Day 70 Do Now

1. Day 70 Do Now A 3-kilogram ball is accelerated from rest to a speed of 10 m/sec. a. What is the ball’s change in momentum? b. What is the impulse? c. If a constant force of 40 newtons is applied to change the momentum in this situation, for how long does the force act?

2. Conservation of Momentum

3. Momentum • Remember the formula for momentum • p = mv • m= mass • v = velocity • p = momentum

4. Momentum • The momentum of an object stays the same, unless an outside force acts upon the object. • The outside force causes the velocity to change because the object accelerates. • Usually the mass of an object remains the same

5. Collision • When two objects collide their momentum changes due to the force of each object on the other. • What happens to their momentum?

6. Collision • The momentum of an object usually changes when it collides with another. • Often, the momentum is transferred to the other object.

7. Conservation of Momentum • The total momentum before a collision equals the total momentum after the collision. • Example • http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/momentum/cbb.html

8. Explanation(mass was added) • The momentum before the collision was only the momentum of the cart since the other object is not moving • The momentum after the collision was the same as before. • The cart went slower only because it had a mass added to it.

9. Another type of collision • http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/momentum/crete.html • This is called an elastic collision (there is no loss of energy) • Notice that the total momentum is the same before the collision as it is after the collision.

10. Demonstration • Write down what happens with each interaction • One standing still, and one moving • Both moving toward each other • Both moving the same direction

11. Equation • The formula for conservation of momentum is as follows: m1v1+m2v2 +… = m1v1’+m2v2’+... m = with a subscript is the mass of each object before the collision v = with a subscript is the velocity of each object before the collision v’ = with a subscript is the velocity of each object after the collision.

12. Example of cons of mom • A marble of mass .06 kg is traveling with a speed of 1.7 m/s. It then strikes a marble of equal mass. The collision is elastic, and the first marble bounces off in the opposite direction with the same speed of the second. • What is the momentum before the collision? • What is the momentum after the collision? • What is the speed of the first marble after the collision?

13. Car Collision • A 2275 kg SUV going 28 m/s rear ends an 875 kg compact going 16 m/s on ice going in the same direction. The two cars stick together. How fast does the wreck move immediately after the collision.

14. We know it’s a conservation problem • Sketch the problem • Show the before and after • Label the cars A and B and show their velocity vectors • Draw a vector diagram for momentum • The length of the momentum vector after the collision equals the sum of the two momentum vectors before the collsion.

15. Knowns mA = 2275 kg vA = 28 m/s mB = 875 Kg vB =16 m/s Unknows Final velocity of the combined cars. V2 = ? The variables

16. The Diagram

17. The Math • P1= P2 • PA1+PB1 = PA2+PB2 • mAvA1 + mBvB1 = mAvA2 + mBvB2 • But they stick together so vA2=vB2=v2 • mAvA1 + mBvB1 = (mA+mB)v2 • Solve for V2 = 25 m/s

18. Explosion Problems • Astronaut Strawberry is at rest in space fires a thruster pistol that expels 0.035 kg of hot gas at 875 m/s. The combined mass of the pistol and astronaut is 84 kg (it’s a heavy suit). How fast and in what direction is the astronaut moving after the firing the pistol

19. Knowns mA = 84 kg mB = 0.035 kg vA1 = vB1 = 0 m/s vB2 = -875 m/s Unknowns Final velocity of Straw vA2 = ? Knowns and …

20. Draw the figure and label

21. The Calculations • P1 = pA1 + pB1 = 0 • PA1+PB1 = PA2+PB2 • 0 = PA2+PB2 • PA2 = -PB2 • mAvA2 = -mBvB2 • vA2 = (- mBvB2)/mA = 0.35 m/s

22. Pool Halls and Momentum • Note when there is a pool hall collision the balls always move at a right angle (when the collision isn’t head on) • Use trig sin and cos to workout the size of the final vectors.

23. Example1 • A 15-kg medicine ball is thrown at a velocity of 20 km/hr to a 60-kg person who is at rest on ice. The person catches the ball and subsequently slides with the ball across the ice. Determine the velocity of the person and the ball after the collision.

24. Before Collision • After Collision • Person • 0 • (60 kg) • v • Medicine ball • (15 kg) • (20 km/hr) = 300 kg • km/hr • (15 kg) • v • Total • 300 kg • km/hr • 300 • 60 • v + 15 • v = 300 • 75 • v = 300 • v = 4 km/hr

25. A 0.150-kg baseball moving at a speed of 45.0 m/s crosses the plate and strikes the 0.250-kg catcher's mitt (originally at rest). The catcher's mitt immediately recoils backwards (at the same speed as the ball) before the catcher applies an external force to stop its momentum. If the catcher's hand is in a relaxed state at the time of the collision, it can be assumed that no net external force exists and the law of momentum conservation applies to the baseball-catcher's mitt collision. Determine the post-collision velocity of the mitt and ball.

26. 0.15 kg • v + 0.25 kg • v = 6.75 kg•m/s • 0.40 kg • v = 6.75 kg•m/s • v = 16.9 m/s

27. A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s. Assuming that momentum is conserved during the collision, determine the velocity of the truck immediately after the collision.

28. 3000*v + 15 000 = 30 000 • 3000*v = 15 000 • v = 5.0 m/s