Step-by-step explanation:
You can solve this with kinematics or with energy.
Using kinematics, given:
v₀ = 600 m/s
v = 200 m/s
Δx = 0.05 m
Find: a
v² = v₀² + 2aΔx
(200 m/s)² = (600 m/s)² + 2a (0.05 m)
a = -3,200,000 m/s²
Given:
v₀ = 600 m/s
v = 0 m/s
a = -3,200,000 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (600 m/s)² + 2(-3,200,000 m/s²) Δx
Δx = 0.05625 m
So the bullet needs to travel an additional 0.00625 m, or 0.625 cm.
Using energy, the work done on the bullet equals the change in kinetic energy.
ΔKE = W
½ mv² − ½ mv₀² = Fd
½ m (200 m/s)² − ½ m (600 m/s)² = F (0.05 m)
(-160,000 m²/s²) m = F (0.05 m)
F/m = -3,200,000 m/s²
½ mv² − ½ mv₀² = Fd
½ v² − ½ v₀² = (F/m) d
½ (0 m/s)² − ½ (600 m/s)² = (-3,200,000 m/s²) d
d = 0.05625 m
