Now consider two tennis balls…… The “System” Internal Forces – exerted by objects within the system External Forces – exerted by objects outside the system Momentum of a system m1 m2 m3 v1 v2 v3 P=total system momentum = p1 + p2 + p3 = m1v1 + m2v2 + m3v3

Solve with G10 understanding???? If each tennis ball has a mass of 1 kg and a speed of 2 m/s, what is the total momentum of the system? Two people are initially at rest on ice skates. After they push off on each other, the boy has a speed of 4.0 m/s. What is the speed of the girl? 75 kg 50 kg Solve with G10 understanding????

Conservation of Momentum If the net external forces on a system are zero, the total system momentum is conserved (remains constant). If a system has no external forces acting on it (eg no friction), then the system is isolated. Total momentum in an isolated system is said to be conserved.

Although interactions within the system may change the distribution of the total momentum among the various bodies in the system, the total momentum does not change. Such interactions can give rise to two general classes of events: a. explosions, in which an original single body flies apart into separate bodies

b. collisions, in which two or more bodies collide and either stick together or move apart, in each case with a redistribution of the original linear momentum.

Conservation of Momentum - Proof Fm = - FM FM Fm M m FmDt = - FMDt M m v2 u2 Dpm = - DpM m(u2 - u1) = -M(v2 - v1) Conservation of Momentum: Blocks have the same speed (v) at closest approach. mu2 - m u1 = - Mv2 + Mv1 Conservation of Momentum Mv1 + m u1 = Mv2 + mu2 p(before) = p(after)

Completely Inelastic Collisions Conservation of Momentum holds true for both types of collisions………… Elastic Collisions The total amount of kinetic energy is conserved Inelastic Collisions The total amount of kinetic energy is not conserved Completely Inelastic Collisions The objects stick together

Perfectly Inelastic Collision #1 vo v = ? M m M m before after Using the conservation of momentum to find the final speed when two blocks stick together.

A cart moving at a speed v collides with an identical stationary cart on an air track, and the two stick together after the collision. What is their velocity after colliding? 2 v v 0.5 v zero

after before v vo M m Find the speed of the bullet and block after impact.

Ex: A speeding bullet of mass 7.50 g is flying through the air at 625 m/s when it strikes and lodges in a wood block of mass 350 g as it rests on the ice. What is the speed of the bullet/block combination immediately after the collision?

Perfectly Inelastic Collision before v1 v2 after m1 m2 vf m1 m2 Find the final speed after a Perfectly Inelastic Collision

Ex: A completely inelastic collision What is the final velocity? Completely inelastic collision

Ballistic Pendulum vo M h m v Ballistic Pendulum Cons. Energy Cons. Momentum vo M m v Ballistic Pendulum Use conservation of momentum and energy.

Ballistic Pendulum Find the speed of the bullet vo M h m v Cons. Momentum vo M m v

Ballistic Pendulum h Cons. Energy M v

K0 = Uf m1v1= ( m1+m2 )v' m1 = 0.012 g m2= 2 kg h = 0.1 m A 12-g bullet is fired into a 2-kg block of wood suspended from a cord. The impact of the bullet causes the block to swing 10-cm above its original level. Find the velocity of the bullet as it strikes the block. m1 = 0.012 g m2= 2 kg h = 0.1 m K0 = Uf m1v1= ( m1+m2 )v'

K0 = Uf = 1.4 m/s m1v1= ( m1+m2 )v' = 234.7 m/s m1 = 0.012 g m2= 2 kg h = 0.1 m K0 = Uf = 1.4 m/s m1v1= ( m1+m2 )v' = 234.7 m/s

A bullet of mass 30g is fired with a speed of 400ms-1 into a sandbag A bullet of mass 30g is fired with a speed of 400ms-1 into a sandbag. The sandbag has a mass of 10kg and is suspended by two ropes so that it can swing. What is the maximum vertical height, h, that the sandbag rises as it recoils with the bullet lodged inside?

Hard! Ex: An elastic collision What will the final velocities be? Using conservation of momentum and conservation of kinetic energy, we can show that: Hard! and Elastic collision, m2 initially at rest