Respuesta :
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
The impulse delivered to the ball when a player bats the ball directly toward the pitcher is,
[tex]I=({10.88\hat i+7.04\hat j})[/tex]
What is impulse force?
Impulse force is the product of average force and the total time for which the force maintained is called the impulse force.
- (a) Determine the impulse (in N · s) delivered to the ball. I = N·s
The initial speed is 19.0 m/s and angle is 40.0° below the horizontal. Thus, the initial speed in both the direction are,
[tex]v_ix=(-19)\cos (40)\\v_ix=-14.6\rm m/s\\v_iy=(-19)\sin(40)\\v_ix=-12.2\rm m/s\\[/tex]
The final velocity is 46.0 m/s at 30.0° above the horizontal. Thus, the initial speed in both the direction are,
[tex]v_fx=46\cos (30)\\v_fx=39.8\rm m/s\\v_fy=46\sin(30)\\v_fx=23\rm m/s\\[/tex]
Change in momentum, In x direction,
[tex]\Delta P_x=P_2-P_1\\\Delta P_x=m(V_{fx}+V_{ix})\\\Delta P_x=0.200(39.8+(-14.6))\\\Delta P_x=10.88\Rm N[/tex]
In y direction,
[tex]\Delta P_y=P_2-P_1\\\Delta P_y=m(V_{fy}+V_{iy})\\\Delta P_y=0.200(23+(-12.2))\\\Delta P_y=7.04\Rm N[/tex]
Hence, the impulse (in N · s) delivered to the ball is,
[tex]I=({10.88\hat i+7.04\hat j})[/tex] N-s.
- (b) The maximum force (in N) on the ball-
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,
For the linearly increasing force, the average force will be divided by 2. Thus, the average force for this three force, in this condition can be given as,
[tex]F=\dfrac{4+0}{2}+20+\dfrac{4+0}{2}\\F=24\rm Ns[/tex]
As, the maximum force is the ratio of impulse force to the average force. Therefore, the maximum force is
[tex]F_{max}=\dfrac{10.88\hat i7.04\hat j}{24\times10^{-3}}\rm N[/tex]
Hence, the impulse delivered to the ball when a player bats the ball directly toward the pitcher is,
[tex]I=({10.88\hat i+7.04\hat j})[/tex]
Learn more about the impulse force here;
https://brainly.com/question/20586658
