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A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 55.0m/s , and it leaves the bat traveling to the left at an angle of 40? above horizontal with a speed of 60.0m/s . The ball and bat are in contact for 1.85ms .Part AFind the horizontal component of the average force on the ball. Take the x-direction to be positive to the rightPart BFind the vertical component of the average force on the ball.

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lcmendozaf

Answer:

A. Fx = -7913.08 N

B. Fy = 3022.84 N

Explanation:

By conservation of the linear momentum:

ΔP = m*(Vf - Vo) = F*Δt

where:

m = 0.145 kg

Vf = [-60*cos(40°), 60*sin(40°)] m/s

Vo = [55, 0] m/s

F = [Fx, Fy]

Δt = 0.00185 s

With these values we can solve for F on both axis.

On the x-axis:

[tex]Fx = m*(Vfx - Vox)/\Delta t = 0.145*(-45.96 - 55)/0.00185 = -7913.08 N[/tex]

On the y-axis:

[tex]Fy = m*(Vfy - Voy)/\Delta t = 0.145*(38.57 - 0)/0.00185 = 3022.84 N[/tex]

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