A bat strikes a 0.145kg baseball. Just before impact, the ball is traveling horizontally to the right at 60.0 m/s , and it leaves the bat traveling to the left at an angle of 40degrees above horizontal with a speed of 65.0 m/s . The ball and bat are in contact for 1.75 milliseconds.

1.) Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right
F{sub}x =

2.) Find the vertical component of the average force on the ball.
F{sub}y =

To solve this problem, we can use the principles of Newton's laws of motion.

1.) Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right.
We can use the impulse-momentum principle, which states that the change in momentum of an object is equal to the impulse experienced by the object. The impulse is the product of the average force and the contact time.

The change in momentum of the ball in the x-direction is given by:
Δp{sub}x = m(v{sub}i{sub}x - v{sub}f{sub}x)

Where:
m = mass of the ball = 0.145 kg
v{sub}i{sub}x = initial velocity of the ball in the x-direction = 60.0 m/s (to the right)
v{sub}f{sub}x = final velocity of the ball in the x-direction (to the left)

We need to find the final velocity of the ball in the x-direction. Since the ball moves in an upward direction at an angle of 40 degrees above the horizontal, the vertical component of its final velocity is given by:
v{sub}f{sub}y = v{sub}f * sinθ

Where:
v{sub}f = magnitude of the final velocity of the ball = 65.0 m/s

So the final velocity of the ball in the x-direction can be found using the horizontal component of the final velocity:
v{sub}f{sub}x = v{sub}f * cosθ

Where:
θ = angle of the final velocity of the ball above the horizontal = 40 degrees

Therefore, we have:
v{sub}f{sub}x = 65.0 m/s * cos(40 degrees)
v{sub}f{sub}x = 65.0 m/s * 0.766 = 49.8 m/s

Now we can calculate the change in momentum in the x-direction:
Δp{sub}x = 0.145 kg * (60.0 m/s - 49.8 m/s)
Δp{sub}x = 0.145 kg * 10.2 m/s
Δp{sub}x = 1.479 kg m/s

Since we know that the impulse is equal to the average force multiplied by the contact time, we can rearrange the impulse-momentum principle equation to solve for the average force:

Impulse = F * t

F = Impulse / t

Where:
Impulse = Δp{sub}x = 1.479 kg m/s
t = contact time = 1.75 milliseconds = 1.75 * 10^-3 s

So, substituting the values into the equation, we get:
F{sub}x = 1.479 kg m/s / (1.75 * 10^-3 s)
F{sub}x ≈ 845.71 N

Therefore, the horizontal component of the average force on the ball is approximately 845.71 Newtons to the right.

2.) Find the vertical component of the average force on the ball.
The change in momentum of the ball in the y-direction is given by:
Δp{sub}y = m(v{sub}i{sub}y - v{sub}f{sub}y)

Where:
v{sub}i{sub}y = initial velocity of the ball in the y-direction = 0 (since it doesn't have any vertical component initially)
v{sub}f{sub}y = final velocity of the ball in the y-direction (upward)

We already found the final velocity of the ball in the y-direction to be:
v{sub}f{sub}y = v{sub}f * sinθ
v{sub}f{sub}y = 65.0 m/s * sin(40 degrees)
v{sub}f{sub}y = 65.0 m/s * 0.642 = 41.83 m/s

Now we can calculate the change in momentum in the y-direction:
Δp{sub}y = 0.145 kg * (0 - 41.83 m/s)
Δp{sub}y = -0.145 kg * 41.83 m/s
Δp{sub}y = -6.06 kg m/s

Similarly, using the impulse-momentum principle equation, we can find the average force in the y-direction:

F{sub}y = Δp{sub}y / t
F{sub}y = -6.06 kg m/s / (1.75 * 10^-3 s)
F{sub}y ≈ -3462.86 N

Therefore, the vertical component of the average force on the ball is approximately -3462.86 Newtons (downward).

To solve this problem, we need to use the principle of conservation of momentum. When two objects collide, the total momentum before the collision is equal to the total momentum after the collision.

1.) Find the horizontal component of the average force on the ball:
Before the collision, the ball is traveling horizontally to the right, so the initial momentum in the x-direction is given by:

p{sub}xi = m*v{sub}xi

where m is the mass of the ball and v{sub}xi is the initial velocity in the x-direction.

p{sub}xi = (0.145 kg)*(60.0 m/s) = 8.7 kg*m/s

After the collision, the ball is traveling to the left at an angle of 40 degrees above the horizontal, so we need to find the horizontal component of the final velocity, v{sub}xf.

v{sub}xf = v{sub}f*cos(theta)

where v{sub}f is the final velocity magnitude (65.0 m/s) and theta is the angle (40 degrees).

v{sub}xf = (65.0 m/s)*cos(40 degrees) = 49.7 m/s

Using the conservation of momentum, we can find the change in momentum, which is equal to the impulse (average force times the time of impact):

Impulse = change in momentum = p{sub}xf - p{sub}xi

The final momentum in the x-direction, p{sub}xf, is given by:

p{sub}xf = m*v{sub}xf

p{sub}xf = (0.145 kg)*(-49.7 m/s) = -7.2 kg*m/s

Impulse = (-7.2 kg*m/s) - (8.7 kg*m/s) = -15.9 kg*m/s

The average force in the x-direction, F{sub}x, can be calculated using the equation:

F{sub}x = Impulse / time

F{sub}x = (-15.9 kg*m/s) / (1.75*10^(-3) s) = -9085 N

Therefore, the horizontal component of the average force on the ball is -9085 N.

2.) Find the vertical component of the average force on the ball:
To find the vertical component of the average force, we need to consider the change in the vertical momentum of the ball.

The initial vertical momentum, p{sub}yi, is given by:

p{sub}yi = m*v{sub}yi

where v{sub}yi is the initial velocity in the y-direction.

Since the ball is initially moving horizontally, the initial velocity in the y-direction is 0 (no vertical component).

p{sub}yi = (0.145 kg)*(0 m/s) = 0 kg*m/s

The final vertical momentum, p{sub}yf, is given by:

p{sub}yf = m*v{sub}yf

where v{sub}yf is the final velocity in the y-direction.

Since the ball leaves the bat at an angle of 40 degrees above the horizontal, we need to find the vertical component of the final velocity, v{sub}yf.

v{sub}yf = v{sub}f*sin(theta)

v{sub}yf = (65.0 m/s)*sin(40 degrees) = 41.7 m/s

The change in vertical momentum is given by:

Impulse = change in momentum = p{sub}yf - p{sub}yi

Impulse = (m*v{sub}yf) - 0

Impulse = (0.145 kg)*(41.7 m/s) = 6.05 kg*m/s

The average force in the y-direction, F{sub}y, can be calculated using the equation:

F{sub}y = Impulse / time

F{sub}y = (6.05 kg*m/s) / (1.75*10^(-3) s) = 3460 N

Therefore, the vertical component of the average force on the ball is 3460 N.

answer