Do Now: A 1500 kg car speeds up from 15 m/s to 30 m/s in 6 seconds.

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Presentation transcript:

Do Now: A 1500 kg car speeds up from 15 m/s to 30 m/s in 6 seconds. How much work did the car do over this time? What was the rate of work done?

A Riddle… What do you get when you combine 2 bikes, two foam sticks and 2 idiots?

Conservation of Momentum A Collision! Conservation of Momentum

Review What were the two types of collisions that we discussed? Elastic and Inelastic What does it mean for momentum to be conserved? Pbefore = Pafter Write the Equation for both an elastic and inelastic collision!!!

ONE MORE THING! Momentum deals with objects in motion, so what type of energy do we deal with? KINETIC ENERGY or energy in motion KE is conserved for an elastic collision. KE is NOT conserved for inelastic collisions. If there is friction then KE will not be conserved in either case.

Station Activity Answer the following questions with your group within the time limit.

Station 1 A 1,000-kg car moving at 5 m/s collides into a 1200-kg car at rest. After the collision the 1000 kg car comes to rest. Calculate the velocity of the 1200kg car after the collision. Calculate the total momentum before and after the collision.

Solution 1 momentum before = momentum after pbefore = pafter

Station 2 A 0.26-kg cue ball moving at 1.2 m/s strikes a stationary 0.17-kg 8 ball. After the collision the cue ball comes to rest. A) Calculate the magnitude of the velocity of the 8 ball after the collision. B) Determine the total momentum after the collision.

Solution 2 Pbefore = Pafter v2’ = 1.8 m/s (0.26 kg)(1.2 m/s) + (1.7 kg)(0) =(0.26 kg)(0) + (0.17kg)(v2’) v2’ = 1.8 m/s Total Momentum = (.17 kg)(1.8 m/s) = 0.31 kg.m/s

Station 3 A 0.1-kilogram bullet is fired horizontally with a velocity of 400m/s into a 14.6-kg wooden block at rest. The bullet is imbedded in the wooden block. Determine the speed of the block after impact. Is KE conserved? Prove your answer with a calculation!

Solution 3 Pbefore = Pafter (0.1 kg)(400 m/s) + (14.6 kg)(0) =(0.1 kg + 14.6 kg)(v’) v’ = 2.72 m/s KE = 0.5mv2 KEBefore = 8000 J KEAfter = 54.4 J KE is not conserved!

Explosion/Recoil What happens to the man and the bullet in this case? How is momentum conserved? What was his momentum before the recoil?

Explosion/Recoil How does direction matter in this example? Two objects held together are PUSHED APART Animation BEGIN WITH A MOMENTUM OF ZERO Direction matters because velocity will be negative going left and positive to the right. This is how it sums up to be zero! How does direction matter in this example?

Example #1 A 100 kilogram cannon has a 5 kilogram cannonball inside it. When the cannonball is fired, what happens to the cannon? If the cannonball has an initial velocity of 20 meters per second, calculate the velocity of the cannon. The cannon will move backward

Summary How is an elastic collision different from an inelastic one? How is momentum conserved for a recoil or explosion situation?