Question

4.
A balloon of mass 100 kg is stationary at a height of 100 m above the
ground. A man is standing on the rope ladder hanging from the bottom of
the balloon. As the man of mass 50 kg climbs up with speed of 0.6 m/s
relative to the rope, the balloon
1) Remains stationary
2) Moves down with a speed of 0.3 m/s
3) Moves up with a speed of 0.4 m/s
4) Moves with an acceleration of 0.1m/s2
​

Answer 1

Hello dear,

◆ Answer -

(2) Moves down with a speed of 0.3 m/s

◆ Explaination -

# Given -

mb = 100 kg

mm = 50 kg

vm = 0.6 m/s

vb = ?

# Solution -

Initially when the man is standing still, total momentum of the system is -

pi = mm.um + mb.ub

pi = 50 × 0 + 100 × 0

pi = 0 kgm/s

Later when the man is climbing up, total momentum of the system is -

pf = mm.vm + mb.vb

pf = 50 × 0.6 + 100 × vb

pf = 30 + 100vb

According to law of conservation of momentum -

pi = pf

0 = 30 + 100vb

vb = -30/100

vb = -0.3 m/s

Therefore, balloon will move down with 0.3 m/s speed.

Thanks dear..