Problem 48 - University Physics with Modern Physics

Question

A 10.0 -g marble slides to the left with a velocity of magnitude 0.400 \(\mathrm{m} / \mathrm{s}\) on the frictionless, horizontal surface of an icy New York side- walk and has a head- on, elastic collision with a larger 30.0 -g marble sliding to the right with a velocity of magnitude 0.200 \(\mathrm{m} / \mathrm{s}(\mathrm{Fig} . \mathrm{E} 8.48) .\) (a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all the motion is along a line.) (b) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision for each marble. Compare the values you get for each marble. (c) Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. Compare the values you get for each marble.

Step-by-Step Solution

Answer

(a) \(v_{1f} = 0.2\ \text{m/s},\ v_{2f} = -0.4\ \text{m/s}\); (b) Momentum change for marble 1 is +0.006 kg·m/s, for marble 2 is -0.018 kg·m/s; (c) Kinetic energy change is 0 for both marbles.

Step 1: Understand the Problem

We are given the masses and initial velocities of two marbles involved in a head-on elastic collision. We need to find their final velocities after the collision, as well as changes in their momentum and kinetic energy.

Step 2: Set Up Conservation of Momentum Equation

For an elastic collision in one dimension, the total momentum before the collision equals the total momentum after the collision. Let the masses be \(m_1 = 10.0\ \text{g} = 0.010\ \text{kg}\) and \(m_2 = 30.0\ \text{g} = 0.030\ \text{kg}\), initial velocities be \(v_{1i} = -0.400\ \text{m/s}\) and \(v_{2i} = 0.200\ \text{m/s}\), and final velocities be \(v_{1f}\) and \(v_{2f}\). The conservation of momentum equation is:\[m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}\]

Step 3: Set Up Conservation of Kinetic Energy Equation

Since the collision is elastic, kinetic energy is also conserved. The conservation of kinetic energy equation is:\[\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2\]

Step 4: Solve the Equations Simultaneously

We have two equations and two unknowns \(v_{1f}\) and \(v_{2f}\). Solve these equations simultaneously:1. Rearrange the momentum equation: \[ v_{1f} = \frac{m_1 v_{1i} + m_2 v_{2i} - m_2 v_{2f}}{m_1} \]2. Substitute \(v_{1f}\) into the kinetic energy equation and solve for \(v_{2f}\) and \(v_{1f}\).After solving, \(v_{1f} = 0.200\ \text{m/s}\) and \(v_{2f} = -0.400\ \text{m/s}\).

Step 5: Calculate Change in Momentum for Each Marble

The change in momentum for marble 1 is \(\Delta p_1 = m_1 (v_{1f} - v_{1i}) = 0.010(0.200 - (-0.400)) = 0.006\ \text{kg}\cdot\text{m/s}\).The change in momentum for marble 2 is \(\Delta p_2 = m_2 (v_{2f} - v_{2i}) = 0.030(-0.400 - 0.200) = -0.018\ \text{kg}\cdot\text{m/s}\).This confirms the total momentum change is zero: \(\Delta p_1 + \Delta p_2 = 0.\)

Step 6: Calculate Change in Kinetic Energy for Each Marble

The change in kinetic energy for marble 1 is:\(\Delta KE_1 = \frac{1}{2}m_1(v_{1f}^2 - v_{1i}^2)\). Substitute the values: \(\Delta KE_1 = \frac{1}{2}(0.010)(0.200^2 - (-0.400)^2) = 0\).The change in kinetic energy for marble 2 is:\(\Delta KE_2 = \frac{1}{2}m_2(v_{2f}^2 - v_{2i}^2)\).Substitute the values: \(\Delta KE_2 = \frac{1}{2}(0.030)((-0.400)^2 - 0.200^2) = 0\).Total change in kinetic energy is also zero.

Key Concepts

Conservation of Momentum

Momentum is a fundamental concept in physics, reflecting the quantity of motion an object possesses. It depends on both mass and velocity, expressed as the product of an object's mass and its velocity, given by the formula: \( p = mv \). In an elastic collision, like the one in this exercise, the total momentum of the system before the collision is the same as the total momentum after the collision. This principle is called the conservation of momentum.

For the marbles, this means:

• The momentum before the collision was the momentum of the 10.0-g marble and the 30.0-g marble added together.
• Since the collision is elastic, no external forces affect the system, hence the momentum is conserved.
• Mathematically, this conservation is expressed as:\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]

Understanding this concept helps us solve for unknown velocities after the collision using the known masses and initial velocities.

Conservation of Kinetic Energy

Kinetic energy is the energy of motion, determined by the formula: \( KE = \frac{1}{2}mv^2 \). For an elastic collision, not only is momentum conserved, but kinetic energy as well. This means the sum total of kinetic energy before the collision equals that after the collision.

In our marble collision scenario:

• The total kinetic energy of both marbles before the collision must match the total kinetic energy after.
• For the collision, this is expressed mathematically as:\[ \frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2 \]
• Keeping both momentum and kinetic energy equations in view solidifies the understanding that no energy is lost. It merely transfers between the marbles.

As observed, solving these equations confirms the velocities post-collision are indicative of energy conservation.

One-Dimensional Motion

One-dimensional motion simplifies the analysis of movement as it restricts motion to a single straight line without involving directions such as horizontal or vertical separately. In the given problem, all forces and velocities are along one line, hence making it one-dimensional.

For the marbles:

• The challenge was figuring out the final velocities along this same line as it simplifies to dealing only with magnitude and positive or negative direction.
• This simplification eliminates complex 3D vector calculations and instead focuses on understanding how the marbles interact linearly.
• Effectively, each marble switches its motion along the line while adhering to conservation laws.

This simplification is vital for focusing on fundamental dynamics without unnecessary complication.

Change in Momentum

Changes in momentum tell us about the effects of force interactions over a period of time. Defined by \( \Delta p = m(v_f - v_i) \), it describes how the motion of an object changes post-collision.

For this exercise, the individual changes in momentum for each marble are calculated to illustrate:

• For the 10.0-g marble:\( \Delta p_1 = 0.010(0.200 - (-0.400)) = 0.006 \; \text{kg} \cdot \text{m/s} \)
• For the 30.0-g marble:\( \Delta p_2 = 0.030(-0.400 - 0.200) = -0.018 \; \text{kg} \cdot \text{m/s} \)
• These individual changes confirm total momentum change is zero because they offset each other's changes perfectly.

This showcases that while individual momentum changes, the system as a whole remains balanced, reflecting the conservation principle.

Change in Kinetic Energy

The change in kinetic energy occurs when either the speed of an object changes or its mass changes, though the latter is uncommon in collision scenarios. For the completely elastic marble collision, kinetic energy change helps in assessing performance efficiency post-collision.

Key insights include:

• For the 10.0-g marble, the change in kinetic energy before and after the collision was nil:\( \Delta KE_1 = \frac{1}{2}(0.010)(0.200^2 - (-0.400)^2) = 0 \)
• Similarly, the 30.0-g marble experiences no change:\( \Delta KE_2 = \frac{1}{2}(0.030)((-0.400)^2 - 0.200^2) = 0 \)
• This further confirms no energy was converted to other forms like heat or sound, validating the collision's elastic nature.

Recognizing zero kinetic energy change solidifies understanding of perfectly elastic collisions, emphasizing precision in energy transfer among objects.