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Chapter 8: Problem 4

In a certain men's track and field event, the shotput has a mass of 7.30 \(\mathrm{kg}\) and is released with a speed of 15.0 \(\mathrm{m} / \mathrm{s}\) at \(40.0^{\circ}\) above the horizontal over a man's straight left leg. What are the initial horizontal and vertical components of the momentum of this shotput?

Short Answer

Expert verified
The horizontal component of the momentum is approximately 83.88 kg路m/s, and the vertical component is approximately 70.37 kg路m/s.

Step by step solution

01

Identify Given Information

We have the mass of the shotput, \( m = 7.30 \, \text{kg} \), and its release speed, \( v = 15.0 \, \text{m/s} \), at an angle \( \theta = 40.0^\circ \) above the horizontal.
02

Calculate Horizontal Component of Velocity

The horizontal component of the velocity is given by the formula \( v_x = v \cdot \cos(\theta) \). Substitute the known values to find \( v_x = 15.0 \cdot \cos(40.0^\circ) \approx 11.49 \, \text{m/s} \).
03

Calculate Vertical Component of Velocity

The vertical component of the velocity is calculated using the formula \( v_y = v \cdot \sin(\theta) \). Substitute the values to find \( v_y = 15.0 \cdot \sin(40.0^\circ) \approx 9.64 \, \text{m/s} \).
04

Calculate Horizontal Component of Momentum

The horizontal component of momentum is \( p_x = m \cdot v_x \). Substitute the mass and horizontal velocity to find \( p_x = 7.30 \times 11.49 \approx 83.88 \, \text{kg} \cdot \text{m/s} \).
05

Calculate Vertical Component of Momentum

The vertical component of momentum is \( p_y = m \cdot v_y \). Substitute the mass and vertical velocity to find \( p_y = 7.30 \times 9.64 \approx 70.37 \, \text{kg} \cdot \text{m/s} \).

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Key Concepts

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

Vertical Component
When discussing the vertical component in physics, especially in projectile motion like shotput, it involves breaking down vectors into smaller parts. For the exercise, the vertical component focuses on the shotput's momentum moving directly upwards or downwards. It doesn't consider any sideways movement. This component is crucial because it determines how high and how long the shotput stays in the air.
To find the vertical component of velocity, use the sine function related to angles in trigonometry:
  • First, recognize the overall velocity and the angle at which the shot is released.
  • Apply the formula: \[ v_y = v \cdot \sin(\theta) \] where \( v_y \) is the vertical component, \( v \) is the total speed, and \( \theta \) is the launch angle.
From there, the vertical component of momentum is another step:
  • It's a product of the mass and the calculated vertical velocity component: \[ p_y = m \cdot v_y \]
Understanding the vertical component helps grasp how forces like gravity affect the projectile.
Horizontal Component
The horizontal component of a projectile like a shotput describes how far it travels along the ground path. This component is crucial in determining the distance covered since it doesn't go up or down but purely horizontal.
For the horizontal component of velocity, you rely on the cosine function, which reflects the angle's role in projecting speed:
  • Identify the initial speed and angle.
  • Use the formula: \[ v_x = v \cdot \cos(\theta) \] where \( v_x \) is the horizontal speed, \( v \) is the overall speed, and \( \theta \) is the angle of projection.
Similarly, to find the horizontal component of momentum:
  • Multiply the mass by this horizontal velocity: \[ p_x = m \cdot v_x \]
This part of momentum remains unchanged (ignoring air resistance) throughout the flight, illustrating how the shotput moves across the field.
Shotput Physics
Shotput physics combines principles of physics with athletic practice. It revolves around launching a heavy spherical object 鈥� the shot 鈥� with maximum distance achieved through optimal force application and angle. The throw involves:
  • Calculating momentum and decomposing it into vertical and horizontal components.
  • Understanding the interplay between speed, angle, and mass to achieve the farthest possible distance.
  • Using a balance of maximum force and optimal angle of projection, usually around 40 to 42 degrees for typical shotput events.
Mastery of shotput physics enables athletes to adjust their techniques for better performance, while showcasing the practical application of trigonometry and momentum in sports.
Vector Decomposition
Vector decomposition refers to breaking a vector into two components: vertical and horizontal. This technique is invaluable in analyzing projectile motion like a shotput throw. Here's how it works:
  • Imagine the entire motion as a single vector at a specific angle; this includes both speed and direction.
  • Decompose this vector along axes: one for horizontal movement and another for vertical.
  • The trigonometric functions sine and cosine link angles to velocities and are essential tools in this process.
By decomposing vectors, you can better predict the behavior of moving objects:
  • Calculate separate components for comprehensive analysis.
  • Use this method across various fields from engineering to sports.
Vector decomposition grants detailed insight into each movement aspect, assisting in precise calculations and predictions of motion patterns.

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Most popular questions from this chapter

A \(4.00-\mathrm{g}\) bullet, traveling horizontally with a velocity of magnitude 400 \(\mathrm{m} / \mathrm{s}\) , is fired into a wooden block with mass 0.800 \(\mathrm{kg}\) , initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 120 \(\mathrm{m} / \mathrm{s} .\) The block shides a distance of 45.0 \(\mathrm{cm}\) along the surface from its initial position. (a) What is the coefficient of kinetic friction between block and surface? (b) What is the decrease in kinetic energy of the bullet? (c) What is the kinetic energy of the block at the instant after the bullet passes through it?

Three odd-shped blocks of chocolate have the following masses and center-of- mass coordinates: (1) \(0.300 \mathrm{kg},(0.200 \mathrm{m}\) \(0.300 \mathrm{m} ) ;(2) 0.400 \mathrm{kg},(0.100 \mathrm{m},-0.400 \mathrm{m}) ;(3) 0.200 \mathrm{kg}\) \((-0.300 \mathrm{m}, 0.600 \mathrm{m}) .\) Find the coordinates of the center of mass of the system of three chocolate blocks.

Three coupled railroad cars roll along and couple with a fourth car, which is initially at rest. These four cars roll along and couple with a fifth car initially at rest. This process continues until the speed of the final collection of railroad cars is one-fifth the speed of the initial three railroad cars. All the cars are identical. Ignoring friction, how many cars are in the final collection?

Changing Mass. An open-topped freight car with mass \(24,000 \mathrm{kg}\) is coasting without friction along a level track. It is raining very hard, and the rain is falling vertically downward. Originally, the car is empty and moving with a speed of 4.00 \(\mathrm{m} / \mathrm{s}\) . What is the speed of the car after it has collected 3000 \(\mathrm{kg}\) of rainwater?

In a volcanic eruption, a 2400 -kg boulder is thrown vertically upward into the air. At its highest point, it suddenly explodes (due to trapped gases) into two fragments, one being three times the mass of the other. The lighter fragment starts out with only horizontal velocity and lands 274 \(\mathrm{m}\) directly north of the point of the explosion. Where will the other fragment land? Neglect any air resistance.
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