Collisions and restitution flashcards

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Impulse
The change in momentum of a particle: I = mv − mu. A vector; unit N s (= kg m s⁻¹). For a constant force F acting for time t, I = Ft.
Momentum (as a vector)
Mass × velocity, mv. It has the direction of the velocity; unit kg m s⁻¹ or N s.
Principle of conservation of linear momentum
If no external impulse acts, the total momentum of a system is the same before and after a collision.
Newton's law of restitution
For a direct impact, speed of separation = e × speed of approach.
Coefficient of restitution e
The constant ratio (speed of separation) ÷ (speed of approach) for two given bodies; 0 ≤ e ≤ 1.
Perfectly elastic impact
e = 1: no kinetic energy is lost.
Inelastic impact (e = 0)
The bodies do not separate: they coalesce and move with a common velocity.
Direct impact
A collision in which the particles move along the line of centres before and after the impact.
Impulse–momentum principle (vector form)
I = mv − mu, applied separately to the i and j components. (No — learn it)
Conservation of momentum for two particles on a line
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (signed velocities). (No — learn it)
Restitution equation with signed velocities
v₂ − v₁ = e(u₁ − u₂), where particle 1 is behind particle 2. (No — learn it)
Particle hitting a fixed smooth wall at right angles
Rebound speed = e × impact speed, direction reversed. (No — learn it)
Kinetic energy lost in a direct impact (general)
½ · m₁m₂/(m₁ + m₂) · (1 − e²) · w², where w is the speed of approach. (No — derive it)
Velocity after an impulse
v = u + I/m. (No — rearrange I = mv − mu)
Momentum triangle for an impulse
Draw mu, then I head to tail from its end; the third side, from the start of mu to the end of I, is mv.
Velocity–time graph of a ball bouncing on the ground (up positive)
Parallel straight lines of gradient −g; at each bounce a vertical jump from −v to +ev.
Before-and-after diagram for a collision
Masses on each particle, a positive direction, and every unknown velocity drawn in the positive direction.
Distance–time graph for two particles after impacts
A further collision happens where the two lines cross; parallel or diverging lines mean no further collision.
Fraction of KE lost against e (fixed masses)
Largest at e = 0 (coalescing), falls as e increases, zero at e = 1.
Why are the impulses on two colliding particles equal and opposite?
Newton's third law: equal and opposite forces act for the same time.
Why can e not be greater than 1?
The kinetic energy lost is proportional to 1 − e²; e > 1 would create kinetic energy with no source.
Is momentum conserved when a ball bounces off a wall?
No: the wall gives the ball an external impulse. Use restitution and I = mv − mu instead.
When do two particles on a line collide again?
When the particle behind has a greater velocity (in the positive direction) than the particle in front.
Equal masses, e = 1, one at rest: what happens?
The moving particle stops and the other moves off with its velocity (they exchange velocities).
Why do bounce heights fall by a factor e²?
Each rebound speed is e times the impact speed, and height is proportional to speed squared.
0.5 kg particle, velocity (2i − j) m s⁻¹, receives impulse (3i + j) N s. New velocity and speed?
v = u + I/m = (2 + 6)i + (−1 + 2)j = 8i + j m s⁻¹; speed √65 = 8.06 m s⁻¹.
Two 2 kg particles; one at 4 m s⁻¹ hits the other at rest, e = ½. Velocities after?
4·2 = 2vA + 2vB and vB − vA = 2 ⇒ vA = 1 m s⁻¹, vB = 3 m s⁻¹ (same direction).
For the 2 kg pair above (4 m s⁻¹ onto rest, e = ½), kinetic energy lost?
Before 16 J, after ½·2·(1² + 3²) = 10 J, so 6 J lost.
0.3 kg ball hits a wall at right angles at 5 m s⁻¹, e = 0.6. Rebound speed and impulse?
Rebound 3 m s⁻¹; impulse 0.3(3 + 5) = 2.4 N s away from the wall.
Ball dropped from 0.9 m bounces to 0.4 m. Find e.
e = √(0.4/0.9) = ⅔ (speeds are √(2g·0.9) and √(2g·0.4)).
Mass m at speed u hits mass 3m at rest, e = ½. Velocities after?
mu = mv₁ + 3mv₂ and v₂ − v₁ = ½u ⇒ mass m: v₁ = −u/8 (rebounds at u/8); mass 3m: v₂ = 3u/8 forwards.
Then 3m hits a wall (coefficient f) and rebounds. For which f does it catch m again?
3fu/8 > u/8 ⇒ f > ⅓.
Restitution sign trap
Use signed velocities in v₂ − v₁ = e(u₁ − u₂); a reversing particle has a negative velocity, so speeds add.
Impulse on a ball that rebounds
I = m(v − u) with opposite signs: 0.2 kg from 6 m s⁻¹ to 3 m s⁻¹ back gives 0.2 × 9 = 1.8 N s, not 0.6 N s.
Answering “show there are no further collisions”
List all final velocities with signs, then compare each neighbouring pair in a sentence.
Kinetic energy lost — common slip
Use total KE before minus total KE after with speeds squared; a negative answer signals an arithmetic error.
Unknown e in a “show that”
Keep e as a letter and use 0 ≤ e ≤ 1 to justify directions, e.g. 2 − e > 0.
Oblique impacts
Not in M2: here every impact is direct, along the line of centres or at right angles to a wall.

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