/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 Two skaters collide and grab on ... [FREE SOLUTION] | 91Ó°ÊÓ

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Two skaters collide and grab on to each other on frictionless ice. One of them, of mass \(70.0 \mathrm{kg},\) is moving to the right at 2.00 \(\mathrm{m} / \mathrm{s}\) , while the other. of mass 65.0 \(\mathrm{kg}\) , is moving to the left at 2.50 \(\mathrm{m} / \mathrm{s}\) . What are the magnitude and direction of the velocity of these skaters just after they collide?

Short Answer

Expert verified
The skaters move together with a velocity of 0.167 m/s to the left.

Step by step solution

01

Understand the conservation of momentum

In a collision on a frictionless surface, the total momentum before and after the collision remains constant. Therefore, the sum of the momenta of both skaters before collision equals their combined momentum after collision.
02

Write the momentum equation before collision

The momentum of the first skater is given by \( m_1 \times v_1 \) and for the second skater by \( m_2 \times v_2 \). For the first skater, \( m_1 = 70.0 \, \text{kg} \) and \( v_1 = 2.00 \, \text{m/s} \). For the second skater, \( m_2 = 65.0 \, \text{kg} \) and \( v_2 = -2.50 \, \text{m/s} \). Note that \( v_2 \) is negative due to the left direction.
03

Calculate total initial momentum

Calculate the initial momentum of both skaters:\[p_{initial} = (70.0 \, \text{kg} \times 2.00 \, \text{m/s}) + (65.0 \, \text{kg} \times -2.50 \, \text{m/s}) = 140.0 \, \text{kg m/s} - 162.5 \, \text{kg m/s} = -22.5 \, \text{kg m/s}.\]
04

Set the final momentum equal to the initial momentum

Let \( v_f \) be the final velocity of both skaters together after collision. Their combined mass is \( m_1 + m_2 = 70.0 \, \text{kg} + 65.0 \, \text{kg} = 135.0 \, \text{kg} \). The equation is:\[135.0 \, \text{kg} \times v_f = -22.5 \, \text{kg m/s}.\]
05

Solve for final velocity

Rearrange the equation for \( v_f \):\[v_f = \frac{-22.5 \, \text{kg m/s}}{135.0 \, \text{kg}} = -0.1667 \, \text{m/s}.\]The negative sign indicates the direction is to the left.

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Key Concepts

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

Inelastic Collisions
Inelastic collisions are a type of collision where two objects collide and move together after the impact. This means their kinetic energy is not conserved as some of it is transformed into other forms of energy, such as heat or sound. However, the total momentum of the system remains the same before and after the collision. This principle is known as the conservation of momentum. In the given problem, the two skaters collide and stick together, making it a perfect example of an inelastic collision.
The moment they grab onto each other, their individual velocities change to become a shared velocity, resulting in a combined mass system moving in a single direction.
Momentum Equation
The momentum equation is crucial for understanding how the motion of objects changes in collisions. Momentum (\(p\)) is calculated as the product of mass (\(m\)) and velocity (\(v\)). The equation for momentum is:
  • \(p = m \cdot v\)
To solve the collision problem, we calculate the momentum of each skater before they collide.
The first skater's momentum is \((70.0 \, \text{kg} \times 2.00 \, \text{m/s})\) and the second skater's momentum is \((65.0 \, \text{kg} \times (-2.50) \, \text{m/s})\). Notice we use a negative velocity for the second skater as they move in the opposite direction.
The total initial momentum is then calculated by summing these values, which provides the total momentum before the collision.
Frictionless Surfaces
Frictionless surfaces are idealized surfaces where no force opposes the movement of objects. This assumption simplifies our calculations, as no external forces like friction alter the momentum of the objects involved.
In our problem, the skaters are on an ice rink, approximating a frictionless surface, allowing us to use the conservation of momentum without adjustment for friction.
This means the total momentum before the collision is the same as the total momentum afterwards, as no energy is lost due to friction. This makes frictionless environments perfect for examining momentum conservation principles.

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