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Particle 1 of mass \(200 \mathrm{~g}\) and speed \(3.00 \mathrm{~m} / \mathrm{s}\) undergoes a onedimensional collision with stationary particle 2 of mass \(400 \mathrm{~g}\). What is the magnitude of the impulse on particle 1 if the collision is (a) elastic and (b) completely inelastic?

Short Answer

Expert verified
(a) 0.8 kg·m/s (b) 0.4 kg·m/s

Step by step solution

01

Understand the Impulse Concept

Impulse is a measure of the change in momentum of an object. It is defined as the product of the average force applied to an object and the time duration over which it is applied: \[J = \Delta p = m \Delta v\] where \( J \) is the impulse, \( \Delta p \) is the change in momentum, \( m \) is mass, and \( \Delta v \) is the change in velocity.
02

Initial Setup for Elastic Collision

In an elastic collision, both momentum and kinetic energy are conserved. First, calculate the initial momentum of particle 1. Initial momentum of particle 1, \( p_{1i} = m_1 v_{1i} = (0.2 \, \text{kg})(3.00 \, \text{m/s}) = 0.6 \, \text{kg}\cdot\text{m/s} \).
03

Calculate Final Velocities for Elastic Collision

Using conservation of momentum and kinetic energy, the final velocities can be computed:\[ v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i} = \frac{0.2 - 0.4}{0.2 + 0.4} \times 3 = -1 \, \text{m/s}\]\[ v_{2f} = \frac{2m_1}{m_1 + m_2} v_{1i} = \frac{2 \times 0.2}{0.6} \times 3 = 2 \, \text{m/s}\]
04

Calculate Impulse in Elastic Collision

The impulse experienced by particle 1 in the elastic collision is:\[J = m_1 (v_{1f} - v_{1i}) = 0.2 \, ( -1 \, - \, 3) = -0.8 \, \text{kg}\cdot\text{m/s}\]The magnitude of the impulse is 0.8 kg·m/s.
05

Initial Setup for Inelastic Collision

In a completely inelastic collision, the two particles stick together and move with the same final velocity. The total initial momentum is conserved:\[p_{initial} = p_{final}\]\[0.2 \, imes \, 3 = (0.2 + 0.4) \, v_f \]
06

Calculate Final Velocity in Inelastic Collision

Calculate the common velocity of both particles after collision using momentum conservation equation:\[0.6 \, = 0.6 \, v_f \]\[v_f = 1 \, \text{m/s}\]
07

Calculate Impulse in Inelastic Collision

The impulse on particle 1 in the inelastic collision is:\[J = m_1 (v_{f} - v_{1i}) = 0.2 \, ( 1 \, - \, 3) = -0.4 \, \text{kg}\cdot\text{m/s}\]The magnitude of the impulse is 0.4 kg·m/s.

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Key Concepts

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

Understanding Elastic Collisions
An elastic collision is a fascinating interaction between particles or objects. In this type of collision, both momentum and kinetic energy are conserved. This means that the total momentum and kinetic energy before the collision will be exactly the same after the collision. During an elastic collision, the two colliding particles bounce off each other, and no energy is lost to the environment.
For example, imagine two billiard balls colliding on a pool table. If the collision is perfectly elastic, the balls will exchange some velocities based on their masses, but the total kinetic energy remains unchanged. To mathematically express this, we use the conservation laws. The conservation of momentum is given by \[ m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \]
where the subscripts \( i \) and \( f \) denote initial and final velocities. The conservation of kinetic energy is expressed as:\[ \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 \]
These two equations help us find out the velocities after the collision, which are crucial in calculating the impulse experienced by the particles involved.
Exploring Inelastic Collisions
In a completely inelastic collision, the participating objects end up sticking together post-collision. While momentum is still conserved in this type of collision, kinetic energy is not. Some of the energy is transformed into other forms, such as thermal energy, due to deformation or bonding during the collision.
Consider two cars crashing into each other and intertwining into one tangled mass post-collision. Here, they continue moving together, albeit at a reduced speed. Mathematically, the conservation of momentum for an inelastic collision is written as:\[ m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f \]
where \( v_f \) is the final velocity of the combined mass. This equation allows us to solve for \( v_f \), which is needed to calculate the impulse on the objects involved.
  • Initial Kinetic Energy: More than final kinetic energy
  • Physical deformation may occur
  • Objects may couple together (e.g., Velcro collision)
Momentum Conservation in Collisions
Momentum conservation is a key principle when studying collisions, whether they are elastic or inelastic. Momentum, defined as the product of an object's mass and velocity, remains constant in isolated systems where no external forces act.
This fundamental concept helps explain why, during collisions, momentum before the impact equals momentum after. The total momentum of all objects involved stays the same, although individual momentums may change.
Let's consider two colliding particles in a frictionless environment. For these particles, the momentum conservation equation is expressed as:\[ m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \]
This expresses the fact that the sum of the initial momentum of the particles before collision matches the sum of their final momentum after collision. This equation, combined with others that represent energy conservation, offers powerful tools to solve collisions in physics.
  • Makes use of both initial and final velocities
  • Helps in analyzing collision problems
  • Base foundation for understanding collision mechanics

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Most popular questions from this chapter

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