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A \(0.30 \mathrm{~kg}\) softball has a velocity of \(15 \mathrm{~m} / \mathrm{s}\) at an angle of \(35^{\circ}\) below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while in contact with the bat if the ball leaves with a velocity of (a) \(20 \mathrm{~m} / \mathrm{s}\), vertically downward, and (b) \(20 \mathrm{~m} / \mathrm{s}\), horizontally back toward the pitcher?

Short Answer

Expert verified
In scenario (a), the change in momentum is significant due to the vertical exit velocity. In scenario (b), it is significant due to the reversal of horizontal direction.

Step by step solution

01

Understand the Problem

We need to calculate the change in momentum of a softball before and after being hit by the bat in two scenarios. The initial conditions are a softball with a mass of 0.30 kg, an initial velocity of 15 m/s at an angle of 35° below the horizontal. In scenario (a), the ball leaves vertically downwards at 20 m/s and in scenario (b), it leaves horizontally back towards the pitcher at 20 m/s.
02

Calculate Initial Momentum Components

Calculate the initial velocity components of the softball. The initial velocity is 15 m/s and is at an angle of 35° below the horizontal.The horizontal component is:\[ v_{ix} = v_i \cos(35°) = 15 \times \cos(35°) \]The vertical component is:\[ v_{iy} = v_i \sin(35°) = 15 \times \sin(35°) \]Using these, the initial momentum components can be calculated by multiplying each by the mass (0.30 kg).
03

Compute Initial Momentum

Calculate the initial horizontal momentum \( p_{ix} \) and vertical momentum \( p_{iy} \):\[ p_{ix} = 0.30 \times 15 \times \cos(35°) \]\[ p_{iy} = 0.30 \times 15 \times \sin(35°) \]
04

Calculate Final Momentum for Scenario (a)

Scenario (a) states that the ball leaves vertically downward at 20 m/s. Thus, the horizontal velocity component is 0 m/s and the vertical velocity is 20 m/s downward.The horizontal momentum is:\[ p_{fx} = 0.30 \times 0 = 0 \]The vertical momentum is:\[ p_{fy} = 0.30 \times (-20) \](Note: The velocity is negative because it is opposite to the original downward direction defined as positive.)
05

Calculate Change in Momentum for Scenario (a)

Compute the change in momentum in both the horizontal and vertical directions.Horizontal change in momentum:\[ \Delta p_x = p_{fx} - p_{ix} = 0 - (0.30 \times 15 \times \cos(35°)) \]Vertical change in momentum:\[ \Delta p_y = p_{fy} - p_{iy} = 0.30 \times (-20) - (0.30 \times 15 \times \sin(35°)) \]Finally, calculate the magnitude of the change in momentum:\[ \Delta p = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2} \]
06

Calculate Final Momentum for Scenario (b)

In scenario (b), the final velocity is 20 m/s horizontally back towards the pitcher. Thus, the vertical velocity component is 0 m/s and the horizontal velocity is -20 m/s (since it is opposite the original direction).Horizontal momentum:\[ p_{fx} = 0.30 \times (-20) \]Vertical momentum:\[ p_{fy} = 0.30 \times 0 = 0 \]
07

Calculate Change in Momentum for Scenario (b)

Compute the change in momentum in both the horizontal and vertical directions for scenario (b).Horizontal change in momentum:\[ \Delta p_x = p_{fx} - p_{ix} = 0.30 \times (-20) - (0.30 \times 15 \times \cos(35°)) \]Vertical change in momentum:\[ \Delta p_y = p_{fy} - p_{iy} = 0 - (0.30 \times 15 \times \sin(35°)) \]Finally, calculate the magnitude of the change in momentum:\[ \Delta p = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2} \]
08

Conclusion

After solving the equations for both scenarios, calculate the total change in momentum for both scenarios and compare. The resulting answers indicate the magnitude of change in momentum for each condition described in parts (a) and (b) of the problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Momentum Components
In the study of physics, momentum is expressed as the product of an object's mass and velocity. To fully understand the change in momentum for objects moving at angles, we need to break down momentum into its components: horizontal and vertical. These components are crucial as they allow us to analyze motion in two dimensions independently.

For a softball moving at an angle, its velocity vector is split into horizontal and vertical components. This is done using trigonometric functions:
  • The horizontal component: computed as the cosine of the angle times the velocity, or \( v_{ix} = v_i \cos(\theta) \)
  • The vertical component: obtained as the sine of the angle times the velocity, or \( v_{iy} = v_i \sin(\theta) \)
When the mass of the object is considered, these components turn into momentum components, exactly like the calculation done for the softball's initial and final momentum.
Calculating Velocity
Velocity is not just about speed; it includes direction. When calculating velocities for different scenarios in momentum problems, it's essential to correctly define the direction. For example, a downward or backward movement would require assigning a negative sign to velocity.

In the given exercise, we start with an initial velocity and then observe changes as the ball is hit. Here's how you can calculate initial velocity components:
  • Initial horizontal velocity component: \( v_{ix} = 15 \cos(35^\circ) \)
  • Initial vertical velocity component: \( v_{iy} = 15 \sin(35^\circ) \)
For scenarios where the ball is hit changing its angle of motion, separate calculations are performed, considering the new direction of the ball, whether vertical or horizontal. Adjusting velocity in calculations is critical to achieving accurate momentum change results.
Scenario Analysis
Scenario analysis is vital for understanding the real-life application of physics problems. In this exercise, two scenarios show different outcomes based on how the ball is hit by the bat.

**Scenario (a):**When the ball leaves vertically downward, its final horizontal velocity is zero. This leads to zero horizontal momentum, simplifying calculations for the horizontal change. The vertical velocity becomes \(-20\, \text{m/s}\) as it goes directly downward.

**Scenario (b):**Conversely, this scenario lets the hit ball move horizontally back towards the pitcher. Here, the vertical velocity becomes zero, and the final horizontal velocity is \(-20\, \text{m/s}\), going in the opposite direction to the original horizontal component.
  • Such as defining directions carefully, where opposing movements clarify why some values are negative.
  • Points out how different direction leads to distinctly different momentum impact.
Scenario-specific analysis helps in identifying these nuances, making sense of changes in momentum.
Components of Vector Resolution
Breaking vectors into components is crucial when dealing with momentum changes, especially with angled motions. Vector resolution is the process of simplifying a vector into horizontal and vertical parts, allowing easier scalar operations.

In physics problems like this, vector resolution helps manage complexities involving angles:
  • Use trigonometric identities (sine and cosine) to determine the vector's component along each axis.
  • Contributes to simplifying momentum calculations by treating each directional component independently.
  • Permits comparisons between initial and final states of the object, ultimately simplifying the momentum change equation.
By understanding vector resolution, you'll more effectively address problems involving changes in direction, ultimately leading to a comprehensive understanding of momentum in multiple dimensions.

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