ENERGY AND MOMENTUM: A Windy Day at the Lake

Q: Wind blows a 43.0 kg raft across a lake at 1.1 m/s northward relative to the water underneath. It carries a passenger whose mass is 38.0 kg. The passenger begins to walk westward at 0.71 m/s. What is the final velocity of the raft relative to the water?

A: We first determine the momentum of the system. Momentum is a product of mass and velocity. This being the case, the combined mass of the raft and passenger must be used in our calculations:

p = mv

p = ( 43.0 kg + 38.0 kg )( 1.1 m/s )

p_s = 89.1 kg*m/s N

When the passenger begins walking westward, the momentum transferred to the raft causes its momentum to shift eastwardly. We must keep this in mind when final calculations are carried out:

p = ( 38.0 kg )( 0.71 m/s )

p_p = 27.0 kg*m/s 

We now use the Pythagorean Theorem to determine the final momentum of the system:

p²_f = p²_s + p²_p

p_f = √( p²_s + p²_p )

p_f = √[ ( 89.1 kg*m/s )² + ( 27.0 kg*m/s )² ]

p_f = √(  kg²*m²/s² )

p_f = 93.1 kg*m/s NE

The angle between the raft and its former northern trajectory is determined via use of trigonometry:

tan θ = opp/adj

tan^-1 ( opp/adj ) = θ

tan^-1 ( 27.0 kg*m/s / 89.1 kg*m/s ) = θ

tan^-1 ( 0.30 ) = θ

θ = 16.7^o 

We now determine the final velocity:

p = mv

( 93.1 kg*m/s NE / 81.0 kg ) = v

v = 1.15 m/s NE