1a i) KE = ½ x mass x speed2 1a i) KE = ½ x 500 kg x 122 (m/s)2 1a i) KE = ½ x 500 kg x 144 (m/s)2 1a i) KE = 36,000 J 1a ii) KE = ½ x mass x speed2 1a ii) KE = ½ x 0.44 kg x 202 (m/s)2 1a ii) KE = ½ x 0.44 kg x 400 (m/s)2 1a i) KE = 88 J

1b KE = ½ x mass x speed2 1b 2 x KE = m x V2 1b 2 x KE = V 2 m 1b 2 x KE = V m 1b 2 x 2 x 36,000 J = 17 m/s 500 kg

2a. i) The elastic potential energy store increases The chemical energy store of the person decreases. 2aii) The elastic potential energy store decreases The kinetic energy store of the object increases. KE = ½ m V 2 2b i) ∆GPE = weight x ∆ height 2b i) ∆GPE = 2. 0 N x 5.0 m 1b 2 x KE = V m 2b i) ∆GPE = 10.0 J 2b ii) ∆GPE = ∆ KE = 10 J 1b 2 x 10 J = 10 m/s 0.2 kg

3a KE = Work done by brakes KE = Fs 360,000 = F 100 3,600 N = F 3b 360,000 J = ½ x m kg x 302 (m/s)2 3b 360,000 J = ½ x m kg x 900 (m/s)2 3b 360,000 J = ½ x m kg x 450 (m/s)2 3b 360,000 = ½ x 800 kg 450

4. ∆ elastic PE = ½ k e2 2b i) ∆ elastic PE = ½ x 250 x 0.212 2b i) ∆ elastic PE = 5.5 J