A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity $100 \mathrm{~ms}^{-} { }^{1}$, 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: $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Conservation of Momentum

A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity

100 ms^{-}^{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 ms^{- 2})

(A)20 m
(B)30 m
(C)40 m✓ Correct
(D)10 m

Answer: (C)

Solution

First find when they meet: the wood falls from the

100

m roof as

100 - \frac{1}{2} g t^{2}

and the bullet climbs from the ground as

100 t - \frac{1}{2} g t^{2}

; equating gives

100 = 100 t

, so they collide at

t = 1

s. At that instant both are at

95

m, which is

5

m below the roof, with the wood moving down at

10

m/s and the bullet up at

90

m/s. Momentum conservation gives the stuck-together speed

v = \frac{0.03 ( - 10 ) + 0.02 ( 90 )}{0.05} = 30

m/s upward, and from

95

m the lump rises a further

\frac{v ^{2}}{2 g} = \frac{900}{20} = 45

m to reach

140

m above the ground. That peak stands

140 - 100 = 40

m above the top of the building.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Conservation of Momentum

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