Respuesta :
Answers:
a) [tex]65.075 kgm/s[/tex]
b) [tex]10.526 s[/tex]
c) [tex]61.82 N[/tex]
Explanation:
a) Impulse delivered to the ball
According to the Impulse-Momentum theorem we have the following:
[tex]I=\Delta p=p_{2}-p_{1}[/tex] (1)
Where:
[tex]I[/tex] is the impulse
[tex]\Delta p[/tex] is the change in momentum
[tex]p_{2}=mV_{2}[/tex] is the final momentum of the ball with mass [tex]m=0.685 kg[/tex] and final velocity (to the right) [tex]V_{2}=57 m/s[/tex]
[tex]p_{1}=mV_{1}[/tex] is the initial momentum of the ball with initial velocity (to the left) [tex]V_{1}=-38 m/s[/tex]
So:
[tex]I=\Delta p=mV_{2}-mV_{1}[/tex] (2)
[tex]I=\Delta p=m(V_{2}-V_{1})[/tex] (3)
[tex]I=\Delta p=0.685 kg (57 m/s-(-38 m/s))[/tex] (4)
[tex]I=\Delta p=65.075 kg m/s[/tex] (5)
b) Time
This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately [tex]1.0 cm=0.01 m[/tex]:
[tex]V_{2}=V_{1}+at[/tex] (6)
[tex]V_{2}^{2}=V_{1}^{2}+2ad[/tex] (7)
Where:
[tex]a[/tex] is the acceleration
[tex]d=0.01 m[/tex] is the length the ball was compressed
[tex]t[/tex] is the time
Finding [tex]a[/tex] from (7):
[tex]a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}[/tex] (8)
[tex]a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}[/tex] (9)
[tex]a=90.25 m/s^{2}[/tex] (10)
Substituting (10) in (6):
[tex]57 m/s=-38 m/s+(90.25 m/s^{2})t[/tex] (11)
Finding [tex]t[/tex]:
[tex]t=1.052 s[/tex] (12)
c) Force applied to the ball by the bat
According to Newton's second law of motion, the force [tex]F[/tex] is proportional to the variation of momentum [tex]\Delta p[/tex] in time [tex]\Delta t[/tex]:
[tex]F=\frac{\Delta p}{\Delta t}[/tex] (13)
[tex]F=\frac{65.075 kgm/s}{1.052 s}[/tex] (14)
Finally:
[tex]F=61.82 N[/tex]
Answers:
a) 65.125 Ns
b) 5.263 * 10^(-4) s
c) 123737.5 N
Explanation:
a) Impulse delivered to the ball F.dt
According to the Impulse-Momentum we have the following:
[tex]F*dt = m*(V_{2} - V_{1})[/tex]
Using the given data we insert in equation above:
[tex]Impulse = 0.685 kg (57 - (-38))\\\\Impulse = 65.1225 Ns[/tex]
Answer: 65.125 Ns
b)
This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 0.01m :
Using the kinematic equations for constant acceleration:
[tex](v_{f})^2 = (v_{i})^2 + 2*a*s[/tex]
Where:
vf = 0 (ball stops on the bat)
vi = 38 m/s
s = compression = 0.01 m
Using the equation above we compute acceleration:
[tex]a = \frac{(v_{f})^2 - (v_{i})^2}{2*s} \\\\a = \frac{0^2 - 38^2}{2*0.01} \\\\a = -72,200 m/s^2[/tex]
Using the acceleration to compute time:
[tex]v_{f} = v_{i} + a*t\\\\t = \frac{v_{f} - v_{i}}{a}\\\\t = \frac{0 - 38}{-72,200}\\\\t = 5.263*10^(-4) s[/tex]
Answer: 5.263*10^(-4) s
c)
According to Newton's second law of motion:
[tex]F_{avg} * dt = Impulse[/tex]
Using answer from part a and b
[tex]F_{avg} = \frac{Impulse}{dt} \\\\F_{avg} = \frac{65.125}{5.263*10^(-4)} \\\\F_{avg} = 123737.5 N[/tex]
Answer: 123737.5 N
