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 =10ms?)
(1) 30 m
(2) 10 m
(3) 40 m
(4) 20 m
Answers
Answer:
40 m
Explanation:
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building
Distance covered in t sec using S = ut + (1/2)at²
a = g = 10 m/s² u = 0 m/s as dropped
S = 5t²
a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 m/s
So distance covered by bullet in t sec
= 100t - (1/2)(10)t²
= 100t - 5t²
Both distance should add up 100 m as height of building
5t² + 100t - 5t² = 100
=> t = 1 sec
Velocity of wood after 1 sec = V = u + at
= 0 + 10(1) = 10 m/s Down ward
Distance covered = 5t² = 5 m
=> Collison happens just 5 m below top of building
Velocity of bullet after 1 sec
= 100 - 10*1
= 90 m/s upward
Momentum before collision = 0.02* 90 - 0.03*10
= 1.8 - 0.3
= 1.5 kgm/s
Momentum after collision = (0.03 + 0.02)V = 0.05V
0.05V = 1.5
=> V = 30 m/s (upwards)
Velocity at top = 0 m/s
a = - 10 m/s²
using V² - U² = 2aS
=> 0² - 30² = 2(-10)S
=> S = 45 m
the maximum height to which the combined system reaches above the top of the building = 45 - 5 = 40 m
Answer:
the correct answer is 40 m.