A 65.0 kg ice skater moving to the right with a velocity of 2.5 m/s throws a 0.15 kg snowball to the right with the velocity of 32.0 m/s relative to the ground.

a. What is the velocity of the ice skater after throwing the snowball? disregard the friction between the skates and the ice.
b. A second skater initially at rest with a mass of 60.0 kg catches the snowball. What is the velocity of the second skater after catching the snowball in a perfectly inelastic collision?

To solve these problems, we can use the principles of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, as long as no external forces are acting on the system.

a. To find the velocity of the ice skater after throwing the snowball, we can consider the system of the ice skater and the snowball as a whole. The momentum before throwing the snowball is given by the equation:

Total momentum before = (Mass of ice skater + Mass of snowball) * Velocity of ice skater before

Total momentum before = (65.0 kg + 0.15 kg) * 2.5 m/s
Total momentum before = 162.87 kg·m/s

After throwing the snowball, the ice skater's velocity changes. Let's assume the velocity of the ice skater after throwing the snowball is v.

Total momentum after = (Mass of ice skater + Mass of snowball) * Velocity of ice skater after

Total momentum after = (65.0 kg + 0.15 kg) * v
Total momentum after = 65.15 kg * v

Since the total momentum before is equal to the total momentum after, we can equate these two expressions:

Total momentum before = Total momentum after
162.87 kg·m/s = 65.15 kg * v

Now, we can solve for v:

v = 162.87 kg·m/s / 65.15 kg
v ≈ 2.50 m/s

Therefore, the velocity of the ice skater after throwing the snowball is approximately 2.50 m/s to the right.

b. In a perfectly inelastic collision, the two objects stick together after the collision, so they have the same final velocity.

For the second part of the problem, the momentum before the collision is the same as in part a:

Total momentum before = (Mass of ice skater + Mass of snowball) * Velocity of ice skater before
Total momentum before = (65.0 kg + 0.15 kg) * 2.5 m/s
Total momentum before = 162.87 kg·m/s

After the collision, the two skaters and the snowball move together as a single mass. Let's assume the final velocity is v'.

Total momentum after = (Mass of ice skater 1 + Mass of ice skater 2 + Mass of snowball) * Velocity after collision
Total momentum after = (65.0 kg + 60.0 kg + 0.15 kg) * v'

Again, since the total momentum before is equal to the total momentum after, we can equate these two expressions:

Total momentum before = Total momentum after
162.87 kg·m/s = (65.0 kg + 60.0 kg + 0.15 kg) * v'

Now, we can solve for v':

v' = 162.87 kg·m/s / (65.0 kg + 60.0 kg + 0.15 kg)
v' ≈ 0.756 m/s

Therefore, the velocity of the second skater after catching the snowball in a perfectly inelastic collision is approximately 0.756 m/s to the right.