Physics Study Notes on Collisions and Rocket Propulsion

8.5 Inelastic Collisions in One Dimension

The location of the impact of the tennis ball on the racquet is crucial.
Important Factors:
- Timing of the impact during the stroke.
- Hitting the "sweet spot" minimizes vibration and impact, enhancing ball velocity and reducing sport injuries (e.g., tennis elbow).
Physics concepts in sports science:
- Momentum
- Rotational motion
- Vibrations

Materials Needed: Racquet
- Suitable racquets include tennis, badminton, or others.
- Procedure:
  - Place the racquet on the floor and stand on the handle.
  - Drop a tennis ball onto the strings from a measured height.
  - Measure the height of the ball's bounce.
  - Have a friend hold the racquet at the handle and repeat the drop.
  - Observe what happens to the friend’s hand during collision and measure bounce height.
- Explain observations and measurements.
Coefficient of Restitution (c):
- A measure of collision elasticity, defined as the ratio of post-collision speed to pre-collision speed.
- Perfectly elastic collisions have c = 1.
- For calculating c:
  c = \left( \frac{h}{H} \right)^{1/2}
  where h = bounce height, H = drop height.
- Determine c in experiments, known value for tennis ball on concrete/wood is c = 0.85.

Scenario Description: Two carts collide inelastically.
Cart 1 (m₁):
- Mass = 0.350 kg
- Initial Velocity (v₁) = 2.00 m/s
Cart 2 (m₂):
- Mass = 0.500 kg
- Initial Velocity (v₂) = -0.500 m/s
After collision, Cart 1 observes a recoil velocity = -4.00 m/s.
Questions:
- (a) Determine final velocity of Cart 2.
- (b) Calculate energy released by the spring during the collision.

Conservation of momentum equation:
m₁ v₁ + m₂ v₂ = m₁ v'₁ + m₂ v'₂
Solve for final velocity of Cart 2 (v'₂): v'₂ = \frac{m₁ v₁ + m₂ v₂ - m₁ v'₁}{m₂}
- Substituting known values:
  v'₂ = \frac{(0.350 kg)(2.00 m/s) + (0.500 kg)(-0.500 m/s) - (0.350 kg)(-4.00 m/s)}{0.500 kg}
  v'₂ = 3.70 m/s

Internal Kinetic Energy Before Collision:
KE_{int} = \frac{1}{2} m₁ v₁² + \frac{1}{2} m₂ v₂²
Substitute values:
KE_{int} = \frac{1}{2}(0.350 kg)(2.00 m/s)² + \frac{1}{2}(0.500 kg)(-0.500 m/s)²
= 0.763 J
After Collision:
KE'_int = \frac{1}{2} m₁ v'₁{}^{2} + \frac{1}{2} m₂ v'₂{}^{2}
= \frac{1}{2}(0.350 kg)(-4.00 m/s)² + \frac{1}{2}(0.500 kg)(3.70 m/s)²
= 6.22 J
Change in Internal Kinetic Energy:
KE'int - KE{int} = 6.22 J - 0.763 J = 5.46 J
Outcome: Cart 2's velocity after collision is positive, indicating rightward movement. Internal kinetic energy increases by 5.46 J, released by the spring.

At the end of this section:

Discuss 2D collisions as an extension of 1D analysis.
Define point masses.
Derive conservation of momentum expressions along x and y-axes.
Describe elastic collisions of equal mass objects.
Determine final velocities given initial velocities and scattering angles.
Examining only 1D collisions (incoming and outgoing velocities lie along one line).
Two-dimensional collisions (e.g., billiard balls) = objects scatter.
Use a coordinate system and resolve motion into components.
Rotation Complication: Objects might rotate; e.g., two ice skaters hooking arms.
For currently considered scenarios, avoid rotation by focusing on point masses (structureless, cannot rotate).

While Fnet = 0 (indicates momentum p is conserved), we simplify analyses for coordinate systems with axes parallel to velocity of incoming particle.

P{1x} + P{2x} = P'{1x} + P'{2x}

Substituted in mass and velocity form:
Initial momentum will focus where m₂ is at rest:
m₁ v₁ = m₁ v'{1}(cos \theta1) + m₂ v'{2}(cos \theta2)

0 = m₁ v{1y} + m{2} v_{2y}

Recall: both particle structures initially in horizontal motion; hence $v{1y}$ and $v{2y}$ = 0.

Along y-axis, we get:
0 = m₁ v{1} sin(\theta1) + m₂ v{2} sin(\theta2)
Since innumerable collisions occur with mass m₂ at rest, 2D collision equations become valuable in analyzing interactions.

0.250-kg object (m₁)
Initially slides towards another stationary object (m₂) with mass of 0.400 kg.
m₁ exits at angle 45° with initial direction; speeds 2.00 m/s before impact, 1.50 m/s post-collision.
Find velocity (magnitude and direction) for m₂ after the collision.
Strategy: As surface is frictionless, and momentum is conserved, apply equations for both x and y axes.

m₁u{1} = m{1}v{1}cos(\theta1) + m{2}v{2}cos(\theta_2)
0 = m₁u{1} sin(\theta1) + m{2}v{2} sin(\theta_2)
Find two unknowns with two equations (focus on either equation from y or x-axis).

Apply ratio yields an equation to simplify:
tan(\theta2) = \frac{m{1} sin(\theta{1})}{m{2}sin(\theta_{2})}
Enter known values:
tan(\theta2) = \frac{(1.50 m/s)(0.7071)}{(1.50 m/s)(0.7071)-2.00 m/s} \theta2 = tan^{-1}(-1.129) ≈ 312°
Solve for v'₂ using previous relationships.

Interesting phenomena arise during elastic collisions with equal mass (e.g., billiard balls).
Internal kinetic energy is conserved during collision.
Mathematical relationships establish conservation of both momentum and kinetic energy for elastic collisions.

Two-dimensional collision experimentation enhances subatomic particle understanding, crucial in nuclear and particle physics.