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Applying a Force

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**Impulse Momentum Conservation of Momentum Collisions**

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**Impulse The product of force and contact time**

Vector quantity, Symbol: J Direction is the same as the net or average force applied Units: N-s, kg-m-s-1

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**Momentum Product of mass and velocity Vector quantity, Symbol: p**

Direction is the same as the velocity Units: kg-m-s-1

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**Large Force, Short Contact Time**

Can you give other examples?

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**Small Force, Long Contact Time**

Airbags Seatbelt

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**Small Force, Long Contact Time**

Catching a baseball Bungee Jumping

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**Example A cricket ball, mass 0.5 kg, was bowled at**

50 m s-1 at a batsman who misreads the ball and the 5 kg bat is knocked out of his 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?

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**Conservation of Momentum**

Total momentum before is equal to total momentum after In a closed system (external forces are negligible)

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**Inelastic Collision (The colliding bodies couple after the collision)**

Only momentum is conserved Perfectly inelastic collision (The colliding bodies couple after the collision)

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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?

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**Elastic Collision Momentum is conserved.**

Kinetic Energies are conserved. (Relative Velocities) are conserved. Analyzing billiard balls Simulation

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Example A 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?

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Example A 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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Momentum Notes. Momentum Momentum ( ρ) – inertia in motion Mass x Velocity ρ = mv measured in kg·m/s.

Momentum Notes. Momentum Momentum ( ρ) – inertia in motion Mass x Velocity ρ = mv measured in kg·m/s.

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