A child on a sledge slides down a steep hill and then travels in a straight line up an ice-covered slope, as illustrated in Fig. 3.1. [Figure 3.1] The sledge passes point A with speed 18ms⁻¹ at time t = 0 and then comes to rest at point B. The child applies a brake to the sledge at point B. The brake does not keep the sledge stationary and it immediately slides back down the slope towards A. The variation with time t of the velocity v of the sledge from t = 0 to t = 24 s is shown in Fig. 3.2. [Figure 3.2]
- (a)State the time taken for the sledge to travel from A to B.[1]
- (b)Determine the displacement of the sledge up the slope from point A at time t = 24 s.[3]
- (c)Show that the acceleration of the sledge as it moves from B back towards A is 0.50 ms⁻².[2]
- (d)(i)The child and sledge have a total mass of 70kg. The component of the total weight of the child and sledge that acts down the slope is 80 N. Determine the frictional force on the sledge as it moves from B towards A,[2]
- (d)(ii)the angle θ of the slope to the horizontal.[2]
- (e)(i)The child on the sledge blows a whistle between t = 4.0s and t = 8.0s. The whistle emits sound of frequency 900 Hz. The speed of the sound in the air is 340ms⁻¹. A man standing at point A hears the sound. Use Fig. 3.2 to determine the initial frequency of the sound heard by the man,[2]
- (e)(ii)describe and explain qualitatively the variation, if any, in the frequency of the sound heard by the man.[1]
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Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [A1] time = 12s
- [C1] distance (up slope) = ½ × 12 × 18 (= 108)
- [C1] distance (down slope) = ½ × 12 × 6 (= 36)
- [A1] displacement from A = 108 – 36 = 72m
- [C1] v = u + at or a = gradient or a = ∆v/(∆)t
- [A1] a = 6/12 = 0.50 (ms⁻²) (other points from the line may be used)
- OR
- (C1) v² = u² + 2as and u = 0 OR v² = 2as
- (A1) a = 6.0² / (2 × 36) = 0.50 (ms⁻²)
- OR
- (C1) s = ut + ½at² and u = 0 OR s = ½at²
- (A1) a = 2 × 36 / 12² = 0.50 (ms⁻²)
- OR
- (C1) s = vt - ½at²
- (A1) a = 2 × (6 × 12 – 36)/12² = 0.50 (ms⁻²)
- [C1] F = 70 × 0.50 (= 35)
- [A1] frictional force = 80 – 35 = 45 N
- [C1] sin θ = 80/(70 × 9.81)
- [A1] θ = 6.7°
- [C1] fo = (900 × 340)/(340 + 12)
- [A1] = 870 Hz
- [B1] speed/velocity (of sledge) decreases and (so) frequency increases
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