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Ch. 09 - Linear Momentum
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 09 - Linear MomentumProblem 42a
Chapter 9, Problem 42a

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?

Verified step by step guidance
1
Convert the mass of the baseball from grams to kilograms. Since 1 gram = 0.001 kilograms, the mass of the baseball is \( m_1 = 144 \times 10^{-3} \, \text{kg} = 0.144 \, \text{kg} \).
Apply the principle of conservation of momentum. The total momentum before the collision equals the total momentum after the collision. The equation is: \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \), where \( m_1 \) and \( m_2 \) are the masses of the baseball and brick, \( v_{1i} \) and \( v_{2i} \) are their initial velocities, and \( v_{1f} \) and \( v_{2f} \) are their final velocities.
Substitute the known values into the momentum equation. The initial velocity of the baseball is \( v_{1i} = 28.0 \, \text{m/s} \), the initial velocity of the brick is \( v_{2i} = 0 \, \text{m/s} \), the final velocity of the brick is \( v_{2f} = 1.10 \, \text{m/s} \), and the masses are \( m_1 = 0.144 \, \text{kg} \) and \( m_2 = 4.85 \, \text{kg} \). The equation becomes: \( (0.144)(28.0) + (4.85)(0) = (0.144)v_{1f} + (4.85)(1.10) \).
Simplify the equation to isolate \( v_{1f} \), the final velocity of the baseball. Rearrange the terms to solve for \( v_{1f} \): \( v_{1f} = \frac{(0.144)(28.0) - (4.85)(1.10)}{0.144} \).
Perform the calculations step by step to find \( v_{1f} \), ensuring that the units are consistent throughout. Note that the result will be the baseball's speed after the collision, and it will have a negative sign if the baseball bounces back in the opposite direction.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Conservation of Momentum

The principle of conservation of momentum states that in a closed system, the total momentum before a collision is equal to the total momentum after the collision. This is crucial for analyzing collisions, as it allows us to set up equations based on the masses and velocities of the objects involved.
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Elastic and Inelastic Collisions

Collisions can be classified as elastic or inelastic. In elastic collisions, both momentum and kinetic energy are conserved, while in inelastic collisions, momentum is conserved but kinetic energy is not. The scenario described involves an inelastic collision since the baseball bounces back, indicating some energy is lost.
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Velocity and Speed

Velocity is a vector quantity that includes both speed and direction, while speed is a scalar quantity that only measures how fast an object is moving. Understanding the difference is essential for solving the problem, as the direction of the baseball's velocity changes after the collision, affecting the final calculations.
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Related Practice
Textbook Question

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.

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Textbook Question

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.

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Textbook Question

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?

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Textbook Question

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.

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Textbook Question

(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?

Textbook Question

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?

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