Question

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 35.0 m/s ; when it leaves the bat, the ball is traveling to the left at an angle of 33.0 ∘ above horizontal with a speed of 59.0 m/s . The ball and bat are in contact for 1.69 ms .
Part A

Find the horizontal and vertical components of the average force on the ball. Let +x be to the right and +y be upward

Answer 1

Answer:

Fx= 7242.6 N

Fy= 2751.47 N

Explanation:  

m=0.145 kg

u= 35 m/s

θ=33°

v=59 m/s

t= 1.69 ms

The change linear momentum in x direction

ΔPx = m ( u + v cosθ)

Now by putting the values

ΔPx = m ( u + v cosθ)

ΔPx = 0.145 ( 35 + 59 cos 33°)

ΔPx =12.24 kg.m/s

The change linear momentum in y direction

ΔPy = m v sinθ

Now by putting the values

ΔPy = m v sinθ

ΔPy = 0.145 x 59 sin 33°)

ΔPy =4.65 kg.m/s

From second law of Newtons

The rate of change of linear momentum is known as force.

$F=\dfrac{dP}{dt}$

The force in x direction

$F_x=\dfrac{dP_x}{dt}$

$F_x=\dfrac{12.24}{1.69\times 10^{-3}}$

Fx= 7242.6 N

The force in y direction

$F_y=\dfrac{dP_y}{dt}$

$F_y=\dfrac{4.65}{1.69\times 10^{-3}}$

Fy= 2751.47 N