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JEE Main 201910 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Morning Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2019 · 10 January · Morning Shift · Physics · Center of Mass, Momentum and Collision

A piece of wood of mass 0.03kg is dropped from the top of a 100m...

A piece of wood of mass 0.03kg0.03kg is dropped from the top of a 100m100m height building. At the same time, a bullet of mass 0.02kg0.02kg is fired vertically upward, with a velocity 100ms−1,100m{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=10ms−2)\left(g = 10m{s}^{- 2}\right)A piece of wood of mass 0.03kg is dropped from the top of a 100m height building. At the same time, a bullet of massA piece of wood of mass 0.03kg is dropped from the top of a 100m height building. At the same time, a bullet of massA piece of wood of mass 0.03kg is dropped from the top of a 100m height building. At the same time, a bullet of mass

Options

  1. A40m40m
  2. B20m20m
  3. C10m10m
  4. D30m30m

Correct answer

A. 40m40m

Step-by-step solution

A piece of wood of mass 0.03kg is dropped from the top of a 100m height building. At the same time, a bullet of mass

Using relative velocity

A piece of wood of mass 0.03kg is dropped from the top of a 100m height building. At the same time, a bullet of mass

arel=0{a}_{rel} = 0

vrel=100{v}_{rel} = 100

Time tt for collision,

100=vrel×t100 = {v}_{rel} \times t

t=100100=1sect = \dfrac{100}{100} = 1sec

vbullet=100−1×10=90m/s{v}_{bullet} = 100 - 1 \times 10 = 90m / s

vparticle=10×1=10m/s{v}_{particle} = 10 \times 1 = 10m / s

S=100×1−12×10×1=95mS = 100 \times 1 - \dfrac{1}{2} \times 10 \times 1 = 95m

A piece of wood of mass 0.03kg is dropped from the top of a 100m height building. At the same time, a bullet of mass

Pi=Pf{P}_{i} = {P}_{f}

⇒90×0.02−10×0.03=0.05V⇒ 90 \times 0.02 - 10 \times 0.03 = 0.05V

V=30m/s⇒h=v22g=90020=45mV = 30m / s ⇒ h = \dfrac{{v}^{2}}{2g} = \dfrac{900}{20} = 45m

So from top of building

45−5=40m45 - 5 = 40m
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