Question

A 45kg skater and a 60kg skater are standing still, holding hands on frictionless ice. They push away from each other in opposite directions. If the 45kg skater was measured traveling at 3.5 m/s, how fast was the 60 kg skater traveling?

Answer 1

Answer:

The 60 kg skater is traveling at 2.63 m/s in the opposite direction from the 45 kg skater.

Explanation:

The velocity of the 60 kg skater can be found by conservation of linear momentum:

$P_{i} = P_{f}$

$m_{a}v_{i_{a}} + m_{b}v_{i_{b}} = m_{a}v_{f_{a}} + m_{b}v_{f_{b}}$  (1)

Where:

$m_{a}$: is the mass of the first skater = 45 kg        

$m_{b}$: is the mass of the second skater = 60 kg        

$v_{i_{a}}$: is the initial speed of the first skater = 0 (he is standing still)

$v_{i_{b}}$: is the initial speed of the second skater = 0 (he is standing still)

$v_{f_{a}}$: is the final speed of the first skater = 3.5 m/s

$v_{f_{b}}$: is the final speed of the second skater =?

By replacing the above values into equation (1) and solving for $v_{f_{b}}$ we have:

$0 = m_{a}v_{f_{a}} + m_{b}v_{f_{b}}$

$v_{f_{b}} = \frac{-m_{a}v_{f_{a}}}{m_{b}} = \frac{-45 kg*3.5 m/s}{60 kg} = -2.63 m/s$

The minus sign is because the 60 kg skater is moving in the opposite direction from the other skater.

Therefore, the 60 kg skater is traveling at 2.63 m/s in the opposite direction from the 45 kg skater.

I hope it helps you!