A66.0 kg person throws a 0.0500 kg snowball forward with a ground speed of 34.0 m/s. a second person, with a mass of 56.0 kg, catches the snowball. both people are on skates. the first person is initially moving forward with a speed of 2.20 m/s, and the second person is initially at rest. what are the velocities of the two people after the snowball is exchanged? disregard the friction between the skates and the ice.

thrower

m/s

catcher

m/s

The velocity of the thrower, vf1 is 2.176m/s

The velocity of the catcher, vf2 is 0.03m/s

Explanation:

Parameters given:

Mass of thrower m1= 66kg

Mass of snowball mb = 0.05kg

Ground speed, speed of football, vfb = 34m/s

Mass of catcher, m2 = 56.0kg

Initial speed of thrower, vi1 = 2.2m/s

To find the final velocity of the thrower, we have to use the law of conservation of momentum in an isolated system,

Δp = 0

pf1 - pi1 = 0

pf1 = pi1

=> m1*vf1 + mb*vfb = m1*vi1 + mb*vib

Since the thrower of the snowball is holding the snowball, they are moving at the same initial velocity i.e. vib = vi1

m1*vf1 + mb*vfb = (m1 + mb)*vi1

m1*vf1 = (m1 + mb)*vi1 – mb*vfb

vf1 = [(m1 + mb)*vi1 - mb*vfb] / m1

vf1 = [2.2*(66 + 0.05) - 0.05*34]/66

vf1 = [145.31 - 1.7]/66

vf1 = 143.61/66

vf1 = 2.176m/s

To find the velocity of the catcher, we also have to use the law of conservation of momentum in an isolated system,

Δp = 0

pf2 - pi2 = 0

pf2 = pi2

=> m2*vf2 + mb*vfb = m2*vi2 + mb*vib

In this case, vfb = vf2 because the catcher and the snowball are moving at the same final velocity,

i.e. (m2 + mb)*vf2 = m2*vi2 + mb*vib

Since the catcher is initially at rest, vi2 = 0

=> (m2 + mb)*vf2 = mb*vib

vf2 = mb*vib/(m2 + mb)

vf2 = 0.05*34/(56 + 0.05)

vf2 = 1.7/56.05

vf2 = 0.03m/s

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down