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

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 40 above horizontal with a speed of 60.0 m/s . The ball and bat are in contact for 1.85 ms .

Part A Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right. Express your answer using two significant figures.

Part B Find the vertical component of the average force on the ball. Express your answer using two significant figures.

Explanation / Answer

This is a problem of impulse and momentum, for the first part we use the expression

I = F t = Pf – Pi bold indicate vectors

Part a)

We ask the component X force, so we wrote this component of the above expression

Fx t = Pfx – Pix (1)

Data

m = 0.145 Kg

Vxi = 45 m/s (right)

Vf= -60 m/s (left)

° = 40°

t = 1.85 ms = 1.85 10-3 s

We look for the components of the final speed with trigonometry

Cos 40 = Vfx/ Vf

Vfx = Vf Cos 40

Vfx = -60 Cos 40

Vfx = -45.96 m/s

Sin 40 = Vfy / Vf

Vfy = Vf Sin 40

Vfy = 60 Sin 40

Vfy = 38.57 m/s

Calculate in 1

Fx = (Pfx – Pix )/t

Fx = 0.145 (-45.96 -45) /1.85 10-3

Fx = - 7.13 103N

Part b) We use the same written expression for the Y axis for the vertical component

Fy = (Pfy – Piy )/t

Piy = 0

Pfy = m Vfy

Fy = (m Vfy – 0 )/t

Fy =0.145 38.57 /1.85 10-3

Fy = 3.02 103N

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