Exampro a-level Physics (7407/7408)


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3.4.1.8-Conservation-of-energy

max 6


(b) (i) conservation of momentum gives (500 × 160)


= 150 v + (350 × 240) (1)
from which v = (−)26(.7) (m s−1) (1)
direction: opposite horizontal direction to larger fragment
[or to the left, or backwards] (1)
3


(ii) initial Ek = ½ × 500 × 1602 (1) (= 6.40 × 106 J)


final Ek = (½ × 350 × 2402) + (½ × 150 × 26.72) (1) (= 1.01 × 107 J)
energy released by explosion = final Ek − initial Ek (1)
= 3.7 × 106 (J) (1)
4
[13]

M27. (a) (i) t = (evidence for correct rearrangement or substitution) (1)
= (correct substitution leading to answer) (1)
(= 3.7 (3.696) (s))
2


(ii) = 41 (m s–1) (1) 2sf (1)


2
(iii) (1) = 36 (1) (m s–1)
2


(iv) (or correct scale drawing) (1)


= 54 (m s–1) (1)
ecf from (ii) (iii) [for scale drawing allow range 53 → 56]
tan θ = (1) or correct alternative
(angle from horizontal =) 42 (°) or correct alternative angle
and clear indication of direction (1)
[for scale drawing allow range 40 → 44 (1)
for scale drawing: quality of construction (1)]
4


(b) (i) (= mgh = 22 × 9.81 × 67) = 14000 (14460) (J) (1)


1
(ii) (G)PE → KE (1)
(KE to) internal/thermal/‘heat’ (energy) (1)
2
[13]


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