Key relationship
J = (1+e)[(uA−uB)·n]/(1/mA+1/mB), for approaching pucks. vA = uA−Jn/mA; vB = uB+Jn/mB; tangential components unchanged.
Physics · Class 11
Collisions in 1D and 2D
An isolated collision conserves vector momentum. Restitution specifies how relative normal velocity changes; e = 1 is elastic and e = 0 removes relative normal motion. For smooth oblique contact only the normal components change, while tangential components remain unchanged. With no approaching normal motion the model applies no impulse.
Open the interactive experiment →Conserve both momentum components.
தனிமையான மோதலில் இரு திசைகளிலும் மொத்த உந்தம் காக்கப்படும். மீட்சிக் குணகம் செங்குத்துத் தொடர்பு திசையில் சார்பு வேகத்தை நிர்ணயிக்கும்; வழுவழுப்பான தொடர்பில் தொடுகோட்டு வேகம் மாறாது.
Key notes, relationships and model limits
Remember
An isolated collision conserves vector momentum. Restitution specifies how relative normal velocity changes; e = 1 is elastic and e = 0 removes relative normal motion. For smooth oblique contact only the normal components change, while tangential components remain unchanged. With no approaching normal motion the model applies no impulse.
Common trap
In a smooth 2D impact, e = 0 means equal normal velocities, not necessarily complete sticking in every direction.
Model limits
Instantaneous smooth point-puck contact; n points from A to B; no external impulse, tangential impulse or spin. e applies only along n. e = 0 gives equal normal velocities; in 2D tangential velocities can differ, so pucks need not stick together.
What you will learn
Solve 1D and oblique 2D collisions using momentum, restitution and impact direction.
Try the experiment
Compare 1D elastic and inelastic presets, then the oblique impact. Play through t = 5 s and rotate the impact normal.
Think of it this way
Smooth billiard balls exchange a push along their line of contact; sideways motion survives that push.
Worked example
Equal 2 kg pucks with incoming x velocities 4 and 0 m/s swap velocities for e = 1. For e = 0 they share 2 m/s in 1D and lose 8 J of kinetic energy.
Why this matters
Separating normal and tangential motion turns a 2D collision into a 1D restitution calculation plus unchanged sideways motion.
Before you start
Vector components · velocity · SI units
Check your understanding
For a smooth oblique impact, which velocity component is unchanged for each puck?
- The normal component
- The tangential component
- Both components always
Read the explanation
The impulse acts along the impact normal; it has no tangential component, so each tangential velocity is unchanged.
Course connection: CBSE XI Chapter 5; NEET/Main Unit 4; JEE Advanced Mechanics. Apply each exam’s stated assumptions.