1. 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) s
Impact force = kgm/s s
Impact force = N

2. 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 = ,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 = ,000 kgm/s = kgm/s2 = 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.

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

4. 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