Question

A 62.0 kg water skier at rest jumps from the dock into a 775 kg boat at rest on the east side of the dock. If the velocity of the skier is 4.50 m/s as she leaves the dock, what is the final velocity of the skier and boat?

Answer 1

Answer:

v = 0.33 m/s

Explanation:

The final velocity of the skier and boat can be calculated by conservation of the lineal momentum:

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}$

Where:

m1: is the mass of the water skier = 62.0 kg

m2: is the mass of the boat = 775 kg

v1i: is the initial velocity of the water skier = 4.50 m/s    

v1f: is the final velocity of the water skier = ?

v2i: is the initial velocity of the boat = 0

v2f: is the final velocity of the boat = ?

Since the final velocity of the skier is the same that the final velocity of the boat, we have:      

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{f} + m_{2}v_{f}$  

$m_{1}v_{1i} + 0 = v_{f}(m_{1} + m_{2})$  

$v_{f} = \frac{62.0 kg*4.50 m/s}{(62.0 kg + 775 kg)}$  

$v_{f} = 0.33 m/s$  

Therefore, the final velocity of the skier and boat is 0.33 m/s.

I hope it helps you!