Thermodynamics and Inelastic Collisions: Conservation of Momentum

Inelastic Collisions: When Kinetic Energy Isn't Conserved 💥

In our previous discussion, we explored momentum and elastic collisions, where kinetic energy (KE) is conserved. Now, let's dive into inelastic collisions, where things get a bit more interesting because KE is not conserved.

What are Inelastic Collisions?

• In an elastic collision, both momentum and kinetic energy are conserved. Think of billiard balls bouncing off each other perfectly. 🎱
• In an inelastic collision, momentum is still conserved, but some kinetic energy is converted into other forms of energy, such as heat, sound, or deformation. Imagine two cars colliding and crumpling. 🚗💥
• A completely inelastic collision is a special case where the colliding objects stick together after the collision.

Understanding the Differences

Example: Two Carts Colliding

Let's illustrate with an example:

Two carts of the same mass (m) are on a table. Cart 1 is moving with velocity v, and cart 2 is at rest. They collide and stick together. What's their final velocity (vf)?

1. Initial Momentum: $p_i = m_1v_1 + m_2v_2 = mv + 0 = mv$
2. Final Momentum: $p_f = (m_1 + m_2)v_f = (m + m)v_f = 2mv_f$
3. Conservation of Momentum: $p_i = p_f$, so $mv = 2mv_f$
4. Final Velocity: $v_f = \frac{v}{2}$

Why Inelastic Collisions Matter

• Many real-world collisions are inelastic, especially at the molecular level.
• We often approximate collisions as elastic for simplicity, but understanding the difference is crucial for accurate modeling.

Exam Focus: Be prepared to identify whether a collision is elastic or inelastic based on changes in kinetic energy. Also, be ready to apply conservation of momentum to solve problems involving both types of collisions.

Example Problems

Let's work through a couple of examples:

Example Problem #1:

Two carts, one with a mass of 5 kg and the other with a mass of 2 kg, collide on a frictionless track. The 5 kg cart is initially moving at a speed of 3 m/s to the right, and the 2 kg cart is initially stationary. After the collision, the 5 kg cart moves to the left at a speed of 1 m/s.

(a) Is this collision elastic or inelastic? Justify your answer. (b) What is the common final velocity of the carts after the collision? (c) What is the initial kinetic energy of the 5 kg cart? (d) What is the final kinetic energy of the 5 kg cart? (e) What is the change in kinetic energy of the 5 kg cart during the collision?

Solution:

(a) Inelastic. The initial KE of the 5 kg cart is $KE_i = \frac{1}{2}mv^2 = \frac{1}{2}(5)(3^2) = 22.5 J$. The final KE of the 5 kg cart is $KE_f = \frac{1}{2}mv^2 = \frac{1}{2}(5)(-1)^2 = 2.5 J$. Since KE is not conserved, the collision is inelastic.

(b) Let's use conservation of momentum: $p_i = p_f$. $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$. $(5kg)(3m/s) + (2kg)(0m/s) = (5kg)(-1m/s) + (2kg)v_{2f}$. Solving for $v_{2f}$ gives us $v_{2f} = 10 m/s$. The final velocity of the 2kg cart is 10 m/s to the right.

(c) Initial KE of the 5 kg cart: $KE_i = \frac{1}{2}(5)(3^2) = 22.5 J$

(d) Final KE of the 5 kg cart: $KE_f = \frac{1}{2}(5)(-1)^2 = 2.5 J$

(e) Change in KE of the 5 kg cart: $ΔKE = KE_f - KE_i = 2.5 J - 22.5 J = -20 J$

Example Problem #2:

A 20 kg bowling ball traveling at a speed of 5 m/s to the left collides with a stationary 10 kg bowling ball on a frictionless alley. After the collision, the 20 kg ball moves at a speed of 2 m/s to the right, and the 10 kg ball moves at a speed of 3 m/s to the left.

(a) Is this collision elastic or inelastic? Justify your answer. (b) What is the change in kinetic energy of the system during the collision?

Solution:

(a) Inelastic. The initial KE of the system is $KE_i = \frac{1}{2}(20)(-5)^2 + \frac{1}{2}(10)(0)^2 = 250 J$. The final KE of the system is $KE_f = \frac{1}{2}(20)(2)^2 + \frac{1}{2}(10)(-3)^2 = 40 + 45 = 85 J$. Since KE is not conserved, the collision is inelastic.

(b) Change in KE of the system: $ΔKE = KE_f - KE_i = 85 J - 250 J = -165 J$

Final Exam Focus

• Key Concepts: Understand the difference between elastic and inelastic collisions. Know that momentum is always conserved, but kinetic energy is only conserved in elastic collisions. 💡
• Problem Solving: Be able to apply conservation of momentum to solve for final velocities in both elastic and inelastic collisions. Practice calculating changes in kinetic energy.
• Real-World Applications: Recognize that most real-world collisions are inelastic.

Practice Questions

Practice Question

Multiple Choice Questions

1. Two objects collide. Which of the following statements is always true? (A) Kinetic energy is conserved. (B) Momentum is conserved. (C) Both kinetic energy and momentum are conserved. (D) Neither kinetic energy nor momentum is conserved.

2. A 2 kg ball moving at 3 m/s collides head-on with a 1 kg ball at rest. After the collision, the 2 kg ball moves at 1 m/s in the same direction. What is the velocity of the 1 kg ball after the collision? (A) 1 m/s (B) 2 m/s (C) 4 m/s (D) 5 m/s

3. In an inelastic collision, what happens to the kinetic energy? (A) It is always conserved. (B) It is always increased. (C) It is always decreased. (D) It is converted into other forms of energy.

Free Response Question

A 3 kg cart moving at 4 m/s to the right collides with a 1 kg cart initially at rest. After the collision, the two carts stick together.

(a) What is the final velocity of the two-cart system after the collision? (2 points) (b) What was the initial kinetic energy of the system? (2 points) (c) What is the final kinetic energy of the system? (2 points) (d) How much kinetic energy was lost in the collision? (2 points) (e) Is this collision elastic or inelastic? (1 point)

Scoring Rubric: (a) Correctly applies conservation of momentum: $m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f$ (1 point). Correctly calculates final velocity $v_f = 3 m/s$ (1 point). (b) Correctly calculates initial kinetic energy: $KE_i = \frac{1}{2}m_1v_{1i}^2 = 24 J$ (2 points). (c) Correctly calculates final kinetic energy: $KE_f = \frac{1}{2}(m_1 + m_2)v_f^2 = 18 J$ (2 points). (d) Correctly calculates the change in kinetic energy: $ΔKE = KE_f - KE_i = -6 J$ (2 points). (e) Correctly identifies the collision as inelastic (1 point).