Part 1: the following questions refer to the motion of a baseball. call towards home plate positive and towards the outfield negative. while being thrown, a net force of 125 n acts on a baseball (mass = 141 g) for a period of 0.04 sec. what is the magnitude of the change in momentum of the ball? part 2: if the initial speed of the baseball is v = 0.0 m/s, what will its speed be when it leaves the pitcher's hand? part 3: when the batter hits the ball, a net force of 1165 n, opposite to the direction of the ball's initial motion, acts on the ball for 0.005 s during the hit. what is the final velocity of the ball?

These are three questions and three answers

1. Question 1.

Part 1:

5 N*s

Explanation:

Data:

F = 125 N
m = 141 g = 0.141 kg
t = 0.04 s

Δp =?

2) Formula

Impulse = F.t =

Impulse = Δp

3) Solution:

=> Δp => F.t

Δp = 125 N * 0.04 s = 5 N.m

2) Question 2

Part 2: If the initial speed of the baseball is v = 0.0 m/s, what will its speed be when it leaves the pitcher's hand?

35.5 m/s positive (toward the homelate)

Explanation:

1) Data:
vi = 0.0 m/s
vf = ?

2) Formula

Δp = mΔv

3) Solution:

Δv = Δp / m

Δv = 5 N*s / 0.141 kg = 35.5 m/s

Δv = vf - vi => vf = Δv + vi = 35.5 m/s + 0 = 35.5 m/s, positive because it is toward the homeplate.

3) Question 3

Part 3: When the batter hits the ball, a net force of 1165 N, opposite to the direction of the ball's initial motion, acts on the ball for 0.005 s during the hit. What is the final velocity of the ball?

- 76.8 m/s (negative because it is towards the field)

Explanation:

1) Data:

F = 1165 N
t = 0.005 s
vf = ?

2) Formulas

I = f.t

I = Δp

Δp = mΔv

3) Solution

I = F.t = 1165 N * 0.005s = 5,825 N*s

Δp = I = 5,825 N*s

Δp = mΔv => Δv = Δp / m = 5,825 N*s / 0.141 kg = 41.3 m/s (negative because it is towards the field)

Δv = vf - vi => vf = Δv + vi = - 41.3 m/s - 35.5 m/s = - 76.8 m/s

It's because 3 if it was 10 it wouldn't look right do it goes up to 4d

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