O-LevelPhysicsForces5054/22 · Oct/Nov 2020 · Q17 Marks

A glass beaker has a mass of 50g. A liquid of density 1.8g/cm³ is poured into the beaker until it reaches the 200 cm³ mark.

(a)Calculate the total mass of the beaker and its contents. mass =[3]
(b)The centre of mass of a metre rule is at the 50 cm mark.
(b)(i)State what is meant by centre of mass.[1]
(b)(ii)The metre rule is placed on a pivot. The tip of the pivot is under the 80cm mark on the rule. The beaker with its contents is then placed at different positions along the rule until the rule is balanced. Fig. 1.1 shows the arrangement with the rule balanced. [Figure 1.1] One side of the beaker is at the 84 cm mark and the other side is at the 92 cm mark. Calculate the mass of the rule. mass =[3]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.

[1] (a) (m =) pV or 1.8 × 200 or 360
[1] (a) (1.8 x 200) + 50 or 360 + 50 or 360 or candidate's mass + 50
[1] (a) 410 g or 0.41 kg
[1] (b)(i) (place / region) where the mass appears to be concentrated or where the object balances
[1] (b)(ii) 88 (cm) seen or (84 + 92)/2 seen or 30 (cm) seen or 8(.0 cm) seen or 164 (g)
[1] (b)(ii) m1gx1 = m2gx2 or F1X1 = F2X2 or W1X1 = W2X2 or (m₁ =) M2X2 / X1 or 410 x (88–80)/30 or 164 (g)
[1] (b)(ii) 110 g or 0.11 kg

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