A helicopter of mass m_1 = 2.5 t hovers in the air. A man of…
A helicopter of mass $m_1 = 2.5\ \mathrm{t}$ hovers in the air. A man of mass $m_2 = 80\ \mathrm{kg}$ climbs a rope ladder hanging from it at a constant speed $v = 0.5\ \mathrm{m/s}$ relative to the ladder. With what velocity, and in which direction, does the helicopter move?
Explanation of Method
A man climbing a ladder under a hovering helicopter.
Step 1: Momentum stays zero
Before he starts, everything is at rest, and lift and gravity balance, so the total momentum stays zero. Let the helicopter move down at $u$. The man's velocity relative to the ground is then $v - u$ upward:
\[ m_2\,(v - u) - m_1 u = 0. \]
Step 2: Helicopter speed
\[ u = \frac{m_2 v}{m_1 + m_2} = \frac{80 \cdot 0.5}{2580}. \]
\[u \approx 0.0155\ \mathrm{m/s} = 1.55\ \mathrm{cm/s} \ \text{downward}\]
Result
\[\boxed{\,u \approx 0.0155\ \mathrm{m/s} = 1.55\ \mathrm{cm/s} \ \text{downward}\,}\]
Remark
About the original solution. It used the man's speed relative to the ladder as if it were relative to the ground, getting $u = m_2 v/m_1 = 0.016\ \mathrm{m/s}$. Since the ladder itself moves down, the man's ground speed is $v - u$, which puts $m_1 + m_2$ in the denominator. The difference is small here (0.0155 vs 0.016) only because the helicopter is much heavier than the man. (This assumes the pilot does not change the lift.)