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

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

Step-by-step 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)} = (-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!