A resultant force F moves an object of mass m through distance s in a straight line. The force gives the object an acceleration a so that its speed changes from initial speed u to final speed v.
- (a)(i)1.the work W done by the force, in terms of a, m and s W = ..................................................................................... [1][1]
- (a)(i)2.the distance s, in terms of a, u and v. s = ..................................................................................... [1][1]
- (a)(ii)Use your answers in (i) to show that the kinetic energy of the object is given by kinetic energy = ½ × mass × (speed)². Explain your working. [2][2]
- (b)A ball of mass 0.040 kg is projected into the air from horizontal ground, as illustrated in Fig. 2.1. [Figure 2.1]
- (b)(i)For the movement of the ball from X to Y, draw a solid line on Fig. 2.1 to show: 1. the distance moved (label this line D) 2. the displacement (label this line S). [2][2]
- (b)(ii)By consideration of energy transfer, determine the height h of point Y above the ground. h = ..................................................................... m [3][3]
- (b)(iii)On Fig. 2.2, sketch the variation of the kinetic energy of the ball with its vertical height above the ground for the movement of the ball from X to Y. Numerical values are not required. [Figure 2.2] [2][2]
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Mark scheme
Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.
- [B1] W = mas
- [B1] s = (v² – u²) / 2a
- [B1] W/work equals energy transferred/gain or change in kinetic energy
- [B1] W (= mas) = ma(v² – u²) / 2a
- [B1] 1. solid curved line drawn from X to Y along path of ball and labelled D
- [B1] 2. solid straight line drawn from X to Y and labelled S
- [C1] (Δ)Ε = mg(Δ)h
- [C1] 4.5 = (0.040 × 9.81 × h) + (½ × 0.040 × 9.5²)
- [A1] h = 6.9 m
- [M1] line with a negative gradient starting from a non-zero value of kinetic energy when the vertical height is zero
- [A1] straight line ends at a non-zero value of kinetic energy when the vertical height is h
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