Question part points submissions used a pitcher throws a 0.200 kg ball so that its speed is 19.0 m/s and angle is 40.0° below the horizontal when a player bats the ball directly toward the pitcher with velocity 46.0 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 = n · s (b) if the force on the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, and then decreases linearly to zero in another 4.00 ms, what is the maximum force (in n) on the ball?

The impulse is (10.88 i^ + 7.04 j^) N s

maximum force on the ball is  (4.53 10 2 i^ + 2.93 102 j ^) N  

Explanation:

In a problem of impulse and shocks we must use the impulse equation

       I = dp = pf-p₀         (1)

       p = m V

With we have vector quantities, let's decompose the velocities on the x and y axes

      V₀ = -19 m / s

      θ₀ = 40.0º  

      Vf = 46.0 m / s

      θf = 30.0º

Note that since the positive direction of the x-axis is from the batter to the pitcher, the initial velocity is negative and the angle of 40º is measured from the axis so it is in the third quadrant

      Vcx = Vo cos θ

      Voy = Vo sin θ

      Vox= -19 cos (40) = -14.6 m/s

      Voy = -19 sin (40) =  -12.2 m/s

      Vfx = 46 cos 30 = 39.8 m/s

      Vfy = 46 sin 30 =  23.0 m/s

   a) We already have all the data, substitute and calculate the impulse for each axis

      Ix = pfx -pfy

      Ix = m ( vfx -Vox)

      Ix = 0.200 ( 39.8 – (-14.6))

      Ix = 10.88 N s

      Iy = m (Vfy -Voy)

      Iy = 0.200 ( 23.0- (-12.2))

      Iy=  7.04 N s

In vector form it remains

       I =  (10.88 i^ + 7.04 j^) N s

   b) As we have the value of the impulse in each axis we can use the expression that relates the impulse to the average force and your application time, so we must calculate the average force in each interval.

         I = Fpro Δt

In the first interval

        Fpro = (Fm + Fo) / 2

With the Fpro the average value of the force, Fm the maximum value and Fo the minimum value, which in this case is zero

         Fpro = (Fm +0) / 2

In the second interval the force is constant

          Fpro = Fm

In the third interval

         Fpro = (0 + Fm) / 2

Let's replace and calculate

         I =  Fpro1 t1 +Fpro2 t2  +Fpro3 t3

         I = Fm/2 4 10⁻³ + Fm 20 10⁻³+ Fm/2 4 10⁻³  

         I = Fm  24 10⁻³ N s

         Fm = I / 24 10⁻³

         Fm = (10.88 i^ + 7.04 j^) / 24 10⁻³

         Fm = (4.53 10² i^ + 2.93 10² j ^) N

maximum force on the ball is  (4.53 10 2 i^ + 2.93 102 j ^) N  

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow