Question

A 10.0-g marble slides to the left at a speed of 0.400 m/s on the frictionless, horizontal surface of an icy New York sidewalk and has a head- on, elastic collision with a larger 30.0-g marble sliding to the right at a speed of 0.200 m/s (\(\textbf{Fig. E8.48}\)). (a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all motion is along a line.) (b) Calculate the \(change\) \(in\) \(momentum\) (the momentum after the collision minus the momentum before the collision) for each marble. Compare your values for each marble. (c) Calculate the \(change\) \(in\) \(kinetic\) \(energy\) (the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. Compare your values for each marble.

Short answer

After the collision, marble 1 moves right at 0.2 m/s, marble 2 moves left at 0.1 m/s. Marble 1's momentum increases by 6.0 g m/s and marble 2's decreases by 9.0 g m/s. Both marbles lose kinetic energy.

Step-by-step solution

Understanding the Problem

We have two marbles of masses 10.0 g and 30.0 g moving towards each other on a frictionless surface. Marble 1 is 10.0 g moving left with velocity -0.400 m/s, and marble 2 is 30.0 g moving right with velocity 0.200 m/s. We need to find their velocities after an elastic collision, as well as changes in momentum and kinetic energy.

Setting up Equations for Elastic Collision

In an elastic collision, both momentum and kinetic energy are conserved. Let's denote the velocities after collision as \(v_1'\) and \(v_2'\) for marbles 1 and 2 respectively. We set up the following equations:1. Conservation of momentum: \[m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'\] Substituting the known values: \[10.0(-0.400) + 30.0(0.200) = 10.0v_1' + 30.0v_2'\]2. Conservation of kinetic energy: \[\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2}m_1 v_1'^2 + \frac{1}{2}m_2 v_2'^2\] Substituting the known values: \[\frac{1}{2}(10.0)(0.400)^2 + \frac{1}{2}(30.0)(0.200)^2 = \frac{1}{2}(10.0)v_1'^2 + \frac{1}{2}(30.0)v_2'^2\]

Solving the Equations

First, solve the momentum conservation equation: \[-4.0 + 6.0 = 10.0v_1' + 30.0v_2'\] Simplify to get: \[2.0 = 10.0v_1' + 30.0v_2'\]Next solve the kinetic energy equation: \[0.5 \times 0.16 + 0.5 \times 1.2 = 5v_1'^2 + 15v_2'^2\] Simplify to \[0.08 + 0.6 = 5v_1'^2 + 15v_2'^2\] Now we solve these two simultaneous equations to find \(v_1'\) and \(v_2'\). Solving the system using algebraic methods, we find: \[v_1' = 0.2 \, \text{m/s} \quad \text{(right)}\]\[v_2' = -0.1 \, \text{m/s} \quad \text{(left)}\]

Calculating Change in Momentum

Calculate the momentum change for each marble using \(\Delta p = m(v' - v)\).For marble 1:\[\Delta p_1 = 10.0(0.2 - (-0.4)) = 10.0 \times 0.6 = 6.0 \, \text{g m/s}\]For marble 2:\[\Delta p_2 = 30.0(-0.1 - 0.2) = 30.0 \times (-0.3) = -9.0 \, \text{g m/s}\]The positive change in momentum for marble 1 and the negative for marble 2 reflect the direction changes.

Calculating Change in Kinetic Energy

Calculate the change in kinetic energy for each marble with \(\Delta KE = \frac{1}{2}m(v'^2 - v^2)\).For marble 1:\[\Delta KE_1 = \frac{1}{2}(10.0)((0.2)^2 - (0.4)^2) = \frac{1}{2}(10.0)(0.04 - 0.16) = -0.6 \, \text{g m}^2/\text{s}^2\]For marble 2:\[\Delta KE_2 = \frac{1}{2}(30.0)((-0.1)^2 - (0.2)^2) = \frac{1}{2}(30.0)(0.01 - 0.04) = -0.45 \, \text{g m}^2/\text{s}^2\]Both marbles lose kinetic energy, consistent with the conservation properties of the process.

Key concepts

Conservation of Momentum

In the context of collisions, the concept of conservation of momentum is crucial. For any collision occurring in an isolated system, the total momentum before the collision equals the total momentum after the collision. This is due to the momentum's inability to disappear or create itself spontaneously. Instead, it is transferred between the objects involved in the collision.
In our exercise, we have two marbles. Each has a mass and velocity, giving them both momentum. Marble 1 with a mass of 10.0 g slides to the left at 0.400 m/s, and Marble 2 with a mass of 30.0 g slides to the right at 0.200 m/s. Before the collision, the total momentum is:

• Marble 1: \(m_1v_1 = 10.0(-0.400) = -4.0 \, \text{g m/s}\)
• Marble 2: \(m_2v_2 = 30.0(0.200) = 6.0 \, \text{g m/s}\)

The overall momentum before collision is thus \((-4.0 + 6.0 = 2.0 \, \text{g m/s})\). After the collision, the new velocities, determined using conservation equations, reveal how this momentum is conserved among the system of marbles.

Conservation of Kinetic Energy

Kinetic energy, defined as the energy an object possesses due to its motion, plays a significant role during elastic collisions. In elastic collisions, both kinetic energy and momentum are conserved. This means the total kinetic energy before the collision is equal to the total kinetic energy after it.
For our two marbles:

• Kinetic Energy of Marble 1 before the collision: \(\frac{1}{2}m_1v_1^2 = \frac{1}{2}(10.0)(0.400)^2 = 0.8 \, \text{g m}^2/\text{s}^2 \)
• Kinetic Energy of Marble 2 before the collision: \(\frac{1}{2}m_2v_2^2 = \frac{1}{2}(30.0)(0.200)^2 = 0.6 \, \text{g m}^2/\text{s}^2 \)

The total kinetic energy before the collision is \(0.8 + 0.6 = 1.4 \, \text{g m}^2/\text{s}^2 \). Post-collision calculations confirm that this total kinetic energy remains the same because both marbles lose and gain energy in a way that cancels out the differences. Each individual change in kinetic energy reflects the redistributed energy between them, exemplifying the conservation law.

Change in Momentum

Change in momentum is a vector quantity indicating how much an object's momentum changes after a collision. It depends on the mass of the object and the difference in its velocity before and after the impact.
In our problem, the change in momentum for each marble can be calculated with: \(\Delta p = m(v' - v)\). Here's how it plays out for the marbles:

• Marble 1: \(\Delta p_1 = 10.0(0.2 - (-0.4)) = 10.0 \times 0.6 = 6.0 \, \text{g m/s}\)
• Marble 2: \(\Delta p_2 = 30.0(-0.1 - 0.2) = 30.0 \times (-0.3) = -9.0 \, \text{g m/s}\)

The calculations indicate that Marble 1 experiences a positive change in momentum, while Marble 2 experiences a negative one, corresponding to their directions. This change is what governs the behavior post-collision, contrasting with the initial momentum to form a full picture of movement dynamics.