Question

In a slow-pitch softball game, a 0.200-kg softball crosses the plate at 17.0 m/s at an angle of 35.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 46.0 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 ĵ 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?

Answer 1

a)
grom the given data

m = 0.200 kg
v1 = 17.0 m/s

v1x = -v1*cos(35) = -17*cos(35) = -13.9 m/s
v1y = -v1*sin(35) = -17*sin(35) = -9.75 m/s

v2x = v2*cos(30) = 46*cos(30) = 39.8 m/s
v2y = v2*sin(30) = 46*sin(30) = 23 m/s

Impulse in x-direction, Jx = change in momentum in x-direction

= m*v2x - m*v1x

= 0.2*39.8 - 0.2*(-13.9)

= 10.7 N.s

Impulse in y-direction, Jy = change in momentum in y-direction

= m*v2y - m*v1y

= 0.2*23 - 0.2*(-9.75)

= 6.55 N.s

Impulse delivered to the ball, J = sqrt(Jx^2 + Jy^2)

= sqrt(10.7^2 + 6.55^2)

= 12.5 N.s <<<<<<<----------------Answer

b) now use,Impulse = area under F-t curve

12.5 = (1/2)*4*10^-3*F_max + 20*10^-3*F_max + (1/2)*4*10^-3*F_max

12.5 = F_max*( (1/2)*4*10^-3 + 20*10^-3 + (1/2)*4*10^-3 )

12.5 = F_max*0.024

==> F_max = 12.5/0.024

= 521 N <<<<<<<----------------Answer