Question

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 40.0 m/s; when it leaves the bat, the ball is traveling to the left at an angle of 30\(^\circ\) above horizontal with a speed of 52.0 m/s. If the ball and bat are in contact for 1.75 ms, find the horizontal and vertical components of the average force on the ball.

Short answer

The average horizontal force is \(-7043.0 \text{ N}\) and vertical force is \(2154.3 \text{ N}\).

Step-by-step solution

Determine Initial and Final Velocity Components

First, let's break down the initial and final velocities of the baseball into their horizontal and vertical components. Initially, the baseball is moving to the right with a velocity of 40.0 m/s, meaning its horizontal velocity is \( v_{ix} = 40.0 \) m/s and its vertical velocity \( v_{iy} = 0 \) m/s as it has no vertical motion. After striking, the baseball is traveling to the left at a 30° angle above the horizontal with a speed of 52.0 m/s. The horizontal component of this velocity is \( v_{fx} = -52.0 \cdot \cos(30^\circ) \approx -45.0 \) m/s (used negative because it moves to the left), and the vertical component is \( v_{fy} = 52.0 \cdot \sin(30^\circ) = 26.0 \) m/s.

Apply Impulse-Momentum Theorem

Impulse-momentum theorem states that the change in momentum equals the impulse applied. The change in horizontal momentum is \( m(v_{fx} - v_{ix}) \) and in vertical momentum is \( m(v_{fy} - v_{iy}) \). Calculate these changes: for horizontal momentum, \[\Delta p_{x} = 0.145 \times (-45.0 - 40.0) = -12.325 \text{ kg.m/s} \], and for vertical momentum, \[\Delta p_{y} = 0.145 \times (26.0 - 0) = 3.77 \text{ kg.m/s} \].

Calculate Average Force Components

The impulse experienced by the ball equals the average force times the time of contact \( F_{avg} \Delta t = \Delta p \). Using this formula, calculate the average horizontal force component: \( F_{avg,x} = \frac{\Delta p_{x}}{\Delta t} = \frac{-12.325}{1.75 \times 10^{-3}} = -7043.0 \text{ N} \), and the average vertical force component: \( F_{avg,y} = \frac{\Delta p_{y}}{\Delta t} = \frac{3.77}{1.75 \times 10^{-3}} = 2154.3 \text{ N} \). Remember that the negative sign in the horizontal force indicates that the force is in the opposite direction of the initial motion.

Key concepts

Average Force

When discussing the average force in the context of the Impulse-Momentum Theorem, the aim is to understand how force is applied over the time during which an object is in contact with another. In this example, the interaction between a baseball and a bat is key to understanding the average force.
The average force is not constant, but it gives a simplified picture of the complex forces during the contact time of 1.75 milliseconds (very brief!). To find this, we use the impulse, which is the product of the average force and the contact time, and relate it to momentum change.

• The impulse experienced by the object is calculated using: \[ F_{\text{avg}} \Delta t = \Delta p \] where \( F_{\text{avg}} \) is the average force and \( \Delta p \) is the change in momentum.
• In the exercise, the average horizontal and vertical force components were calculated by rearranging the above equation to solve for \( F_{\text{avg}} \).

Understanding the average force helps model real-world dynamic interactions by simplifying the calculations while still capturing the essence of the event.

Velocity Components

Breaking down velocity into components is crucial for solving problems where angles are involved, such as when the baseball changes direction after being struck by the bat. Velocity components help visualize and calculate individual directions of an object's motion.
In physics, velocity is often split into horizontal and vertical components. These components form the legs of a right triangle, with the resultant velocity being the hypotenuse.

• The horizontal component of velocity (\( v_x \)) is found using \( v_x = v \cdot \cos(\theta) \), where \( \theta \) is the angle of motion relative to the horizontal.
• The vertical component (\( v_y \)) is found with \( v_y = v \cdot \sin(\theta) \).

For our baseball: after impact, the initial horizontal velocity was positive (to the right), while the final horizontal velocity became negative (to the left), indicating a direction change. The vertical velocity, initially zero, increased, reflecting the upward motion after the hit. Decomposing velocity into these components helps in accurately computing the momentum changes, which are essential to determining the forces involved.

Momentum Change

Momentum change is a central concept in the Impulse-Momentum Theorem, where momentum of an object is altered due to applied force within a period. Momentum is defined as the product of an object's mass and its velocity, \( p = mv \).
In the given exercise, the change in momentum is used to calculate the average force experienced by the baseball.

• The change in momentum in a specific direction is given by: \( \Delta p = m(v_{f} - v_{i}) \), where \( m \) is the mass, \( v_{f} \) is the final velocity, and \( v_{i} \) is the initial velocity in that direction.
• For horizontal movement, this involves a switch from a rightward 40.0 m/s to a leftward 52.0 m/s, resulting in significant momentum change.
• Vertically, the baseball starts at 0 m/s and moves upwards at a speed influenced by the angle and speed change.

Understanding momentum change is crucial for evaluating how forces exerted over time can alter an object’s motion. This provides insights into not just the magnitude of forces but also their directional impact on the motion involved.