Impulse and Momentum Review

1. Impulse and Momentum Review … This is what’s on the test

2. Basic Relations , momentum • Momentum: a vector that is m•v • Find the momentum of a 40kg bike that is going 10m/s South • If the momentum of a go cart is 2500kg•m/s east when it is going 10m/s east, what is its mass? • If the momentum of a 100kg go cart is 2500kg•m/s east, find its velocity

3. Basic Relations , momentum • Find the momentum of a 40kg bike that is going 10m/s South = 400kg•m/s South • If the momentum of a go cart is 2500kg•m/s east when it is going 10m/s east, what is its mass? 250kg • If the momentum of a 100kg go cart is 2500kg•m/s east, find its velocity 25m/s East

4. Basic Relations , momentum • If the momentum of a go cart is 2500kg•m/s east, what would its momentum be if it doubled its speed? • If the momentum of a go cart is 2500kg•m/s east, what would the momentum of another go cart be if its mass is 25% greater but it has the same velocity?

5. Basic Relations, momentum • If the momentum of a go cart is 2500kg•m/s east, what would its momentum be if it doubled its speed? Double, 5000kg•m/s east • If the momentum of a go cart is 2500kg•m/s east, what would the momentum of another go cart be if its mass is 25% greater but it has the same velocity? 25% greater, 3250kg•m/s

6. Basic Relations, impulse • Impulse = F•t • Find the impulse of a 4000N average force to the west that acts for 50s. • Find the impulse of a 4000N average force to the west that acts for half the time as above. • Will the change in momentum be the same, greater, or less for the 50s impulse?

7. Basic Relations, impulse • Impulse = F•t • Find the impulse of a 4000N average force to the west that acts for 50s. = 200,000N•s west • Find the impulse of a 4000N average force to the west that acts for half the time as above. = 100,000N•s west • Will the change in momentum be the same, greater, or less for the 50s impulse? Greater (2x greater)

8. Basic Relations, impulse is the change in momentum • Impulse = F•t = change inmomentum • If (and only if) the momentum change is the same, force will decrease if time increases. • An object has 400kg•m/s of momentum to the north when a 10N force to the north is applied for 2s. Find the change in momentum and the final momentum. • An object has 400kg•m/s of momentum to the north when a 10N force to the south is applied for 2s. Find the change in momentum and the final momentum.

9. Basic Relations, impulse is the change in momentum • An object has 400kg•m/s of momentum to the north when a 10N force to the north is applied for 2s. Find the change in momentum and the final momentum. = 20N*s north; 420kg*m/s north • An object has 400kg•m/s of momentum to the north when a 10N force to the south is applied for 2s. Find the change in momentum and the final momentum. = 20N*s south; 380kg*m/s north

10. Basic Relations, impulse is the change in momentum • An object has 400kg•m/s of momentum to the north when a 10N force to the east is applied for 20s. Find the change in momentum and the final momentum. • An object has 400kg•m/s of momentum to the north . Find the impulse required to stop it.

11. Basic Relations, impulse is the change in momentum • An object has 400kg•m/s of momentum to the north when a 10N force to the east is applied for 20s. Find the change in momentum and the final momentum. Pythag = 44.7m/stan-1 (40/20) = 63 deg North of E

12. Basic Relations, impulse is the change in momentum • An object has 400kg•m/s of momentum to the north. Find the impulse required to stop it. Dmv = - 400kg•m/s = -400N*s = 400N*s south

13. Impulse and Momentum at an angle • An object has 400kg•m/s of momentum 30deg east of south. Find the components in the north and east. • An object has 400kg•m/s of momentum 30deg east of south when a 20N force is applied to it for 20s at 30deg east of south. Find its final momentum and velocity.

14. Impulse and Momentum at an angle • An object has 400kg•m/s of momentum 30deg east of south. Find the components in the north and east. • 400 sin(30deg) = 200kg•m/seast • 400 cos(30deg) = 346.4kg•m/seast • Vel = (400kg•m/s)/mass 30deg east of south.

15. Impulse and Momentum at an angle • An object has 400kg•m/s of momentum 30deg east of south. Find the components in the north and east. • 400 sin(30deg) = 200kg•m/seast • 400 cos(30deg) = 346.4kg•m/seast

16. Impulse and Momentum at an angle • An object has 400kg•m/s of momentum 30deg east of south. If it strikes another object that is at rest, what must the total momentum be after the collision in the north and east directions? What condition are we assuming for this to be true?

17. Impulse and Momentum at an angle • Since momentum must be conserved (if there are no outside forces), it’s the same as before • 400 sin(30deg) = 200kg•m/seast • 400 cos(30deg) = 346.4kg•m/seast

18. Conservation of Momentum The total linear momentum of an isolated system remains constant. An isolated system is one for which the vector sum of the external forces acting on the system is zero. All of the momentum before an event must be the same after the event. m1v1f + m2v2f= m1v1i + m2v2i

19. Conservation of Momentum • If a 80kg guy is rolling on his skates very fast towards his girlfriend at 8m/s while she is rolling towards him at 2m/s. Find their final velocity if they hold on to one another after the collision assuming that she is only half his mass. • m1v1f + m2v2f= m1v1i + m2v2i • 8okg*v1f + 40kg*v2f= 80kg*8m/s+ 40kg*(-2m/s) • Since they move together, v1f = v2f = vf

20. Conservation of Momentum 8okg*v1f + 40kg*v2f= 80kg*8m/s+ 40kg*(-2m/s) • Since they move together, v1f = v2f = vf • 8okg*vf + 40kg*vf= 80kg*8m/s+ 40kg*(-2m/s) • 8okg*v1f + 40kg*v2f= 560kg*m/s • 120kg(v1f) = 560kg*m/s • 4.67m/s in the boyfriend’s initial direction (which was made the pos direction above)

21. Conservation of Momentum If he pushed her fast enough so that he stops, how fast would the girlfriend have to be going? • m1v1f + m2v2f= m1v1i + m2v2i • 8okg*0m/s+ 40kg*v2f= 80kg*8m/s+ 40kg*(-2m/s) • 0+ 40kg*v2f= 560kg*m/s • v2f= (560kg*m/s)/40kg*v2f= 14 m/s in the orig direction of the boyfriend

22. Conservation of Momentum If a superball is 0.4kg and going 25m/s east before striking a wall and returning at 25m/s, find the force that the wall exerted if the time of contact was 0.20 sec. Dmv = F*t = 0.4kg(-25m/s – 25m/s) = F*t = 0.4kg(-50m/s) = -20kg*m/s = F*t -20kg*m/s = F*(0.20s) • F = -100N

23. Conservation of Momentum • The drawing shows a collision between two pucks on an air hockey table. Puck A has a mass of 0.25 kg and is moving along the x-axis with a velocity of 5.5 m/s. It makes a collision with puck B, which has a mass of 0.5 kg and is initially at rest. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of (a) puck A and (b) puck B.

24. p initial, total = 5.5(.25) = 1.375kg*m/s to the right. • After collision in the y directionmvAsin(65) -mvBsin(37) = 0 • After collision in the x direction mvAcos(65)+mvBcos(37) = 1.375kg*m/s right. • After collision in the y direction.25vA(.9063)+.5vB(-.60) = 0; • vB = 0.753vA. • After collision in the x direction.25vA(.4226)+.5vB(.7986) = 1.375kg*m/s r • .25vA(.4226)+.5(0.753vA)(.7986) = 1.375kg*m/s r • 0.4063vA = 1.375kg*m/s rt • vA = 3.4m/s; Now plug back into vB = 0.753vA. = 0.753(3.4m/s) • vB = 0.753vA. = 0.753(3.4m/s) = 2.55m/s

25. Impulse again • If two objects collide, Newton’s 3rd law tells us that the forces must be equal but opposite directions. Common sense tells us that if A is in contact with B for some time, t, then B is also in contact with A for time, t. • Therefore, the impulse between two interacting bodies will also be equal but opposite (if there are no outside forces)

26. Test Taking Skills • Tip 1: Take everything one step at a time and include units for each step. • Tip 2: Watch the signs/directions • Tip 3: When adding vectors, don’t forget to add head-to-tail • Tip 4: If you put the correct answer, don’t change it!!  • Tip 5: Ask yourself if the answer makes sense.