Question

A bat strikes a 0.145- kg 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 55.0 m/s . The ball and bat are in contact for 1.85 ms. Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right. Find the vertical component of the average force on the ball..

Answer 1

Given Data.

Mass of ball m = 0.145 kg

Initial Speed of ball v1 = 45 m/s

Speed of ball after strike v2 = 55 m/s

The angle of the path of the ball after strike with horizontal = 40 degree

Contact time of ball and bat t = 1.85 ms

We can solve the above problem with the help of newton's second law.

Where Net force FN = (P2 - P1)/t

Where P1 = momentum before strike P2 = momentum after strike t = time of contact.

To apply the law we have to resolve the velocities of ball in vertical and horizontal directions. Therefore Component of velocities are as follow

As initially ball was moving in right direction horizontally therefore

V1x = V1 = 45 m/s

After strike velocity of ball makes an angle of 40 degree with horizontal. Therefore

V2x = -V2cos40 = -42.132 m/s ......( as ball is moving in left direction so negative sign is used)

V2y = V2sin40 = 35.353 m/s

As P = MV = M(Vx + Vy)

therefore, P1 = 0.145*(45)x = 6.525x

P2 = 0.145*( -42.132x + 35.353y) = -6.109x + 5.126y

By Newton's second law

FN =  [-6.109x + 5.126y - (6.525x)]/(1.85 * 10-3) = (-12.634x + 5.126y)/(1.85 * 10-3)

= -6829.18x + 2770.81y

Therefore,

Horizontal component of force Fx = 6829.18 N to the left

Vertical component of force Fy= 2770.81 N upward