A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 50.0 m/s, and it leaves the bat traveling to the left at an angle of 30° above horizontal with a speed of 65.0 m/s. If the ball and bat are in contact for 1.75 ms, find the horizontal and vertical components of the average force on the ball.

Solution 12E Step 1: Data given Mass of the ball m = 0.145 kg Velocity before touching bat v = 50 m/s 1 Velocity after touching bat v 2 65.0 m/s 0 Angle of ball = 30 Time of contact t = 1.75 ms We need to find the horizontal and vertical component of average force on the ball But we have momentum of the ball as Before touching the bat 1= m × v 1 1= 0.145 kg × 50 m/s 1= 7.25 Force is obtained as F = d/dt 1 F = 7.25/1.75 ms 1 F 1 4.14 N Hence the we have the force of the before touching the bat as 4.14 N After touching the bat = m × ( v + v ) 2 2x 2y v2x = v 2 sin v 2x= 65 m/s × sin 30 v 2x= 32.5 m/s v = v × cos 2y 2 v = 65 m/s × cos 30 2y v 2y= 56.29 m/s Thus we have 2 0.145 kg × ( 32.5 m/s + 56.29 m/s) = 12.9 2 Hence we have force as F 2 d/dt F 2 12.9/1.75 ms F 2 7.37 N Hence we get the average force as 7.37 N