Chapter 6

When objects collide their motion changes and this is the result of a concept called momentum. Momentum = mass x velocity p = mv kgm/s or Ns

The reason the momentum changes motion in a collision is that a force is applied to the object for an amount of time and that’s called impulse Impulse = Force x time F t = ∆p ∆p = mv f - mv o

If a 50 g tennis ball is hit by a racket with 18 N of force for s and the ball was at rest, how fast is it going after the collision? F t = mv f – mv o (18N)(0.095s) = (0.050kg)vf – 0 vf = 34 m/s

Conservation of momentum Momentum is always conserved p o = p f So if you jump, 56kg off a 15kg raft with a velocity of 5m/s, how fast is the raft going?

p o = p f (m 1 +m 2 )v o = m 1 v 1f + m 2 v f2 (71kg)(0) = (56kg)(5m/s) + (15kg)(v f2 ) -(56kg)(5m/s) = (15kg)(v f2 ) v f2 = -18 m/s

The best way to do this is to start with p o = p f And then write each momentum on it’s side. If there are two separate objects then each has it’s own (not always combined) m 1 v 1o +m 2 v 2o = m 1 v 1f + m 2 v f2

Elastic Collision (perfectly) KE 0 = KE f and p o = p f Inelastic Collisions KE are not equal but p o = p f KE = ½ mv 2

If your 3500 Kg truck is hit elastically by another 3000 Kg truck travelling 20 m/s of and the second truck bounces back with a speed of 1.8 m/s, how fast are you moving after the collision. Assume it’s straight on. And do both the momentum equation and the kinetic energy equation to verify your answer.

p o = p f m 1 v 1o +m 2 v 2o = m 1 v 1f +m 2 v 2f 3500Kg(0) kg(20m/s) = 3500kg(v 1f ) kg(-1.8m/s) v 1f =

KE 0 = Ke f ½ m 1 v 1o 2 + ½ m 2 v 2o 2 = ½ m 1 v 1f 2 + ½ m 2 v 2f ½ 3000kg(20m/s) 2 = ½ 3500kg(v 1f ) 2 + ½ 3000kg(-1.8m/s) 2 v 1f =