Bullet Embedding — Block MotionHardQuestion 30555
Question
A bullet (0.05 kg, 400 m/s) embeds in a block (1 kg). Block+bullet slide 2 m on a rough surface (μ = 0.3, g = 10). Initial speed of block after impact:
- A19.05 m/sCorrect
- B18.2 m/s
- C20 m/s
- D15.38 m/s
Correct answer
19.05 m/s
Explanation
Combined speed from momentum: v = 0.05×400/1.05 ≈ 19.05 m/s. Then energy dissipation by friction gives sliding distance of about (19.05²)/(2×3) ≈ 60.7 m, not 2 m — so friction calculation uses v²/(2μg) = 19.05²/6 ≈ 60 m, inconsistent with 2 m given. The question likely intends the initial combined speed to be found from the sliding distance: v = √(2μgd) = √(2×0.3×10×2) = √12 ≈ 3.46 m/s, which gives bullet speed = 1.05×3.46/0.05 ≈ 72.7 m/s.