Question

A baseball pitcher delivers a fastball that crosses the plate at an angle of \(7.25^{\circ}\) below the horizontal and a speed of \(88.5 \mathrm{mph}\). The ball (of mass \(0.149 \mathrm{~kg}\) ) is hit back over the head of the pitcher at an angle of \(35.53^{\circ}\) above the horizontal and a speed of \(102.7 \mathrm{mph}\). What is the magnitude of the impulse received by the ball?

Short answer

In this exercise, a baseball is pitched at 88.5 mph and hit back at 102.7 mph, with the pitched ball being 7.25 degrees below the horizontal and the hit ball being 35.53 degrees above the horizontal. By converting the speeds to m/s and calculating the initial and final velocity components of the ball, applying the impulse-momentum theorem, and using Pythagorean theorem to find the magnitude of the impulse, we determine that the ball received an impulse of 4.67 kg·m/s.

Step-by-step solution

Convert speeds from mph to m/s

Since we are given the speeds in mph, let's first convert them to m/s, which is the standard unit for speed in physics. To do this, we'll use the conversion factor \(1 \mathrm{mph} \approx 0.44704 \mathrm{m/s}\): Pitched speed: \(88.5 \mathrm{mph} \times 0.44704 \frac{\mathrm{m/s}}{\mathrm{mph}} = 39.57 \mathrm{m/s}\) Hit speed: \(102.7 \mathrm{mph} \times 0.44704 \frac{\mathrm{m/s}}{\mathrm{mph}} = 45.91 \mathrm{m/s}\)

Determine the initial velocity components

To find the initial velocity components of the ball, we need to use the angle \(7.25^{\circ}\) below the horizontal and the initial speed of \(39.57 \mathrm{m/s}\): Initial horizontal component: \(v_{1x} = 39.57 \cos(7.25^{\circ}) = 38.94 \mathrm{m/s}\) Initial vertical component: \(v_{1y} = 39.57 \sin(-7.25^{\circ}) = -4.97 \mathrm{m/s}\) The initial velocity vector components are \((38.94, -4.97)\) m/s.

Determine the final velocity components

Now, we'll use the angle \(35.53^{\circ}\) above the horizontal and the final speed of \(45.91 \mathrm{m/s}\) to find the final velocity components: Final horizontal component: \(v_{2x} = 45.91 \cos(35.53^{\circ}) = 37.47 \mathrm{m/s}\) Final vertical component: \(v_{2y} = 45.91 \sin(35.53^{\circ}) = 26.47 \mathrm{m/s}\) The final velocity vector components are \((37.47, 26.47)\) m/s.

Calculate the change in momentum

Now that we have the initial and final velocity components, we can find the change in momentum for each component. The ball's mass is given as \(0.149 \mathrm{kg}\), so we use the following formula: Change in momentum \(\Delta p = m\Delta v = m(v_2 - v_1)\) Change in horizontal momentum: \(\Delta p_x = 0.149(37.47 - 38.94) = -0.219 \mathrm{kg\cdot m/s}\) Change in vertical momentum: \(\Delta p_y = 0.149(26.47 + 4.97) = 4.665 \mathrm{kg\cdot m/s}\) The change in momentum vector components are \((-0.219, 4.665)\) kg\(\cdot\)m/s.

Calculate the magnitude of the impulse

Finally, we'll use the Pythagorean theorem to find the magnitude of the total impulse: Impulse magnitude \(I=\sqrt{(\Delta p_x)^2 + (\Delta p_y)^2} = \sqrt{(-0.219)^2 + (4.665)^2} = 4.67 \mathrm{kg\cdot m/s}\) So, the magnitude of the impulse received by the ball is \(4.67 \mathrm{kg\cdot m/s}\).

Key concepts

Momentum Change

When we discuss momentum change in physics, we're essentially exploring the variation of an object's momentum, which is a product of its mass and velocity. For objects in motion, like the baseball in the exercise, momentum change occurs when an external force, like the bat's impact, results in a change of speed or direction.

Mathematically, the momentum change is expressed as \( \Delta p = m(v_2 - v_1) \), where \( m \) is the mass, \( v_1 \) is the initial velocity, and \( v_2 \) is the final velocity. In vector form, this calculation considers each component of velocity, both horizontal and vertical, since velocity is a directional quantity. In the solution steps, the momentum change is calculated for each orthogonal direction separately, indicating how both the speed and direction of the ball are altered upon being hit.

Velocity Components

Velocity components play a crucial role in analyzing motion in two dimensions. The full velocity of an object is split into horizontal and vertical components when we're dealing with angles or projectile motion. These components are separately determined by using trigonometric functions: sine for the vertical and cosine for the horizontal, based on the given angle.

In the exercise at hand, we calculate the components initially when the ball is pitched and finally when it is hit back. For instance, the horizontal velocity component after the ball is hit is found using \( v_{2x} = V \cos(\theta) \), where \( V \) is the velocity magnitude and \( \theta \) is the angle with respect to the horizontal plane. Understanding these components is critical because they allow us to analyze the motion in two dimensions and to eventually calculate the momentum changes in each direction.

Unit Conversion

Unit conversion is fundamental in physics to ensure that equations and calculations are consistent and comparable. Since the standard unit of speed in physics is meters per second (m/s), and our exercise provides speeds in miles per hour (mph), we must convert these values.

We do this by using the conversion factor \( 1 \mathrm{mph} = 0.44704 \mathrm{m/s} \). By multiplying the given mph value by this factor, we switch the unit to m/s. Without proper unit conversion, the numerical results would be meaningless and could not be compared to standard values or used in formulas that are defined with specific units.

Pythagorean Theorem in Physics

The Pythagorean theorem may be familiar from geometry as a way to relate the lengths of sides in a right triangle. However, this theorem also has a significant place in physics, especially when calculating the resultant of orthogonal vectors, like changes in momentum or forces. In our problem, the change in horizontal and vertical momenta are perpendicular components, similar to the sides of a right triangle.

The magnitude of the total momentum change is therefore computed as the hypotenuse, using the Pythagorean theorem: \( I = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2} \). In essence, this calculation gives us the 'straight-line' momentum change the ball experiences despite it moving in a two-dimensional trajectory. The theorem is an elegant method for deriving the magnitude of vectors from their orthogonal components.