1. In Motion! Momentum and collisions

2. Momentum affects collisions

3. Conservation of momentum Unless an external force acts upon an object or system of objects, momentum is conserved (Newton’s first law restated)

4. Let’s look at the example of two kids on skates facing each other (one boy, and one girl). The girl’s mass is 55 kg, and the boy’s is 60.0 kg. The push causes the boy to slide backwards at 1.0 m/s. How fast does the girl slide? The girl gives the boy a little push, they each move backwards (Newton’s third law).

5. REMEMBER: Momentum is conserved (there is no external force, like wind causing their momentum to change). Momentum before the push = momentum after the push! P boy + P girl = 0 ( originally they have no motion, therefore no momentum) After the push, the momentum of the system (the boy and the girl) is conserved (it still adds up to zero) P boy + P girl = 0m boy v boy + m girl v girl = 0 (60)(-1) + (55)(v girl ) = 0 -60 + 55 v girl = 060= 55v girl v girl = 60/55 = 1.1 m/s

6. Practice: Two ice skaters, a woman (mass=54 kg), and a man (mass= 88 kg) are initially facing each other on smooth ice where friction is negligible. Starting from rest they push off from each other. The woman moves away with a velocity of 2.5 m/s. Find the velocity of the man.

7. Solution Given: m m = 88 kg,m w = 54 kg, v w = 2.5 m/s v m = ? P m + P w = 0 P m = - P w m m (v m ) = - (m w (v w ) ) 88(v m ) = -(54)(2.5) v m = -1.5 m/s

8. Momentum is conserved in all situation. The total (sum) momentum of the system before the collision is equal to the total (sum) momentum of the system after the collision.

9. Conservation of Momentum Examples Elastic and Inelastic Collisions Remember momentum is conserved! P system before = P system after Inelastic Collisions (objects join together) Example: A sports car (mass=825 kg) travelling at 12 m/s crashes into a slower moving truck (mass = 1225kg) which is moving at 6.9 m/s in the same direction. If the two vehicles remain locked together (inelastic), how fast will they be going? Given: m c =825 kg, m t = 1225 kg, v car before = 12m/s, v truck before = 6.9 m/s v car&truck after =?

10. Before CollisionAfter Collision Pc + Pt = P car before = (825)(12) = 9900 kg∙m/s P truck before = 1225 (6.9) =8452.5 kg∙m/s P system before = 9900 + 8452.5 = 18352.5 kg∙m/s P (c & t together) = 18352.5 kg∙m/s = m (c & t) ∙ v (c & t) 18352.5 = (825+1225)v 18352.5 = (2050)v v = 18352.5/2080 v = 8.9 m/s + =

11. Elastic Example Example: A 125 g ball travelling at 55 cm/s to the right collides head–on with another ball at rest (m=215g). If the collision is totally elastic, calculate the velocity of the 125g ball after the collision if the 215g ball travels to the right at 17 m/s.

12. Given: InitialFinal m 1 = 125 g =.125 kg m 2 = 215 g =.215 kg m 1 = 125 g =.125 kg m 2 = 215 g =.215 kg v 1 = +55 cm/s =.55 m/s v 1 = ? v 2 = 0v 2 = 17 m/s

13. Momentum of system before collision = momentum of system after collision p 1i + p 2i = p 1f + p 2f p 1i = m 1 (v 1i ) = (.125)(.55) = 0.06875 kg m/s p 2i = m 2 v 2i = 0 p 1f = (.125) (v 1f ) p 2f = (.215)(17) = 3.655 kg m/s p 1i + p 2i = p 1f + p 2f 0.06875 + 0 = (.125) (v 1f ) + 3.655 v 1f = -29 m/s

14. Quick Quiz A 125 g ball travelling at 5 m/s to the right collides head–on with another ball at rest (m=200g). If the collision is totally elastic, calculate the velocity of the 125g ball after the collision if the 200g ball travels to the right at 17 m/s.

15. Breaking Distance and Friction Braking Distance Braking distance is proportional to the square of the velocity d v 2 This means if you are going twice as fast, it takes 4 X the distance to come to a stop! Braking distance also depends on the friction between the two surfaces (high friction, short breaking distance and vice-versa)