Collisions and Impact

Dr. Alyssa Vanderlinden

The collision of two bodies over a small time interval during which the two bodies exert large forces on each other.
The behavior of two objects following an impact depends on momentum and the nature of the impact.
There are two types of impacts:
1. Elastic Impact
2. Inelastic (Plastic) Impact

Definition: The velocities of the two bodies after impact are the same as their relative velocities before impact.
Energy Transfer: No kinetic energy is lost in elastic impacts.
Example: The impact of a bouncy ball with a hard surface approaches perfect elasticity.

Elastic impact can also explain scenarios where the velocity of a moving object is transferred to a stationary object without any loss of kinetic energy.

Definition: At least one of the bodies deforms and does not regain its original shape; the bodies do not separate after impact.
Energy Loss: Kinetic energy is lost during inelastic impacts, typically converted to heat or sound.
Example: A ball of clay dropped on a surface demonstrates inelastic impact behavior.

Principle: In the absence of external forces, the total momentum of a given system remains constant, described mathematically as:
$m_1v_1 = m_2v_2$
Application: This principle assists in calculating the velocity after a collision.
During an inelastic collision, the sum of the two objects' momentums before the collision is equal to their single combined momentum after the collision.

Scenario: A 90 kg hockey player traveling with a velocity of 6 m/s collides head-on with an 80 kg player traveling at 7 m/s. The two players entangle and continue traveling together.
Known Variables:
- Mass of player 1, $m_1=90\operatorname{kg}$
- Velocity of player 1, m1= 6 m/s
- Mass of player 2, $m_2=80\operatorname{kg}$
- Velocity of player 2, v2= -7 m/s
Collision Analysis:
$m_{ ext{total}} = m_1 + m_2 = 90 ext{ kg} + 80 ext{ kg} = 170 ext{ kg}$
Momentum Before Collision:
$m_1V_1 + m_2V_2 = (90 ext{ kg})(6 ext{ m/s}) + (80 ext{ kg})(-7 ext{ m/s})$
$= 540 ext{ kg m/s} - 560 ext{ kg m/s} = -20 ext{ kg m/s}$
After Collision:
$(-20 ext{ kg m/s}) = (170 ext{ kg})(v)$
- Solving for $v$ gives:
  $v = rac{-20 ext{ kg m/s}}{170 ext{ kg}} = 0.12 ext{ m/s}$
- The velocity is in the direction of the original velocity of the 80 kg player.

Definition: Coefficient of Restitution describes the elasticity of an impact and the interaction between two bodies during the impact. It is not descriptive of any single object or surface.
Characteristics:
- Unitless number between 0 and 1:
- Closer to 1: More elastic behavior.
- Closer to 0: More inelastic behavior.

Two Methods to Calculate CoR:
1. Based on Velocities
- Involves the velocities of two moving bodies before and after impact.
- Newton's formulation: $|v_1 - v_2| = e |u_1 - u_2|$
  - Where:
    - $e$ is CoR
    - $u_1, u_2$ are velocities before impact
    - $v_1, v_2$ are velocities after impact
1. Based on Heights
- Involves the impact between one moving body and a stationary body, as the velocity of a stationary body is considered to be zero.
- Formula:
  e = \frac{v2 - v1}{u1 + u2}, where e is the coefficient of restitution, v1 and v2 are the final velocities of the two bodies, and u1 and u2 are their initial velocities.

Increases in impact velocity and temperature can increase CoR:
- For example, higher incoming ball or implement velocity can increase the CoR between the implement and the ball.
- Manufacturers of sports equipment must adhere to strict rules regarding the stiffness of their equipment, which affects CoR.
- Different types of balls reflect different CoR characteristics.

Problem: A basketball is dropped from a height of 2 m onto a gymnasium floor. If the coefficient of restitution between the ball and the floor is 0.9, how high will the ball bounce?
Known Variables:
- Drop Height ( $h_d$ ) = 2 m
- CoR ( $e$ ) = 0.9
Calculation:
$e = rac{h_b}{h_d} <br>ightarrow 0.9 = rac{h_b}{2 ext{ m}}$
- Rearranging Gives:
  $h_b = 0.9 imes 2 ext{ m} = 1.8 ext{ m}$