Question

Please help me with this question ​

Answer 1

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