Question

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)

Answer 1

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