A pitcher throws a 0.200 kg ball so that its speed is 17.0 m/s and angle is 40.0° below the horizontal when a player bats the ball directly toward the pitcher with velocity 48.0 m/s at 30.0° above the horizontal. Assume î to be along the line from the batter to the pitcher and ĵ to be the upward vertical direction. (Express your answers in vector form.) (a) Determine the impulse (in N · s) delivered to the ball. I

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24-10-2019
Physics

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A pitcher throws a 0.200 kg ball so that its speed is 17.0 m/s and angle is 40.0° below the horizontal when a player bats the ball directly toward the pitcher with velocity 48.0 m/s at 30.0° above the horizontal. Assume î to be along the line from the batter to the pitcher and ĵ to be the upward vertical direction. (Express your answers in vector form.) (a) Determine the impulse (in N · s) delivered to the ball. I

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moya1316 moya1316 · Answer 1 · 2019-10-28T10:42:36+01:00

Answer:

I = (5.71 i ^ + 6.99j ^) N s

Explanation:

The concept of impulse is very useful for problems where shocks occur, the impulse is equal to the change of moment of the system

I = Δp = m [tex]v_{f}[/tex] -m v₀

Let's fix an xy coordinate system, where y is vertical and we decompose the velocities

Initial

sin (-40) = [tex]v_{oy}[/tex] / vo

cos (-40) = v₀ₓ / Vo

[tex]v_{oy}[/tex] = 17.0 sin (-40)

[tex]v_{oy}[/tex] = -10.93 m / s

Vox = 17 cos (-40)

Vox = 13.02 m / s

Final

sin 30 = [tex]v_{fy}[/tex] / vf

cos 30 = [tex]v_{fx}[/tex] / Vf

[tex]v_{fy}[/tex] = 48.0 sin30

[tex]v_{fy}[/tex] = 24.0 m / s

[tex]v_{fx}[/tex] = 48 cos 30

[tex]v_{fx}[/tex] = 41.57 m / s

Let's calculate the momentum on each axis

X axis

Iₓ = m [tex]v_{fx}[/tex] -m [tex]v_{ox}[/tex]

Iₓ = 0.200 (41.57 - 13.02)

Iₓ = 5.71 N s

Axis y

[tex]I_{y}[/tex] = m [tex]v_{fy}[/tex] - m [tex]v_{oy}[/tex]

[tex]I_{y}[/tex] = 0.2000 (24 - (-10.93))

[tex]I_{y}[/tex] = 6.99 N s

I = (5.71 i ^ + 6.99j ^) n s

onyebuchinnaji onyebuchinnaji · Answer 2 · 2022-04-12T16:29:44+02:00

The impulse delivered to the ball is (-6.128i + 0.94j) N.s.

Impulse delivered to the ball

The impulse received by the ball is determined by applying the principle of conservation of angular momentum as shown below;

J = ΔP

Initial momentum of the ball from the pitcher to the batter is calculated as follows;

postive x - direction

[tex]P_x_i = mvcos\theta[/tex]

θ is angle of projection above the horizontal = 40 + 90 = 130⁰

[tex]P_x_i = 0.2 \times 17 \times cos(130)\\\\P_x_i = -2.185 \ kgm/s[/tex]

Final momentum of the ball from the batter to the pitcher is calculated as follows;

negative x - direction

[tex]P_x_f = -mv cos\theta[/tex]

[tex]P_x_f = - 0.2 \times 48 \times cos(30)\\\\P_x_f = -8.313 \ kgm/s[/tex]

Change in momnetum of the ball in x - direction

[tex]J_x = \Delta P_x = -8.313 - (-2.185)\\\\J_x = \Delta P_x = -6.128 \ kgm/s[/tex]

Resultant change in momentum of the ball

[tex]\Delta P = m(v - u)\\\\\Delta P = 0.2(48 - 17)\\\\\Delta P = 6.2 \ kgm/s[/tex]

Change in momnetum of the ball in y - direction

[tex]\Delta P = \sqrt{\Delta P_x^2 + \Delta P_y^2} \\\\\sqrt{\Delta P^2 - \Delta P_x^2 }= \Delta P_y \\\\\sqrt{(6.2)^2 - (-6.128)^2} = \Delta P_y\\\\0.94 \ kgm/s = \Delta P_y[/tex]

Impulse received by the ball

J = (-6.128i + 0.94j) kgm/s = (-6.128i + 0.94j) N.s

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