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QJEEM19Phc03f: JEE Main 2019 Physics - Class 11 JEEM19Phc03f Solution: A piece of wood of mass 0.03 kg is dropped from the top of a…

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)

Multiple Choice Options

  • A 30 m
  • B 40 m
  • C 20 m
  • D 10 m

Step-by-Step Verified Solution

JEE Main 2019 (Online) 10th January Morning Slot Physics - Center of Mass and Collision Question 98 English Explanation 1
Time taken for the particles to collide,

t = fVrel=100100=1{f \over {{V_{rel}}}} = {{100} \over {100}} = 1 sec

Speed of wood just before collision = gt = 10 m/s

& speed of bullet just before collision v-gt

= 100 −- 10 = 90 m/s

Now, conservation of linear momentum just before and after the collision -

−- (0.02) (1v) + (0.02) (9v) = (0.05)v

⇒\Rightarrow    150 = 5v

⇒\Rightarrow    v = 30 m/s

Max. height reached by body h = v22g{{{v^2}} \over {2g}}

JEE Main 2019 (Online) 10th January Morning Slot Physics - Center of Mass and Collision Question 98 English Explanation 2

h = 30×302×10{{30 \times 30} \over {2 \times 10}} = 45m

∴\therefore   Height above tower = 40 m

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