Momentum and Collisions Chapter 6

Chapter Organization Momentum of a single object Momentum between objects or collisions

Single Object Momentum Momentum (r) = mv Momentum (r) – kg m/s or N s Mass (m) – kg Velocity (v) – m/s

Change in Momentum Change in momentum = D r r = m D v r = m(vf –vi) r = mvf – mvi r = rf - ri D r = Impulse = F t

Single Object Momentum-EX A car weighing 15,680 N and moving at 20.0 m /s is acted upon by 640 N force until it is brought to a halt. What is the mass and weight of the car? Weight = 15,680 N m = W/g = 1568 kg What is the change in momentum? Dr = m D v = 1568 kg (0 – 20.0 m /s) = -31,360 kg m /s (momentum is decreasing as it slows down so it is negative)

Single Object Momentum-EX What is the initial momentum ? ri = m vi = 1568 kg (20.0 m /s) ri =31,360 kg m/s What is the final momentum? rf = m vf = 1568 kg (0 m /s) rf = 0 kg m/s

Single Object Momentum-EX What is the impulse of the car? Dr = rf - ri = 0 kg m /s – 31,360 kg m /s Dr = impulse = -31,360 kg m /s How long does the braking force act on the car to bring it to a halt? Dr = F t t = Dr / F t= -31,360 kg m /s / - 640 N = 49 s

Conservation of Momentum m1v1 + m2 v2 = m1 v1’ + m2v2’ Subscript 1 is object one Subscript 2 is object two ‘ is called prime Prime is used to indicate after the collision

Three types of collisions “Nonsticky” “Sticky” “Newton’s Third”

Nonsticky Same direction EX: A plastic ball of mass 0.200kg moves with a velocity of 0.30 m/s. This plastic ball collides with a second plastic ball of mass 0.100kg that is moving along the same line at a velocity of 0.10 m/s. After the collision, the velocity of the 0.100kg ball is 0.26 m/s. What is the velocity of the first ball? 0.2 ( 0.3) + 0.1 (0.1) = 0.2 (v1’) + 0.1 (0.26 ) 0.06 + 0.01 = 0.2 (v1’) + 0.026 0.07 – 0.026 = 0.2 (v1’) + 0.026 – 0.026 0.044 / 0.2 = 0.2 (v1’) /0.2 0.22 m/s = v1’

Nonsticky Opposite direction Sign rule: 1st object is right &positive and left is negative EX: A 0.50kg ball traveling at 6.0 m/s collides head-on with a 1.00kg moving in the opposite direction at a velocity of -12.0 m/s. The 0.50kg moves away at -14 m/s after the collision. Find the velocity of the 1.00 kg ball after the collision. .5(6) + 1(-12) = .5(-14 ) + 1(v2’) -9 = -7 + 1(v2’) -2 = 1(v2’) -2 m/s = v2’

Sticky Sticky means that the two objects stick together when they collide and fly off together at the same speed. You still use m1v1 + m2 v2 = m1 v1’ + m2v2’ but v1’ = v2’

Example of a Sticky m1v1 + m2 v2 = (m1 +m2) v’ Moving at 20.0 m/s, a car of mass 700Kg collides with a stationary truck of mass 1400 kg. If the two vehicles interlock as a result of the collision, what is the velocity of the car-truck system? 700kg(20m/s) + 1400kg(0m/s)= (700+1400) v’ 14000 kg m/s = 2100kg ( v’) 14000 kg m/s/ 2100kg = 2100kg ( v’) / 2100 kg 6.67 m/s = v’

Newton’s 3rd Law Problems m1v1 + m2 v2 = m1 v1’ + m2v2’ Where v1 = v2 = 0 Ex: A 40.0 kg projectile leaves a 2000 kg launcher with a velocity of 800 m/s. What is the recoil velocity of the launcher? 2000(0) +40 (0) = 2000 (v1’)+ 40 (800) 0 = 2000 (v1’)+ 32,000 -32,000 = 2000 (v1’) -32,000 /2000 = 2000 (v1’) / 2000 -16 m/s = v1’