Abat strikes a 0.145-kg baseball. just before impact, the ball is traveling horizontally to the right at a speed of 40.0 m/s , and it leaves the bat traveling to the left at an angle of 40 above horizontal with a speed of 65.0 m/s . the ball and bat are in contact for 1.85 msa- 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. b- find the vertical component of the average force on the ball. the positive y-axis points upwards. express your answer using two significant figures.

A. Average horizontal force on the ball is 2.36N

B. Average vertical force on the ball is 1.80N

Explanation:

given

mass of ball M1= 0.45kg

velocity of ball V1= 40m/s

velocity of bat V2= 65m/s

mass of bat= ?

assume g = 10m/s2

∅ = 40

to calculate for mass of the bat, we will use the newton second law of motion which states that momentum in A= momentum in B

M1 * V1 = M2 * V2

inputing the parameters to get M2

40x0.45 = M2 x 65

M2 = 0.28kg

mass of bat= 0.28kg

the average horizontal force on the ball is calculated using the resultant force methods

for inclined force to the horizontal = M2 *g * cos∅

for the horizontal force is = M1 * g

Average horizontal force = -M2 *g * cos∅ + M1 * g

                                          =-(0.28 * 10 * cos 40) + (0.45 * 10)

                                          =-2.145 + 4.5 = 2.36N

B. the average vertical force on the ball is calculated using the resultant force methods

for inclined force to the vertical = M2 *g * sin∅

the ball did not move in vertical direction

Average vertical force = M2 *g * sin∅

                                          =(0.28 * 10 * sin 40)

                                          =1.80N