Elastic Collisions: Oblique Impact

Elastic Collisions: Oblique Impact

Oblique impact problems involve objects colliding at angles — a ball bouncing off a wall at an angle, or two snooker balls colliding off-centre. The key principle is to resolve into components parallel and perpendicular to the line of impact.

Newton's Experimental Law (Restitution)

For a direct collision between two objects:

Relative speed of separation = e × Relative speed of approach

where e is the coefficient of restitution (0 ≤ e ≤ 1).

Value of e	Type of collision
e = 1	Perfectly elastic (KE conserved)
0 < e < 1	Inelastic (some KE lost)
e = 0	Perfectly inelastic (objects coalesce)

Oblique Impact with a Fixed Surface

When a ball hits a smooth wall at angle α to the wall (or equivalently, at angle (90° − α) to the normal):

Resolve into two components:

• Parallel to wall: unchanged (no friction, so no impulse in this direction)
• Perpendicular to wall (normal): speed is multiplied by e and direction reverses

If the ball approaches with speed u at angle α to the wall:

• Component parallel to wall: u cos α → u cos α (unchanged)
• Component perpendicular to wall: u sin α → eu sin α (reversed)

Rebound speed: v = √(u²cos²α + e²u²sin²α) = u√(cos²α + e²sin²α)

Rebound angle β to the wall: tan β = eu sin α / (u cos α) = e tan α

So: tan β = e tan α

This is a key result: the rebound angle satisfies tan β = e tan α, where both angles are measured from the wall.

Kinetic Energy Lost

KE lost = ½mu²sin²α(1 − e²)

This is the energy lost in the normal component; the parallel component contributes no loss.

Oblique Impact Between Two Spheres

When two smooth spheres collide, the line of centres (the line joining their centres at the moment of impact) determines the direction of impact.

Resolution principle:

• Along the line of centres: apply conservation of momentum AND Newton's law of restitution
• Perpendicular to line of centres: each sphere's velocity component is unchanged (smooth spheres exert no friction on each other)

Setting Up the Problem

Step 1: Identify the line of centres at the moment of collision.

Step 2: For each sphere, resolve its velocity into:

• u_n = component along the line of centres ("normal" component)
• u_t = component perpendicular to the line of centres ("tangential" component)

Step 3: Apply to the normal components only:

• Conservation of momentum: m₁u₁_n + m₂u₂_n = m₁v₁_n + m₂v₂_n
• Restitution: v₂_n − v₁_n = e(u₁_n − u₂_n) [separation speed = e × approach speed]

Step 4: The tangential components are unchanged:

• v₁_t = u₁_t and v₂_t = u₂_t

Step 5: Recombine to find each sphere's final velocity (magnitude and direction).

Worked Example: Two Sphere Collision

Two smooth spheres A (mass 3 kg, velocity 8 m/s) and B (mass 1 kg, at rest) collide. The velocity of A makes an angle of 30° with the line of centres. e = 0.5.

Normal components (along line of centres):

• u_A_n = 8 cos 30° = 4√3 m/s
• u_B_n = 0

Conservation of momentum: 3(4√3) + 1(0) = 3v_A_n + v_B_n

12√3 = 3v_A_n + v_B_n ... (1)

Restitution: v_B_n − v_A_n = 0.5(4√3 − 0) = 2√3

v_B_n = v_A_n + 2√3 ... (2)

From (2) into (1): 12√3 = 3v_A_n + v_A_n + 2√3 → 4v_A_n = 10√3

v_A_n = 5√3/2 ≈ 4.33 m/s, v_B_n = 5√3/2 + 2√3 = 9√3/2 ≈ 7.79 m/s

Tangential components:

• v_A_t = u_A_t = 8 sin 30° = 4 m/s
• v_B_t = u_B_t = 0

Final velocities:

• A: speed = √(v_A_n² + v_A_t²) = √(75/4 + 16) = √(139/4) ≈ 5.89 m/s at angle arctan(4/(5√3/2)) ≈ 24.8° from line of centres
• B: speed = 9√3/2 ≈ 7.79 m/s along the line of centres

Successive Bounces

A ball bouncing repeatedly between two walls, or on a floor, loses speed in the normal direction by factor e each bounce. After n bounces:

• Normal speed = eⁿ × original normal speed
• Parallel speed unchanged (smooth surfaces)
• Height of nth bounce = e^(2n) × original height (for vertical bouncing)

Loss of Kinetic Energy in Oblique Collisions

For two-sphere oblique collisions, KE is only exchanged/lost in the normal direction:

KE lost = (m₁m₂)/(2(m₁+m₂)) × (1 − e²) × (u₁_n − u₂_n)²

The tangential components contribute no loss.

Exam Tips

• Always clearly define the line of centres and resolve velocities into normal and tangential components before doing anything else
• The tangential components are unchanged for smooth surfaces — this is the simplest part, don't overcomplicate it
• In the restitution equation, make sure you use separation speed / approach speed, and get the signs right (both measured along the same direction)
• tan β = e tan α for wall bounces — learn this result, it saves time
• Check your answer is physically reasonable: the rebound angle should be shallower than the approach angle (unless e = 1)
• For repeated bouncing problems, use geometric series for total distance or time
• Conservation of momentum applies along the line of centres only (perpendicular impulse is zero for smooth spheres)