10.6 Impact forces Q1. a. Impact force = change in momentum impact time Impact force ( decreases) = change in momentum (stays the same) impact time ( increases ) b. Impact force = ∆ momentum ∆ time Impact force = (0.12 kg x 18 m/s) - ( 0.12kg x 0m/s) 0.0003 s Impact force = 2.16 kgm/s 0.0003 s Impact force = 7200 N

Q2. ai. Impact force = ∆ momentum ∆ time Braking force = (800 kg x 30m/s) - (800 kg x 0m/s) ∆ time Braking force = (800 kg x 30m/s) - (800 kg x 0m/s) 6.0 s Braking force = 24,000 kgm/s = 4,000 kgm/s2 = 4,000 N 6.0 s aii. Braking force = (800 kg x 30m/s) - (800 kg x 0m/s) 30 s Braking force = 24,000 kgm/s = 800 kgm/s2 = 800 N 30 s b. If the vehicle stops in a shorter time, of less than a second, the change in momentum has to occur in a short time so the impact force is much greater.

Q3. a) For collisions: p before = p after (2000 kg x 12 m/s) + ( 10,000kg x 0 m/s) = 12,000 kg x Vcom m/s 24,000 kgm/s = 12,000 kg x Vcom m/s 24,000 kgm/s = 12,000 kg x Vcom m/s 24,000 kgm/s = 12,000 kg x Vcom m/s 2m/s = Vcom bi. Deceleration of van = ∆ velocity = 12m/s - 2 m/s = 33.3 m/s/s ∆ time 0.3 s bii. Change in momentum of van = m∆v = 2000 kg x 10m/s = 20,000 kgm/s biii. Impact force on van = ∆ mv = 20,000 kgm/s = 67,000 N ∆ time 0.3 s

Q4 . Play grounds have cushioned surfaces to reduce the risk of injury. A low impact force reduces the risk of injury. A low impact force is achieved by increasing the impact time . The momentum change remains the same value. Impact force = ∆ momentum ∆ time