A65kg person throws a 0.045kg snowball forward with a ground speed of 30m/s. a second person, with a mass of 60kg, catches the snowball. both people are on skates. the first person is initially moving forward with a speed of 2.5 m/s, and the second person is initially at rest. what are the velocities of the two people after the snowball is exchanged? disregard friction between the skates and the ice.

v₁ = 2.48m/s, v₂ = 0.02m/s

Explanation:

Momentum p must be conserved. p = mv

1) First person throwing the snow ball. The momentum before the throw:

p = (65kg + 0.045kg) * 2.5 m/s

The momentum after the throw:

p = 65kg * v₁ + 0.045kg * 30m/s

Solving for the velocity v₁ of person 1:

v₁ = ((65kg + 0.045kg) * 2.5 m/s - 0.045kg * 30m/s) / 65kg = 2.48m/s

2) Second person catching the ball. The momentum before the catch:

p = 0.045kg * 30m/s + 60kg * 0m/s

The momentum after the catch:

p = (60kg + 0.045kg) * v₂

Solving for velocity v₂ of person 2:

v₂ = 0.045kg * 30m/s / (60kg + 0.045kg) = 0.02 m/s

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures

i believe the answer is electric potential.

explanation: