Question

A 0.14 kg baseball moves horizontally with a speed of 34 m/s toward a bat. After striking the bat the ball moves vertically upward with 40% of its initial speed. Find the direction and magnitude of the impulse delivered to the ball by the bat.

° (measured from the initial direction of the ball)

kg·m/s

Answer 1

We know,

impulse= force*time= mv_final - mv_initial

here, m= 0.14kg and V_initial = -34m/s and V_final = 0.4*34 =13.6m/s

it will be helpful for us if we use vector components,

Hence, IMPULSE= (f_xi + f_yj)*t = mv_final - mv_initial = 0.14*13.6*j - (-34)*0.14*i

on, compairing we get

f_x*t= 34*0.14 and f_y*t= 0.14*13.6

magnitude = [(f_x*t)² + (fy*t)²]^1/2 ( that is square root of f_x and f_y)

hence, magnitude = 5.76 N

and direction= tan alpha= f_y/f_x = 0.4

alpha= tan^-1 0.4

alpha = 21.80 degree

hence. magnitude= 5.76 N and direction= 21.80 degree upwards