1. Cart collision lab 3 groups Purpose to find relationship between
Mass of moving, mass of stationary, initial velocity of moving all compared to final velocity of stationary cart
Need to be consistent – what factors?
Graph: your independent variable vs velocity
All on whiteboards – share with class on Monday
- each group of 3-4 make a whiteboard

2. Ch 5 Impulse and Linear Momentum
Pre AP physics

3. Impulse and linear momentum
How does jet propulsion work?
How can you measure the speed of a bullet?
Would a meteorite collision significantly change Earth's orbit?
© 2014 Pearson Education, Inc.

4. Where we are headed:
To use Newton's laws effectively, we need to know the forces that objects exert on each other.
If two cars collide, we don't know the force that one car exerts on the other during the collision.
When fireworks explode, we don't know the forces that are exerted on the pieces flying apart.
This chapter introduces a new approach that helps us analyze and predict mechanical phenomena when the forces are not known.
© 2014 Pearson Education, Inc.

5. 5.1 Mass in a closed system is conserved (stays constant)

6. 5.2 Linear momentum ()
I. General
A. defn – mass of an object times its velocity
b. eqn:  = mv
c. unit: kg . m/s
d. it’s a vector - has magnitude and direction
e. total momentum of system adds up to net
ƿnet = m1v1 + m2v2 + m3v3 +…..= ∑ mv

7. II. Two-Particle Collisions
A. Law of conservation of momentum
In a closed, isolated system, momentum does not change. So momentum lost by an object in a collision, is gained by something else.
eqn: Ai + Bi = Af + Bf
Or: m1v1i + m2v2i = m1v1f + m2v2f

8. mA = 1875kg, mB = 1025 kg, vAi = 23m/s, vBi = 17m/s
. EX : A 1875 kg car going 23 m/s rear-ends a 1025 kg compact car going 17 m/s on ice in the same direction. The two cars stick together. How fast do the two cars move together after the collision?
mA = 1875kg, mB = 1025 kg, vAi = 23m/s, vBi = 17m/s
Ai + Bi = Af + Bf
mA vAi + mB vBi = mA vAf + mB vBf
since cars stick after collision, same vf; so
mA vAi + mB vBi = (mA + mB)vf
(1875kg) (23m/s) + (1025kg) (17m/s) = (1875kg kg) vf
vf = 21 m/s

9. III. Impulse (Ft)
A. defn – the product of the average force on an object and the time interval over which it acts
B. eqn: Ft = mv
C. unit : N . s
D. can be found by taking the area of force-time graph

10. Impulsive Forces Usually high magnitude, short duration.
Suppose the ball hits the bat at 90 mph and leaves the bat at 90 mph, what is the magnitude of the momentum change?
What is the change in the magnitude of the momentum?

11. area under curve Impulse (J) on a graph F(N) t (ms) 3000 2000 1000 1 21 2 3 4
t (ms)

12. IV. Impulse
B. the impulse of an object is equal to object’s final momentum minus the object’s initial momentum
C. eqn: Ft = f - i
D. signs (+/-) are important for impulse and mom.

13. V. Using the Impulse-Momentum to save lives
A. A large change in mom occurs only if large impulse
B. A large impulse can result from:
a large force acting of a short period of time OR
a small force acting over a long period of time
C. An airbag reduces force by increasing time interval and it exerts the force over a larger area of the person’s body

14. D. EX: A 2200 kg vehicle traveling at 94 km/h (26 m/s) can be stopped in 21 s by gently applying the brakes. It can be stopped in 3.8 s if the driver slams on the brakes, or in 0.22 s if it hits a concrete wall. What average force is exerted on the vehicle in each of these steps?
m = 2200 kg, vi = +26 m/s, vf = 0 m/s t = 21 s, 3.8 s, s. Find F for each time.
Ft = f - I so need to find  first
i = mvi
= (2200kg) (26 m/s)
i = 5.6 x 104 kg. m/s
f = mvi = 0

15. Ft = f - i
= x104 kg.m/s
F (21 s) = -5.6x104 kg.m/s
Fgentle braking = x 103 N
F(3.8 s) = -5.6x104 kg.m/s
Fslam = -1.5 x 104 N
F (0.22 s) = -5.6x104 kg.m/s
Fconcrete = -2.6 x105 N

16. Bellringer 12/23
Hobbes, the stuffed tiger, has a mass of 31.8 kg. Calvin, the little boy, has a mass of 25.1 kg.
In a game of football, Hobbes runs at Calvin at 7.22 m/s. Calvin is running at Hobbes.
(a) If the two collide, stick together, and are then at rest, what was Calvin’s initial velocity?
(b) What impulse is exerted on Hobbes by Calvin?
(c) What impulse is exerted on Calvin by Hobbes?
(d) If the collision occurred in seconds, Then what force was exerted on Hobbes?

17. More BR
Find velocity and impulse overall

18. more

20. Bellringer 1/5
Find the Impulse for each situation. The mass of the car is 1000kg

21. VI. External versus internal forces
A. External forces: forces coming from outside the system of particles whose momentum is being considered.
External forces change the momentum of the system.
B. Internal forces: forces arising from interaction of particles within a system.
Internal forces cannot change momentum of the system.

22. An external force in golf
The System
The club head exerts an external impulsive force on the ball and changes its momentum.
The acceleration of the ball is greater because its mass is smaller.

23. An internal force in pool
The System
The forces the balls exert on each other are internal and do not change the momentum of the system.
Since the balls have equal masses, the magnitude of their accelerations is equal.

24. VII. Recoil Guns and cannons “recoil” when fired.
This means the gun or cannon must move backward as it propels the projectile forward.
The recoil is the result of action-reaction force pairs, and is entirely due to internal forces. As the gases from the gunpowder explosion expand, they push the projectile forwards and the gun or cannon backwards.

25. A. pushes result in both things moving
Recoil
A. pushes result in both things moving
B. rocket movement in space due to release of gases, internal force “pushes” rocket forward
C. system: rocket and gas
1. if both at rest, then momentum = 0
2. afterwards since high velocity of gas backwards, the jet pushes forward
D. Thrust – force exerted by fuel on rocket

26. EX: An astronaut at rest in space fires a thruster pistol that expels 35g of hot gas at 875m/s. The combined mass of astronaut and pistol is 84 kg. How fast and in what direction is the astronaut moving after firing the pistol?
mA = 84kg, mB = 0.035kg, vAi = 0m/s, vBf = -875m/s

27. i = 0 m/s (all at rest)
I = f
0 = mAvAf + mBvBf
0 = (84kg) (vAf) + (0.035 kg)(-875 m/s)
vAf = 0.36 m/s

28. Bellringer 1/8
Draw vector for this scenario: A red ball moving straight east hits a blue ball moving straight north. They stick together after collision and move 30° NE.

29. Terms Elastic collision – bounce off each other
Ex: rubber ball bounces off floor
Ex: golf club causes ball to move away from it
Inelastic collision – stick together
Perfect inelastic collision – stick together and go same velocity (we will do these types)
Ex: clay ball sticks to floor when dropped
Ex: bullet strikes block of wood and is embedded

30. Momentum and collisions in 2-d (dimension)
A. still have conservation of momentum
In these collisions, the individual momenta can change but the total momentum does not.
B. more than one direction (pool)
C. break prob into x and y components

31. Ex: A 1325 kg car, C, is moving north at 27
Ex: A 1325 kg car, C, is moving north at 27.0 m/s, collides with a 2165 kg car, D, moving east at 11.0 m/s. The two cars are stuck together. In what direction and with what speed do they move after the collision?
mc = 1325kg, mD =2165kg, vciy = 27.0m/s,
vDix = 11.0 m/s, vfx? Vfy? ?

32. Initial mom:
ci = mcvci,y
= (1325 kg) (27.0 m/s)
ci = 3.58 x 104 kg.m/s (north)
Di = mDvDix
(2165kg) (11.0 m/s)
Di = 2.38 x 104 kg.m/s (east)
which = final mom too (for x and y)

33. Using triangle diagram:
f = (f, x)2 + (f,y)2
f = (2.38 x 104 kg.m/s)2 + (3.58 x 104)2
f = 4.30 x 104 kg.m/s
solve for 
tan-1 = fy / fx
tan-1 = 3.58 x 104kgm/s
2.38x104 kg m/s
 = 56.4

34. final speed:
f = (mc + mD) vf
4.30x104kg m/s = (1325 kg kg) vf
vf = 12.3 m/s

35. PARTNER WORK:
Peregrine falcons often grab their prey from above while both falcon and prey are in flight. A falcon, flying at 18m/s, swoops down at 45° angle from behind a pigeon flying horizontally at 9.0m/s. The falcon has a mass of 0.80kg and the pigeon has a mass of 0.36kg. What are the speed and the direction (angle) of the falcon (now holding the pigeon) immediately after impact?

37. X component: pxi = pxf
mfvfxi + mpvpxi = (mf + mp)vfx
mf(-vf cosθ) + mp (-vpxi) = (mf + mp)vfx
(0.80kg) (-18m/s cos 45) + (0.36kg) (-9m/s)
= (0.80kg kg) vfx
vfx = -11.6m/s

38. Y component pyi = pyf
mfvfyi + mpvpyi = (mf + mp)vfx
mf(-vf sinθ) + 0 = (mf + mp)vfx
(0.80kg) (-18m/s) (sin 45) = (0.80kg kg) vfy
Vfy = -8.78m/s
Vectors so make triangle of final velocities
tan θ = vyf / vxf
tan θ = m/s / -11.6m/s
θ = 37°

39. Magnitude of final velocity can be found with Pythagorean theorem
v2 = vfx2 + vfy2
v2 = (-11.6m/s)2 + (-8.78m/s)2
v = 14.5m/s