Question

A 42 kg surfer jumps off the back of a 22 kg surfboard that is moving forward with a velocity of 5.2 m/s. After the surfer leaves the surfboard, it moves forward with a velocity of 6.1 m/s. With what velocity did the surfer leave?

Answer 1

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.

$m_{1}*v_{1}+m_{2}*v_{2}=m_{1}*v_{1}^{'}+ m_{2}*v_{2}^{'}$

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.

$(42*5.2)+(22*5.2)= (42*X)+(22*6.1)\\42*X = 198.6\\x = 4.72[m/s]$