Collisions - Collision in Two Dimensions

A collision in two dimensions, also called an oblique or non-head-on collision, occurs when at least one object's initial velocity is NOT along the line connecting the centers of the two objects at impact. In such collisions, we need to carefully analyze the motion using two perpendicular directions. The first direction is along the line of impact (where forces act during collision), and the second is along the common tangent (where no force acts). By studying these collisions, we can predict how objects will move after impact, which is important in understanding real-world phenomena like ball sports, vehicle crashes, and particle physics. These types of collisions follow the same fundamental principles as head-on collisions but require mathematical treatment in two dimensions.

A collision in two dimensions (oblique collision) is defined as: a collision where the direction of at least one initial velocity is NOT along the line of impact, requiring analysis using two mutually perpendicular directions—the line of impact and the common tangent at the point of contact.

Oblique or non-head-on collision.

To solve for the unknowns in a two-dimensional collision, the following equations are used and solved simultaneously:

1. Apply Conservation of Linear Momentum along the Line of Impact: This provides the first equation relating the components of the initial and final velocities along this line
   m₁u₁cosα₁ + m₂u₂cosα₂ = m₁v₁cosβ₁ + m₂v₂cosβ₂     ---(4.12)

2. Apply Conservation of Momentum along the Common Tangent: Since there is no force along the common tangent (perpendicular to the line of impact), the momentum of each individual body is conserved along this line
   m₁u₁sinα₁ = m₁v₁sinβ₁     ---(4.13)
   and m₂u₂sinα₂ = m₂v₂sinβ₂     ---(4.14)

3. Apply the Coefficient of Restitution (e): The coefficient of restitution is applied along the line of impact, providing a fourth equation
   e = -\[\left(\frac{\mathrm{v}_2\mathrm{cos}\beta_2-\mathrm{v}_1\mathrm{cos}\beta_1}{u_2\mathrm{cos}\alpha_2-u_1\mathrm{cos}\alpha_1}\right)=\frac{\mathrm{v}_2\mathrm{cos}\beta_2-\mathrm{v}_1\mathrm{cos}\beta_1}{u_1\mathrm{cos}\alpha_1-u_2\mathrm{cos}\alpha_2}\]     ---(4.15)

4. Solve the System of Equations: Equations (4.12), (4.13), (4.14), and (4.15) are then solved for the four unknowns: $v_1$, $v_2$, $\beta_1$, and $\beta_2$.

Bullets Hitting a Fixed Surface

Problem: Bullets of mass 40 g each are fired from a machine gun at a rate of 5 per second towards a firmly fixed hard surface of area 10 cm². Each bullet hits normal to the surface at 400 m/s. The coefficient of restitution for the collision is 0.75. Calculate the average force and average pressure experienced by the surface.

Solution:

Step 1: Identify given values

• Mass per bullet: m = 40 g = 0.04 kg
• Initial velocity: u₁ = 400 m/s
• Coefficient of restitution: e = 0.75
• For fixed surface: u₂ = v₂ = 0
• Area: A = 10 cm² = 10⁻³ m²
• Firing rate: 5 bullets/second

Step 2: Calculate final velocity using the coefficient of restitution
e = \[\frac{v_1-v_2}{u_2-u_1}=\frac{v_1-0}{0-400}\]

0.75 = \[\frac{v_{1}}{-400}\]

v₁ = −300 m/s

(Negative sign indicates bullet rebounds in the opposite direction)

Step 3: Calculate momentum change per bullet
$Momentum\; transferred\; to\; surface\; per\; collision=m(u1-v1)$

p = 0.04(400 − (−300)) = 0.04 × 700 = 28 N

Step 4: Calculate momentum received per second
\[\frac {dp}{dt}\] = 28 × 5 = 140 N

This equals the average force experienced by the surface.

Step 5: Calculate average pressure
P = \[\frac {F}{A}\] = \[\frac {140}{10^-3}\] = 1.4 × 10⁵ N/m²

Answer:

• Average Force: 140 N
• Average Pressure: 1.4 × 10⁵ N/m² (approximately 1.4 times atmospheric pressure)