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II A 20 g ball of clay is thrown horizontally at \(30 \mathrm{m} / \mathrm{s}\) toward a \(1.0 \mathrm{kg}\) block sitting at rest on a frictionless surface. The clay hits and sticks to the block. a. What is the speed of the block and clay right after the collision? b. Use the block's initial and final speeds to calculate the impulse the clay exerts on the block. c. Use the clay's initial and final speeds to calculate the impulse the block exerts on the clay. d. Does \(\vec{J}_{\text {blockonclay }}=-\vec{J}_{\text {clayon block }} ?\)

Short Answer

Expert verified
The speed of the block and clay right after the collision is calculated from the law of conservation of momentum. The impulse the clay exerts on the block and the impulse the block exerts on the clay are both calculated using the change in velocities and these impulses are equal in magnitude and opposite in direction, in accordance with Newton's third law of motion.

Step by step solution

01

Calculate the final speed

Use the law of conservation of momentum which states that the total momentum before the collision is equal to the total momentum after the collision. Let the final velocity be \(v\). \[m_{\text{clay}} \cdot v_{\text{clay}} + m_{\text{block}} \cdot v_{\text{block}} = (m_{\text{clay}} + m_{\text{block}}) \cdot v\] Substitute the given values into the equation: \(0.020 \mathrm{kg} \cdot 30 \mathrm{m/s} + 1.0 \mathrm{kg} \cdot 0 \mathrm{m/s} = (1.0 \mathrm{kg} + 0.020 \mathrm{kg}) \cdot v\). Solve this equation to find \(v\).
02

Calculate the impulse exerted by clay on the block

The impulse can be calculated using the formula: \(\text{Impulse} = \text{mass} \times \text{change in velocity}\). Here, the final velocity of the block is obtained from step 1 and the initial velocity is 0 (since it's initially at rest). Therefore, the impulse exerted by the clay on the block is: \(\text{Impulse}_{\text{clay}} = m_{\text{block}} \times (v_{\text{final}} - v_{\text{initial}})\).
03

Calculate the impulse exerted by the block on the clay

Using a similar methodology as in step 2, calculate the impulse exerted by the block on the clay. The initial velocity of the clay is given and the final velocity is obtained from step 1. So, \(\text{Impulse}_{\text{block}} = m_{\text{clay}} \times (v_{\text{final}} - v_{\text{initial}})\).
04

Verify Newton's third law

By Newton's third law, the action and reaction forces (or impulses in this case) are equal and opposite. Therefore, \(\vec{J}_{\text {blockonclay }}=-\vec{J}_{\text {clayon block }}\). This can be verified by confirming that the magnitude of the impulses from Step 2 and Step 3 are the same, only opposite in direction.

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Key Concepts

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

Impulse in Physics
Impulse is a fundamental concept in physics, closely related to the conservation of momentum. It quantifies the effect of a force acting over a period of time and is calculated by the equation:
\[\begin{equation} \text{Impulse} = \text{Force} \times \text{Time} \text{interval} \end{equation}\]
Alternatively, since force is the rate of change of momentum, impulse can also be expressed as the change in momentum of an object: \[\begin{equation} \text{Impulse} = \text{Change in momentum} \end{equation}\]
In the context of the given problem, the impulse exerted on the block by the clay and vice versa is an essential concept to understand the dynamics of the collision. When calculating impulse, we can consider the mass of the object and how much its velocity changes during the interaction, using the formula: \[\begin{equation} \text{Impulse} = \text{mass} \times (v_{\text{final}} - v_{\text{initial}}) \end{equation}\]
Understanding this relationship is crucial for solving problems involving collisions and interactions in closed systems, where no external forces are present.
Newton's Third Law
Sir Isaac Newton's third law of motion states that for every action, there is an equal and opposite reaction. This principle underpins interactions between objects and can be applied to a variety of scenarios, from the grand scale of astronomical forces to the detailed analysis of a simple collision between two objects on a frictionless surface.
In the context of the given problem on calculating impulses during a collision, Newton's third law allows us to assert that the impulse the clay exerts on the block (the action) is equal in magnitude and opposite in direction to the impulse that the block exerts on the clay (the reaction). This is represented mathematically by: \[\begin{equation} \textbf{J}_{\text{block on clay}} = - \textbf{J}_{\text{clay on block}} \end{equation}\]
This equation not only shows the balance of forces involved in the collision but also reaffirms the conservation of momentum within the system. By recognizing this law, we can predict how different bodies in a collision will react to the forces exerted upon them.
Collision in Physics
A collision, in physics, refers to an event where two or more objects come into contact with each other, exerting forces and potentially transferring energy. Collisions can be classified as elastic or inelastic, depending on whether kinetic energy is conserved in the process.
In an elastic collision, both momentum and kinetic energy are conserved, leading to objects bouncing off each other. In contrast, an inelastic collision, like the one described in the exercise with the clay and block, is characterized by the objects sticking together and kinetic energy being transformed, but not conserved.
Analyzing collisions involves applying conservation laws, such as the conservation of momentum, to calculate the change in velocity and therefore determine the outcome. In the given problem, the clay sticks to the block, indicating an inelastic collision. As they move together as one unit after the collision, their combined momentum remains equal to the initial momentum of the system, governed by the equation:\[\begin{equation} m_{\text{clay}} \cdot v_{\text{clay}} + m_{\text{block}} \cdot v_{\text{block}} = (m_{\text{clay}} + m_{\text{block}}) \cdot v_{\text{final}} \end{equation}\]
Understanding the nuances of collisions is crucial, as they form the basis for numerous applications in areas ranging from automotive safety to sports science.

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Most popular questions from this chapter

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