Question

A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the right at 50.0 m/s , and it leaves the bat traveling to the left at an angle of 30.0 above horizontal with a speed of 65.0 m/s. A. What is the horizontal component of the impulse the bat imparts to the ball? B. What is the vertical component of the impulse the bat imparts to the ball? C. If the ball and bat are in contact for 1.75 , find the horizontal component of the average force on the ball. D. If the ball and bat are in contact for 1.75 , find the vertical component of the average force on the ball.

Answer 1

Answer:

a)    Iₓ = 15.41 N s

, b)   I_{y} = 4.75 N s

, c)  Fₓ = 26.97 N

, d)  F_{y} = 2.71 N

Explanation:

The momentum is

        I = F t = ΔP

a) the horizontal component of the momentum is

      Iₓ = Dpₓₓₓx = m vₓₓ - m v₀

      v₀ₓ = 50 m / s

Let's use trigonometry for the final velocity components

     Sin 30 = / v

     Cos 30 = vₓ / v

      v_{y} = v sin 30

      v_{y} = 65 sin 30

      v_{y} = 32.5 m / s

      vₓ = v cos 30

       vₓ = 65 cos 30

       vₓ = 56.29 m / s

The momentum is

         Iₓ = 0.145 (-50) - 0.145 56.29

         Iₓ = 15.41 N s

b) the vertical component of the impulse is

      I_{y} = m v_{y} - m v_{oy}

      I_{y} = 0.145 32.5 -0

      I_{y} = 4.75 N s

c) the contact time is t = 1.75 s, let's look for the average horizontal force

     Fₓ = Iₓ / t

     Fₓ = 15.41 / 1.75

     Fₓ = 26.97 N

d)    F_{y} = I_{y} / t

      F_{y} = 4.75 / 1.75

      F_{y} = 2.71 N