A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 55.0 m/s, and it leaves the bat traveling to the left at an angle of 30 degrees above horizontal with a speed of 60.0 m/s. The ball and bat are in contact for 1.65 ms.

Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right. Use two significant figures

Respuesta :

Answer:

Fx=9399.39 iN

Explanation:

[tex]m_{b}=0.145 kg\\v_{b1}=55 \frac{m}{s}\\v_{b2}=60\frac{m}{s}\\\alpha=30[/tex]

Δt=1.65ms ⇒[tex]1.65x10^{-3} s[/tex]

Momentum of the motion is

[tex]p=m*v[/tex]

so:

∑F= p2-p1/Δt

[tex]p1=0.145kg*55 i\\p1=7.975 i[/tex]

[tex]p2=0.145*(-60*cos(30)i+60*(sen(30)j)\\p2=0.145*(-51.96i+30j)\\p2=-7.534i+4.35j[/tex]

[tex]F= \frac{(-7.534i+4.35j)-(7.975)}{1.65x10^{-3}s}\\Fx=\frac{15.509i}{1.65x^{-3}s} \\Fx=9399.39i N\\Fy=\frac{4.35j}{1.65x^{-3}s} \\Fy=2636.36j N[/tex]

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