The Man M weights 150 lb and jumps onto the boat B, which is originally at rest. If he has a horizontal component of velocity 3 ft/s just before he enters the boat, determine the weight of the boat if it has a velocity of 2 ft/s once the man enters it.

m₂ = 75 lb

Explanation:

Applying the law of conservation of momentum in horizontal direction to the system of man and boat.

Initial momentum of the System = Final Momentum of the System

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where,

m₁ = weight of man = 150 lb

m₂ = weight of boat = ?

u₁ = initial speed of man = 3 ft/s

u₂ = initial speed of boat = 0 ft/s

v₁ = final speed of man = 2 ft/s (same as boat)

v₂ = final speed of boat = 2 ft/s

Therefore,

(150 lb)(3 ft/s) + (m₂)(0 ft/s) = (150 lb)(2 ft/s) + (m₂)(2 ft/s)

450 lb ft/s - 300 lb ft/s = (m₂)(2 ft/s)

m₂ = (150 lb ft/s)/(2 ft/s)

m₂ = 75 lb

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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