In a slow-pitch softball game, a 0.200-kg softball crosses the plate at 18.0 m/s
ID: 1582046 • Letter: I
Question
In a slow-pitch softball game, a 0.200-kg softball crosses the plate at 18.0 m/s at an angle of 40.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 50.0 m/s at 30.0° above the horizontal. (Take positive î to be the horizontal direction from the plate toward center field and take positive to be the upward vertical direction.) (a) Determine the impulse delivered to the ball. (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 on the ball?
Explanation / Answer
mass, m=0.2 kg
speed, u1=18 m/sec at 40 degrees
speed, u2=50 m.sec at 30 degrees
a)
components of momentum before hitting the ball
P1x=m*u1*cos(40)
=0.2*18*cos(40)
=2.76 kg*m/sec
and
P1y=-m*u1*sin(40)
=-0.2*18*sin(40)
=-2.31 kg*m/sec
components of momentum after hitting the ball
P2x=-m*u2*cos(30)
=-0.2*50*cos(30)
=-8.66 kg*m/sec
and
P2y=m*u2*sin(30)
=0.2*50*sin(30)
=5 kg*m/sec
impulse=change in momemtum,
=[(P2x-P1x)i + (P2y-p1y)j]
=(-8.66-2.76)i + (5-(-2.31))j
=-(11.42)i+(7.31)j kg.m/sec
b)
area under the graph is impulse,
I = 1/2*F*t1+ F*t2+ 1/2*F*t3
I=(1/2*4*10^-3+20*10^-3 + 1/2*4*10^-3)*F
-(11.42)i+(7.31)j=(0.024)*F
====> F=(-11.42)i+(7.31)/(0.024)
F=-(475.83)i + (304.58)j
Fmax=-(475.83)i + (304.58)j N