Law of Conservation of Momentum (LCM)

Two Particle Collision Particle A traveling east collides with particle B traveling west. 3rd Law says that the force from A on B is equal to the Force from B on A. FB on A = - FA on B If the forces are the same and they act for the same amount of time, the impulses should be equal. JA = -JB A B

JA = -JB so FA∆t = - FB∆t Therefore mA ∆v = - mB ∆v If this is true than ∆pA = - ∆pB The magnitude of the ∆pA and ∆pB are the same but the direction must be opposite.

LCM The Law of Conservation of Momentum states: “ The total momentum in an isolated system does not change and remains constant.” An isolated system means that no other outside forces are acting on it. This does not exist on earth.

LCM Momentum cannot be created or destroyed but it can be transferred. Momentum Before = Momentum After can be Collision Explosion Any other interaction

Calculating LCM ∆pA = - ∆pB finalA – initialA = -(finalB – initialB) Final will be denoted by ‘ p’A – pA = - (p’B – pB) p’A + p’B = pA + pB Total initial momentum = Total final momentum. ***************************************** mAv’A + mBv’B = mAvA + mBvB

continued For any collision or interaction of 2 objects: Total momentum of the system is the same before and after the collision In an isolated system momentum is transferred and one’s gain is equal to the other’s loss.

Collisions Elastic: the two objects collide and then separate after transferring momentum. Inelastic: two object collide and stick together after transferring momentum. Examples

Elastic Collision mAvA + mBvB = mAv’A + mBv’B A .5kg cart traveling 10m/s collides with a 1kg cart at rest. After the collision the .5kg cart bounces back with a velocity of 4m/s. What is the final velocity of the 1kg cart? 7m/s

Inelastic Collision mAvA + mBvB = mAv’A + mBv’B Or mAvA + mBvB = (mA + mB)v’ A 3kg cart traveling at 4m/s collides and sticks to a 2kg cart traveling 2m/s in the opposite direction. How fast are they moving after the collision? v’ = 1.33m/s