Answer:
The surfer leave the surferboard with a velocity of 4.72[m/s]
Explanation:
This problem is related to the Conservation of Momentum, and it can be calculated using the following equation.
[tex]m_{1}*v_{1}+m_{2}*v_{2}=m_{1}*v_{1}^{'}+ m_{2}*v_{2}^{'} \\[/tex]
Where:
m1 = mass of the surfer = 42[kg]
v1 = velocity of the surfer before jumping = 5.2 [m/s]
m2 = mass of the surfboard = 22 [kg]
v2 = velocity of the surfboard before jumping = 5.2 [m/s]
Now after jumping
m1 = mass of the surfer = 42[kg]
v1' = velocity of the surfer after jumping = x
m2 = mass of the surfboard = 22 [kg]
v2' = velocity of the surfboard after jumping = 6.1 [m/s]
Now replacing in the equation.
[tex](42*5.2)+(22*5.2)= (42*X)+(22*6.1)\\42*X = 198.6\\x = 4.72[m/s][/tex]
