Collisions and Conservation Laws in Physics: Study Notes

Collisions and Conservation Laws

Introduction

This section explores the physics of collisions, focusing on the conservation of momentum and energy. The example problem involves a block sliding down a half-pipe and colliding with another block, illustrating key principles of inelastic collisions and energy transformations.

Conservation of Energy: Block Sliding Down a Ramp

• Potential Energy (PE): The energy stored due to an object's position. For an object of mass m at height h, the gravitational potential energy is given by: $PE = mgh$

• Kinetic Energy (KE): The energy of motion. For an object of mass m moving at velocity v: $KE = \frac{1}{2}mv^2$

• Energy Conservation (No Friction): As the block slides down, potential energy is converted to kinetic energy: $mgh = \frac{1}{2}mv^2$

• Solving for Velocity at Bottom: $v = \sqrt{2gh}$

• Example: For a block of mass 3 kg released from a height of 2 m: $v = \sqrt{2 \times 9.8 \times 2} = 6.26\ \text{m/s}$

Collisions: Types and Conservation of Momentum

• Momentum (p): The product of mass and velocity. For an object: $p = mv$

• Law of Conservation of Momentum: In the absence of external forces, the total momentum before a collision equals the total momentum after the collision: $m_1v_1 + m_2v_2 = (m_1 + m_2)v_f$

• Completely Inelastic Collision: A collision in which the colliding objects stick together after impact. Kinetic energy is not conserved, but momentum is.

• Example: A 3 kg block (moving at 6.26 m/s) collides with a stationary 1.5 kg block. After the collision, both move together. The final velocity is: $v_f = \frac{m_1v_1}{m_1 + m_2} = \frac{3 \times 6.26}{4.5} = 4.2\ \text{m/s}$

Kinetic Energy Loss in Inelastic Collisions

• Kinetic Energy Before Collision: $KE_{\text{initial}} = \frac{1}{2}m_1v_1^2$

• Kinetic Energy After Collision: $KE_{\text{final}} = \frac{1}{2}(m_1 + m_2)v_f^2$

• Energy Lost: The difference between initial and final kinetic energy is the energy lost (often as heat or deformation): $\Delta KE = KE_{\text{final}} - KE_{\text{initial}}$

• Example Calculation: $KE_{\text{initial}} = \frac{1}{2} \times 3 \times (6.26)^2 = 58.8\ \text{J}$ $KE_{\text{final}} = \frac{1}{2} \times 4.5 \times (4.2)^2 = 39.7\ \text{J}$ $\Delta KE = 39.7 - 58.8 = -19.1\ \text{J}$ 19.1 J of kinetic energy is lost in the collision.

Summary Table: Key Quantities in the Example

Quantity	Value	Formula Used
Initial velocity of $m_1$ at bottom	6.26 m/s	$v = \sqrt{2gh}$
Final velocity after collision	4.2 m/s	$v_f = \frac{m_1v_1}{m_1 + m_2}$
Kinetic energy lost	19.1 J	$\Delta KE = KE_{\text{final}} - KE_{\text{initial}}$

Key Takeaways

• In a completely inelastic collision, objects stick together and move with a common velocity after impact.

• Momentum is always conserved in collisions (if no external forces act), but kinetic energy is only conserved in elastic collisions.

• Energy lost in inelastic collisions is transformed into other forms, such as heat or deformation.