Question

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the...

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 45.0 m/s , and it leaves the bat traveling to the left at an angle of 35 ∘ above horizontal with a speed of 65.0 m/s . The ball and bat are in contact for 1.65 ms .

A. Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right (two sig figs)

B. Find the vertical component of the average force on the ball. (Two sig figs)

a)

Initial horizontal component of the momentum is

Pix = mv

Pix = 0.145 kg x 45 m/s = 6.525 kg.m/s

final horizontal component of the momentum is

Pfx = mvf

Pfx = -0.145 kg x 65 m/s x cos(35) = -7.72 kg.m/s

horizontal component of average force on ball is

Fx = Px / t

Fx = Pfx - Pix / t

Fx = ( -7.72 kg.m/s - 6.525 kg.m/s) / (1.65 x 10^-3 s)

Fx = -8.63 x 10^3 N

direction of force is towards left

b)

Initial vertical component of the momentum is

Pix = mv

Pix = 0

final verticle component of the momentum is

Pfy = mvf

Pfy = 0.145 kg x 65 m/s x sin(35) = 5.41 kg.m/s

vertical component of average force on ball is

Fy = Py / t

Fy = Pfy - Piy / t

Fy = ( 5.41 kg.m/s - 0 kg.m/s) / (1.65 x 10^-3 s)

Fy = 3.28 x 10^3 N

direction is upward

direction of force is towards left