applying a force. impulse momentum conservation of momentum collisions momentum

Applying a Force

ImpulseMomentum

Conservation of MomentumCollisions

Momentum

ImpulseThe product of force and contact time

Vector quantity, Symbol: JDirection is the same as the net or

average force appliedUnits: N-s, kg-m-s-1

Momentum

Product of mass and velocityVector quantity, Symbol: p

Direction is the same as the velocity

Units: kg-m-s-1

Large Force, Short Contact Time

Can you give other examples?

Small Force, Long Contact TimeAirbags

Seatbelt

Small Force, Long Contact TimeCatching a baseball

Bungee Jumping

Example

A cricket ball, mass 0.5 kg, was bowled at50 m s-1 at a batsman who misreads theball and the 5 kg bat is knocked out ofhis hands, the ball rebounds at 25 ms-1.

What is the change in momentum of the ball?If the bat was in contact with the ball for 2.0

ms, how much force did the batsman apply on the ball?

Conservation of Momentum

Total momentum before is equal to total momentum after

In a closed system (external forces are negligible)

Inelastic CollisionOnly momentum is conserved

Perfectly inelastic collision(The colliding bodies couple after

the collision)

Example

A railway wagon travelling at 1.0 m s–1 catches up with and becomes coupled to another wagon travelling at 0.5 m s–1 in the same direction. The faster moving wagon has 1.7 times the mass of the slower one. Immediately after impact, what is the speed of the coupled wagons?

Elastic Collision

Momentum is conserved.Kinetic Energies are conserved.

(Relative Velocities) are conserved.Analyzing billiard balls

Simulation

ExampleA trolley of mass 1 kg rolls along a level,

frictionless ramp at a speed of 6 m s-1. It collides with a second trolley of mass 2 kg which is initially at rest. The first trolley rebounds at a speed of 2 m s-1.Find, by conservation of momentum, the

velocity of the second trolley after the collision.

Compare the kinetic energy before and after the collision. Is the collision elastic?

ExampleA ball of mass 0.20 kg is dropped from a

height of 3.2 m onto a flat surface which it hits at 8.0 m s-1. It rebounds to 1.8 m. (g = 9.8 m s-2)What is the rebound speed just after impact?What is the change in energy of the ball?What momentum change has the ball between

just touching the surface and leaving it?

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