Problem 68 - College Physics

Question

A fellow student states that the total momentum of a three-particle system \(\left(m_{1}=0.25 \mathrm{~kg}, m_{2}=0.20 \mathrm{~kg},\right.\) and \(m_{3}=0.33 \mathrm{~kg}\) ) is initially zero. He calculates that after an inelastic triple collision the particles have velocities of \(4.0 \mathrm{~m} / \mathrm{s}\) at \(0^{\circ}, 6.0 \mathrm{~m}\) at \(120^{\circ},\) and \(2.5 \mathrm{~m} / \mathrm{s}\) at \(230^{\circ},\) respec- tively, with angles measured from the \(+x\) -axis. Do you agree with his calculations? If not, assuming the first two answers to be correct, what should be the momentum of the third particle so the total momentum is zero?

Step-by-Step Solution

Answer

No, his calculations are incorrect. Adjust the components of particle 3's momentum to ensure the total momentum is zero.

Step 1: Understand Conservation of Momentum

For a system where the initial total momentum is zero, the total momentum after the collision must also be zero. This means the vector sum of the momenta of all three particles must be zero.

Step 2: Calculate Momentum Components

Calculate the momentum components for each particle in the x and y directions using the formula: \[p_{x_i} = m_i imes v_i imes \cos(\theta_i) \p_{y_i} = m_i imes v_i imes \sin(\theta_i)\]where \(m_i\) is the mass, \(v_i\) is the velocity, and \(\theta_i\) is the angle.

Step 3: Momentum of Particle 1

The momentum components for particle 1 (mass = 0.25 kg, velocity = 4.0 m/s, angle = 0°):\[p_{x_1} = 0.25 \times 4.0 \times \cos(0°) = 1.0 \text{ kg m/s}\p_{y_1} = 0.25 \times 4.0 \times \sin(0°) = 0 \text{ kg m/s}\]

Step 4: Momentum of Particle 2

The momentum components for particle 2 (mass = 0.2 kg, velocity = 6.0 m/s, angle = 120°):\[p_{x_2} = 0.2 \times 6.0 \times \cos(120°) = -0.6 \text{ kg m/s}\p_{y_2} = 0.2 \times 6.0 \times \sin(120°) = 1.04 \text{ kg m/s}\]

Step 5: Momentum of Particle 3

The momentum components for particle 3 (mass = 0.33 kg, velocity = 2.5 m/s, angle = 230°):\[p_{x_3} = 0.33 \times 2.5 \times \cos(230°) = -0.63 \text{ kg m/s}\p_{y_3} = 0.33 \times 2.5 \times \sin(230°) = -0.53 \text{ kg m/s}\]

Step 6: Check Total Momentum

Add up the x and y components of momentum for all particles:Total x-momentum: \[p_{x}\,(\text{total}) = 1.0 - 0.6 - 0.63 = -0.23 \text{ kg m/s}\]Total y-momentum: \[p_{y}\,(\text{total}) = 0 + 1.04 - 0.53 = 0.51 \text{ kg m/s}\]

Step 7: Adjusting Particle 3's Momentum

For total momentum to be zero, the x and y components for the total system must both be zero. Since the current total momentum is not zero, we disagree with the calculations. To correct this, adjust the momentum of the third particle so that:\[p_{x_3}^\prime = -0.4 \text{ kg m/s}\p_{y_3}^\prime = -0.51 \text{ kg m/s}\]

Step 8: Calculate New Velocity or Angle for Particle 3

To achieve the adjusted momentum components for particle 3, compute its new velocity or angle based on possible modifications while maintaining consistent mass.

Key Concepts

Inelastic Collision

Inelastic collisions are exciting events in physics because they illustrate how objects behave when they collide and stick together or interact in some way that their kinetic energy is not conserved. In this type of collision, while kinetic energy may be lost—as energy might get transformed into heat, sound, or deformation—the total momentum of the system is conserved.

Let's take a closer look at these phenomena in the context of our three-particle system. Initially, we had zero total momentum, meaning the particles were balanced in such a way that their collective movement was canceled out. An inelastic collision doesn't change this balance of momentum, even though the individual particles might end up moving in different directions. This is because momentum, a vector quantity, considers both magnitude and direction, which is a key factor in our calculations.

Vector Components

Understanding the role of vector components is crucial in momentum calculations, especially when dealing with angles and directions. Any vector, like velocity or momentum, can be broken down into two perpendicular components: usually the horizontal (x-axis) and vertical (y-axis) components.

For example, when a particle moves in a particular direction at a certain angle, we deduce its x and y components using trigonometric functions - cosine for the x-component and sine for the y-component. Essentially, if you have a vector at an angle \( \theta \), you calculate its components as \( p_{x} = m \times v \times \cos(\theta) \) and \( p_{y} = m \times v \times \sin(\theta) \). This decomposition is fundamental for solving physics problems where vectors are involved, as it allows us to analyze and compute the effects in each direction independently.

Momentum Calculation

Momentum calculations form the backbone of solving physics problems involving collisions. Momentum is defined as the product of an object's mass and its velocity. Since velocity is a vector, momentum too has components along different axes.

In our exercise, the momentum of each particle was calculated separately using their respective velocities and angles. The x and y components were derived as follows:

• Particle 1: Its momentum along the x-axis was calculated using its given angle of \(0^{\circ}\), which pointed purely along the x-axis, resulting in \(1.0\, \text{kg m/s}\) with no y-component.
• Particle 2: With an angle of \(120^{\circ}\), it had an x-component of \(-0.6\, \text{kg m/s}\) and a y-component of \(1.04\, \text{kg m/s}\).
• Particle 3: It ended up with \(-0.63\, \text{kg m/s}\) for the x-component and \(-0.53\, \text{kg m/s}\) for the y-component, calculated similarly through its angle \(230^{\circ}\).

Adding these components allowed us to determine the total momentum for each direction, showing discrepancies that needed addressing to maintain total zero momentum for the system.

Physics Problem Solving

Successfully solving physics problems often involves meticulous working through calculations and rechecking assumptions. When dealing with the conservation of momentum, especially in a system with multiple moving parts like our three-particle system, accuracy is everything.

In practice, solving such problems requires:

• Breaking down vectors into their components for easy summation.
• Ensuring that the total momentum in all directions adheres to conservation laws.
• Identifying any potential discrepancies and systematically pinpointing where recalibration is needed.

In this case, the sum of x and y components highlighted the error, prompting a revision of particle 3's momentum to restore equilibrium. Calculating the exact values necessitated modifying either its velocity or its angle. This careful step-wise analytical approach is essential in physics to achieve accurate results.