Answer:
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
