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A bullet of mass 0.02 kg moving horizontally at 400 m/s strikes and…
Question
A bullet of mass 0.02 kg moving horizontally at 400 m/s strikes and embeds itself in a stationary block of mass 1.98 kg that rests at the foot of a smooth incline on a frictionless surface. Immediately after the collision the bullet-plus-block combination slides up the smooth incline. Taking g = 10 m/s^2, find the maximum vertical height (in metres) the combination rises above its starting level.
✓ Verified answer: 0.8checked by our engine — not a guess
Step-by-step solution
Momentum conservation in the perfectly inelastic collision: v = m v0/(m + M) = 0.02*400/2.0 = 4 m/s.
On the smooth incline, KE converts fully to PE: (1/2)v^2 = g h, so h = v^2/(2g) = 16/20 = 0.8 m.
Final answer0.8
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