are drawn to a scale of 1 cm : 100 cm.

Figure 3.1, a polar bear and a person each standing on their own patch of thick ice, drawn to a scale of 1 cm to 100 cm

Table 3.1 contains information about the polar bear and the person.

Table 3.1

	polar bear	person
mass / kg	590	84
area in contact with ice / m $^{2}$	1.12	0.020

Take the weight of 1.0 kg to be 9.8 N (acceleration of free

f a l l = 9.8

m/s

^{2}

3(a)(i)Forces Free Body DiagramsEasy1 mark

Calculate the weight of the polar bear.

Answer

3(a)(ii)Density PressureEasy2 marks

Calculate the pressure that the polar bear exerts on the ice.

Answer

3(a)(iii)Density PressureMedium2 marks

Use the numbers to complete each equation. Each number may be used once, more than once or not at all.

mass of polar bear =

\times

mass of person area of polar bear in contact with ice =

\times

area of person in contact with ice pressure exerted by person =

\times

pressure exerted by polar bear

3(b)Density PressureEasy2 marks

Under the ice is sea water with a density of 1030 kg/m

^{3}

. Calculate the change in pressure

Δ p

between the top of the sea water and a depth of 2.50 m.

Answer