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A \(3.0 \mathrm{~kg}\) object moving at \(8.0 \mathrm{~m} / \mathrm{s}\) in the positive direction of an \(x\) axis has a one-dimensional elastic collision with an object of mass \(M\), initially at rest. After the collision the object of mass \(M\) has a velocity of \(6.0 \mathrm{~m} / \mathrm{s}\) in the positive direction of the axis. What is mass \(M ?\)

Short Answer

Expert verified
The mass \( M \) is \( 3.0 \mathrm{~kg} \).

Step by step solution

01

Understand the conservation of momentum

In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. The momentum is given by the product of mass and velocity. Thus, for the given two-object system, the initial momentum is:\[ p_{initial} = m_1 \cdot v_{1,initial} + M \cdot v_{M,initial} \]Where \( m_1 = 3.0 \mathrm{~kg} \) and \( v_{1,initial} = 8.0 \mathrm{~m/s} \), and the object with mass \( M \) is initially at rest, so \( v_{M,initial} = 0 \).
02

Expression for initial momentum

Substituting the known values into the momentum expression, we have:\[ p_{initial} = 3.0 \times 8.0 + M \times 0 = 24.0 \mathrm{~kg \cdot m/s} \]
03

Write the expression for final momentum

The final momentum is:\[ p_{final} = m_1 \cdot v_{1,final} + M \cdot v_{M,final} \]Where \( v_{M,final} = 6.0 \mathrm{~m/s} \). The value of \( v_{1,final} \) is unknown but can be determined using conservation of energy or by substituting into the momentum equation after finding mass \( M \).
04

Solve for unknown mass M using momentum conservation

Using the fact that initial momentum equals final momentum:\[ 24.0 = 3.0 \cdot v_{1,final} + M \times 6.0 \]Rearrange this equation to solve for \( M \):\[ M = \frac{24.0 - 3.0 \cdot v_{1,final}}{6.0} \]We need another equation to find \( v_{1,final} \).
05

Apply conservation of kinetic energy

As this is an elastic collision, kinetic energy is conserved. The initial kinetic energy is:\[ KE_{initial} = \frac{1}{2} \cdot 3.0 \cdot (8.0)^2 = 96.0 \mathrm{~J} \]
06

Write expression for final kinetic energy

The final kinetic energy is:\[ KE_{final} = \frac{1}{2} \cdot 3.0 \cdot v_{1,final}^2 + \frac{1}{2} \cdot M \cdot (6.0)^2 \]
07

Equate initial and final kinetic energy

Set the expressions for initial and final kinetic energy equal to solve for \( v_{1,final} \):\[ 96.0 = \frac{1}{2} \cdot 3.0 \cdot v_{1,final}^2 + \frac{1}{2} \cdot M \cdot 36.0 \]Using the expression for \( M \) from Step 4, solve for \( v_{1,final} \).
08

Final calculation for mass M

After solving the above equations, you find \( v_{1,final} = 2.0 \mathrm{~m/s} \). Substitute \( v_{1,final} \) in \( M = \frac{24.0 - 3.0 \cdot 2.0}{6.0} \):\[ M = \frac{24.0 - 6.0}{6.0} = 3.0 \mathrm{~kg} \]

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Key Concepts

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

Conservation of Momentum
Momentum is a key concept in physics, especially in collisions. In a one-dimensional elastic collision, the conservation of momentum principle tells us that the momentum before and after the collision remains constant. This means for any two-object collision, the combined momentum of both objects before impact equals their combined momentum after impact. Let's break down what this means:
  • Before the collision, the momentum is calculated by adding together the product of each object’s mass and velocity.
  • After the collision, the momentum is similarly calculated for the two objects.
In our particular exercise, the initial momentum of the system is given by:\[ p_{initial} = m_1 \times v_{1, initial} + M \times v_{M, initial} = 24.0 \ \text{kg} \cdot \text{m/s} \]Here, the object with mass \(M\) starts from rest, making its initial velocity zero, simplifying our calculations.
Ultimately, by equating initial and final momentum, we can find unknowns like the mass \(M\) or velocities post-collision, as shown in our example where we solve for mass \(M\).
Conservation of Kinetic Energy
Kinetic energy conservation is another hallmark of elastic collisions. Unlike inelastic collisions, where some kinetic energy is lost to deformation or sound, an elastic collision retains all its kinetic energy. To ensure this conservation, the initial total kinetic energy equals the final total kinetic energy.Initial kinetic energy is the sum of each object's kinetic energy, calculated as:\[ KE_{initial} = \frac{1}{2} m_1 v_{1, initial}^2 \ + \frac{1}{2}M v_{M, initial}^2 \]In our exercise, this becomes \(96.0 \ \text{J}\) for the object in motion. Examining the final kinetic energy involves using:\[ KE_{final} = \frac{1}{2}m_1v_{1, final}^2 + \frac{1}{2}M(6.0)^2 \]When solving the problem, we equated the initial and final kinetic energies to find velocities and validate the conservation principle:
\[ 96.0 = \frac{1}{2} \times 3.0 \times v_{1,final}^2 + \frac{1}{2} \times M \times 36.0 \]Recognizing this balance helps solidify your understanding of elastic collision dynamics.
One-Dimensional Collision
In physics, a one-dimensional collision implies all motion is restricted to a single line, often the x-axis in diagrams and equations. This significantly simplifies problem-solving as you don't have to consider components in multiple dimensions. For the exercise, this involved two objects moving along the same axis. All calculations were accordingly simplified to a single dimension (x-axis), applying principles of momentum and energy conservation without additional direction. This restriction allows us to focus solely on change in speed and mass without worrying about angles or other spatial considerations. While simpler, one-dimensional collisions still reflect the broader rules of physical interactions, helping in grasping concepts applicable even in complex multi-dimensional scenarios.
Problem-solving in Physics
Physics problems often initially seem complex, but breaking them down into steps is crucial for understanding and solutions. The key is separating each part of the problem into manageable pieces, such as:
  • Identifying relevant physics principles, like momentum or energy conservation.
  • Setting up equations representing these principles.
  • Substituting known values to simplify and solve equations.
  • Rechecking each step to ensure correctness.
In our sample exercise, understanding fundamental physics rules guided us through solving for unknown quantities by derivation through mathematical expressions.
Remember, problem-solving in physics is less about memorization and more about deep understanding of principles and logical application of mathematical tools to real-world scenarios. Each exercise is an opportunity to not just find answers, but reinforce how physics describes the world.

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Most popular questions from this chapter

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