Answer all the questions in the spaces provided. (a) Complete Table 1.1 by stating whether each of the quantities is a vector or a scalar. Table 1.1 quantity vector or scalar acceleration power work (b) The variation with time t of the velocity v of an object is shown in Fig. 1.1. [Figure 1.1]
- (a)Complete Table 1.1 by stating whether each of the quantities is a vector or a scalar. Table 1.1 quantity vector or scalar acceleration power work[2]
- (b)(i)Determine the acceleration of the object from time t = 0 to time t = 4.0s. acceleration = ..................................................... ms⁻²[2]
- (b)(ii)Determine the distance moved by the object from time t = 0 to time t = 4.0s. distance = ..................................................... m[2]
- (c)(i)Define force.[1]
- (c)(ii)The motion represented in Fig. 1.1 is caused by a resultant force F acting on the object. On Fig. 1.2, sketch the variation of F with time t from t = 0 to t = 12.0s. Numerical values of F are not required. [Figure 1.2][3]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2021). The mark or mark type is shown in brackets.
- [B2] (a) acceleration: vector work: scalar power: scalar Three correct scores 2 marks. Two correct scores 1 mark.
- [C1] (b)(i) a = (v – u) / t or a = gradient or a = ∆v/ (△)t e.g. a = (1.40 - 0.70) / 4.
- [A1] (b)(i) = 0.18 m s-2
- [C1] (b)(ii) distance =0.5 × (0.70 + 1.40) × 4.0 or (0.70 × 4.0) + (0.5 × 0.70 × 4.0)
- [A1] (b)(ii) = 4.2 m
- [B1] (c)(i) (force equal to) rate of change of momentum
- [B1] (c)(ii) horizontal line starting from t = 0 and ending at t = 4.0 s at a positive value of F
- [B1] (c)(ii) horizontal line starting from t = 4.0 s and ending at t = 8.0 s at F = 0
- [B1] (c)(ii) horizontal line starting from t = 8.0 s and ending at t = 12.0 s at a negative value of F and the magnitude of F is larger than from t = 0 to 4.0 s
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