A baseball of weight 2.3 kg is moving horizontally at a speed of 150km/h when it is struck by the bat. It leaves the bat in a direction at an angle 35 degree above its incident path and with a speed of 180km/h.

a). Find the impulse of the force exerted on the ball.

b). Assuming the collision lasts for 1.5ms. What is the average force?

c). Find the change in momentum of the bat?

To tackle this problem, we need to break it down into manageable parts. We will calculate the impulse of the force exerted on the baseball, the average force during the collision, and the change in momentum of the bat. Let's go through each part step by step.

Impulse Calculation

Impulse is defined as the change in momentum of an object when a force is applied over a period of time. It can also be calculated as the product of the average force and the time duration of the force application. The formula for impulse (J) is:

J = Δp = p_final - p_initial

Step 1: Calculate Initial Momentum

The initial momentum (p_initial) of the baseball can be calculated using the formula:

p_initial = m * v_initial

Where:

• m = mass of the baseball = 2.3 kg
• v_initial = initial velocity in m/s = 150 km/h = 150 * (1000 m / 3600 s) = 41.67 m/s

Now substituting the values:

p_initial = 2.3 kg * 41.67 m/s = 95.83 kg·m/s

Step 2: Calculate Final Momentum

The final momentum (p_final) is calculated similarly, but we need to consider the angle and the final speed:

First, convert the final speed to m/s:

v_final = 180 km/h = 180 * (1000 m / 3600 s) = 50 m/s

Next, we find the components of the final velocity:

• v_final_x = 50 * cos(35°)
• v_final_y = 50 * sin(35°)

Calculating these components:

v_final_x ≈ 50 * 0.8192 ≈ 40.96 m/s

v_final_y ≈ 50 * 0.5736 ≈ 28.68 m/s

Now we can calculate the final momentum:

p_final = m * v_final = 2.3 kg * (40.96 m/s, 28.68 m/s) = (94.21 kg·m/s, 65.00 kg·m/s)

Step 3: Calculate Change in Momentum

Now we can find the change in momentum:

Δp = p_final - p_initial

Since the initial momentum is only in the x-direction:

Δp_x = 94.21 kg·m/s - 95.83 kg·m/s = -1.62 kg·m/s

Δp_y = 65.00 kg·m/s - 0 = 65.00 kg·m/s

The total change in momentum can be expressed as a vector:

Δp = (-1.62, 65.00) kg·m/s

Impulse of the Force

The impulse exerted on the ball is equal to the change in momentum:

J = Δp = (-1.62, 65.00) kg·m/s

Average Force Calculation

To find the average force (F_avg) exerted on the baseball during the collision, we use the formula:

F_avg = J / Δt

Where Δt is the duration of the collision, given as 1.5 ms (or 0.0015 s).

Calculating the average force:

F_avg_x = -1.62 kg·m/s / 0.0015 s ≈ -1080 N

F_avg_y = 65.00 kg·m/s / 0.0015 s ≈ 43333.33 N

Change in Momentum of the Bat

Assuming the bat is initially at rest and the collision is perfectly elastic, the change in momentum of the bat will be equal in magnitude and opposite in direction to the change in momentum of the ball:

Δp_bat = -Δp_ball = (1.62, -65.00) kg·m/s

Thus, the change in momentum of the bat is:

Δp_bat = (1.62, -65.00) kg·m/s

In summary, we have calculated the impulse exerted on the baseball, the average force during the collision, and the change in momentum of the bat, providing a comprehensive understanding of the dynamics involved in this scenario.