A piece of wood of mass 0.03 k g is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 k g is fired vertically upward, with a velocity 100 m s - 1 , from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is: g = 10 m s - 2

Question

A piece of wood of mass 0.03 k g is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 k g is fired vertically upward, with a velocity 100 m s - 1 , from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is: g = 10 m s - 2, 40 m, 20 m, 10 m, 30 m

MARKS App · Accepted Answer

Using relative velocity a r e l = 0 v r e l = 100 Time t for collision, 100 = v r e l × t t = 100 100 = 1 s e c v b u l l e t = 100 - 1 × 10 = 90 m / s v p a r t i c l e = 10 × 1 = 10 m / s S = 100 × 1 - 1 2 × 10 × 1 = 95 m P i = P f ⇒ 90 × 0.02 - 10 × 0.03 = 0.05 V V = 30 m / s ⇒ h = v 2 2 g = 900 20 = 45 m So from top of building 45 - 5 = 40 m

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