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?
2.78 m/s to the east
2.78 m/s to the west
0.360 to the west
0.360 to the east

Answer 1

Answer:

The final velocity of the skier and boat is 0.33 m/s to the east.

Explanation:

We can find the final velocity of the skier by conservation of linear momentum:

$m_{s}v_{s_{i}} + m_{b}v_{b_{i}} = m_{s}v_{s_{f}} + m_{b}v_{b_{f}}$

Where:  

$m_{s}$: is the mass of the water skier = 62.0 kg

$m_{b}$: is the mass of the boat = 775 kg

$v_{s_{i}}$: is the initial velocity of the skier = 4.50 m/s (as she leaves the dock)

$v_{b_{i}}$: is the initial velocity of the boat = 0 (it is at rest)

$v_{s_{f}}$: is the final velocity of the skier =?

$v_{b_{f}}$: is the final velocity of the boat =?

Since the final velocity of the skier is the same that the velocity of the boat ($v_{f}$) we have:

$m_{s}v_{s_{i}} + 0 = v_{f}(m_{s} + m_{b})$

[tex]v_{f} = \frac{m_{s}v_{s_{i}}}{m_{s} + m_{b}} = \frac{62.0 kg*4.50 m/s}{62.0 kg + 775 kg} = 0.33 m/s                  

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

I hope it helps you!