Skip to main content
Ch. 09 - Linear Momentum
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 09 - Linear MomentumProblema 42b
Capitolo 9, Problema 42b

A 144-g baseball moving 28.0 m/s strikes a stationary 4.85-kg brick resting on small rollers so it moves without significant friction. After hitting the brick, the baseball bounces straight back, and the brick moves forward at 1.10 m/s. Find the total kinetic energy before and after the collision.

Guida verificata passo dopo passo
1
Convert the mass of the baseball from grams to kilograms. Since 1 gram = 0.001 kilograms, the mass of the baseball is \( m_1 = 144 \times 10^{-3} \, \text{kg} \).
Calculate the initial kinetic energy of the baseball using the formula \( KE = \frac{1}{2} m v^2 \). Here, \( m_1 = 0.144 \; \text{kg} \) and \( v_1 = 28.0 \; \text{m/s} \).
Determine the initial kinetic energy of the brick. Since the brick is stationary before the collision, its initial velocity \( v_2 = 0 \; \text{m/s} \), so its initial kinetic energy is zero.
Calculate the final kinetic energy of the baseball after the collision. Use the same formula \( KE = \frac{1}{2} m v^2 \), but substitute the final velocity of the baseball (which is not given directly in the problem but can be determined if needed).
Calculate the final kinetic energy of the brick after the collision using \( KE = \frac{1}{2} m v^2 \), where \( m_2 = 4.85 \; \text{kg} \) and \( v_2 = 1.10 \; \text{m/s} \). Add the final kinetic energies of the baseball and the brick to find the total kinetic energy after the collision.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
10m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = 1/2 mv², where m is the mass and v is the velocity. In this scenario, both the baseball and the brick have kinetic energy before and after the collision, which must be calculated to analyze the system's energy changes.
Video consigliato:
Percorso guidato
06:07
Intro to Rotational Kinetic Energy

Conservation of Momentum

The principle of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on it. In this collision, the momentum before the impact must equal the momentum after the impact, allowing us to relate the velocities of the baseball and the brick to find their respective speeds post-collision.
Video consigliato:
Percorso guidato
05:58
Conservation Of Momentum

Elastic vs. Inelastic Collisions

Collisions can be classified as elastic or inelastic based on whether kinetic energy is conserved. In this case, since the baseball bounces back and the brick moves forward, it suggests an inelastic collision where some kinetic energy is transformed into other forms of energy, such as sound or heat, which must be accounted for when calculating total kinetic energy before and after the collision.
Video consigliato:
Percorso guidato
08:56
Intro To Elastic Collisions
Pratica correlata
Domanda del libro di testo

The force on a bullet along the barrel of a firearm is given by the formula F = [740 ― (2.3 x 10⁵ s⁻¹ ) t] N over the time interval t = 0 to t = 3.0 x 10⁻³ s. Plot a graph of F versus t for t = 0 to t = 3.0 ms. Use the graph to estimate the impulse given the bullet.

1664
views
Domanda del libro di testo

A 144-g baseball moving 28.0 m/s strikes a stationary 4.85-kg brick resting on small rollers so it moves without significant friction. After hitting the brick, the baseball bounces straight back, and the brick moves forward at 1.10 m/s. What is the baseball’s speed after the collision?

1265
views
Domanda del libro di testo

A bullet of mass m = 0.0010 kg embeds itself in a wooden block with mass M = 0.999 kg, which then compresses a spring (k = 140 N/m) by a distance 𝓍 = 0.050 m before coming to rest. The coefficient of kinetic friction between the block and table is μ = 0.50. What fraction of the bullet’s initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block?

1021
views
Domanda del libro di testo

The distance between a carbon atom (m = 12 u) and an oxygen atom (m = 16 u) in the CO molecule is 1.13 x 10⁻¹⁰ m. How far from the carbon atom is the center of mass of the molecule?

2022
views
Domanda del libro di testo

(II) A pendulum consists of a mass M hanging at the bottom end of a massless rod of length ℓ, which has a frictionless pivot at its top end. A mass m, moving horizontally as shown in Fig. 9–44 with velocity v, impacts M and becomes embedded. What is the smallest value of v sufficient to cause the pendulum (with embedded mass m) to swing clear over the top of its arc?

1798
views
Domanda del libro di testo

Croquet ball A moving at 4.3 m/s makes a head-on collision with ball B of equal mass initially at rest. Immediately after the collision, ball B moves forward at 3.0 m/s. What fraction of the initial kinetic energy is lost in the collision?

1449
views