1. Use this as extra support for the concepts of momentum, impulse and collisions.

2. Linear Momentum  a vector quantity describing the product of an object’s mass and velocity. Momentum is represented by a lowercase “p” Mathematically, momentum is represented: p = mv Momentum (kgm/s) = (Ns) Mass (kg) Velocity at a given moment (m/s) Don’t believe me? Check for yourself… Ns = kgm/s recall that N = kgm/s/s so if you multiply N x s you will get (kgm/s/s) x s = kgm/s

3. Question: Consider applying a net force (F) to an object as shown. If this force is applied for a time frame of Δt, then the box will accelerate to some resulting (final) velocity. If the time frame is doubled, what happens to the resulting velocity? F Obviously, the resulting velocity will increase (because it experiences an acceleration for a longer period of time.)

4. Impulse is a vector quantity describing the product of the (average) force (applied to an object) and the time over which that force is applied. Impulse is represented by a capital “J” Mathematically, impulse is represented: J = F Δt Impulse (Ns) Force applied (N) Time during which the force is applied (s)

5. Objects respond to an impulse (J) by changing their momentum (p). * A large impulse will produce a large response (i.e. a baseball will move faster if subjected to a larger impulse) to a given mass….BUT…the same impulse will have a different effect on different masses (i.e. That same impulse applied to a bowling ball will not result in the same speed as the baseball had). J J

6. We know that F = maand So F = m Ft = m(v f – v o ) J = mv f - mv o J = p f – p o = Δp  Impulse-Momentum Theory We already stated that the quantity Ft is impulse (J). We can also rewrite the right side of the equation giving us: The quantity mass x velocity is known as momentum. This gives us:

7. This is a HUGE DEAL! Think about it…. Consider that you get a job as a stuntman (or stunt woman as it might be) in the next Marvel Comic movie. You will be filming a scene in which you crash a speeding car into a wall. The director asks you, “would you rather perform the stunt by actually crashing into a brick wall or would you rather crash into a Styrofoam replica placed in front of a bunch of hay bales?”

8. …obviously you would choose the Styrofoam replica. You intuitively know that the Styrofoam/hay combination will provide a “softer” stop than the actual brick wall. But, did you ever stop to think why? The “why” is because of the impulse-momentum theorem.

9. Let’s assign some values. Let us say that the 1100-kg car is initially traveling east at 30 m/s. No matter which object the car hits, it will stop in both cases. So the change in momentum is the same for both cases…. v o =30 m/s v f = 0 m/s m = 1100-kg Δp = p f – p o Δp = mv f – mv o = -33000 kgm/s FOR BOTH THE BRICK WALL AND THE STYROFOAM WALL.

10. The change in momentum will be -33000 kgm/s for either case. But HOW they stop will indicate whether a large or small force was responsible. The brick wall would stop the car VERY QUICKLY…let’s say it brings the car to rest in 2.1-seconds. The Styrofoam replica would stop the car SLOWLY….let’s say it brings the car to rest in 5.8-seconds. Because the TIME of IMPACT is different, the FORCE applied will be different as well.

11. Hitting the Brick Wall…. Δp = -33000 kgm/s t = 2.1 s F = ? Δp = Ft -33000 = F(2.1) F = -15714.3 N Hitting the Styrofoam…. Δp = -33000 kgm/s t = 5.8 s F = ? Δp = Ft -33000 = F(5.8) F = -5689.7 N The Shorter the time frame, the greater the force. The Longer the time frame, the weaker the force. THIS IS A BIG DEAL! NOTE – the negative signs on the forces indicate the direction of the force.

12. Before we get into the law, let’s look at a few definitions: Internal Forces  forces that act within the system  objects within the system exert forces on each other External Forces  forces exerted on the objects (of the system) by external agents. System  the collection of objects being studied.  an isolated system is one for which all external forces are neglected.

13. A collision is an interaction between masses in which there is a transfer of momentum and energy. For ALL collisions, (total) momentum is ALWAYS conserved! There are two MAIN categories of collisions (but three types if you consider a sub-group). Elastic Collisions Inelastic Collisions Perfectly inelastic collisions

14. Collisions are classified according to changes in the kinetic energy. 1) elastic collisions total kinetic energy IS CONSERVED! This is the defining characteristic.  the objects will move independently after they collide (this, however, does not make it elastic) 2) inelastic collisions total kinetic energy is NOT conserved! This is the defining characteristic.  the objects may move independently after they collide (this, however, does not make it inelastic) 3) Completely (or perfectly) inelastic collisions Because it is an inelastic collision, KE is not conserved. What makes it special is that the objects will stick together and move as one object after colliding. (We will be able to represent this mathematically.)

15. A cue ball (mass m A = 0.400kg) moving with speed v A = 1.80m/s strikes ball B, initially at rest, of mass m B = 0.500kg. As a result of the collision, the cue ball travels (in the same direction) with a speed of 0.20 m/s. A) What is the velocity of ball B? B) Is this collision elastic or inelastic?

16. Cue Ball (A) m A = 0.40-kg V Ao = 1.80 m/s V Af = 0.20 m/s Eight Ball (B) m B = 0.50-kg V Bo = 0 m/s V Bf = ? m A v Ao + m B v Bo = m A v Af + m B v B f m A v Ao - m A v Af = m B v Bf (m A v Ao – m A v Af ) / m B = v Bf 1.28 m/s= v Bf 0 Momentum is conserved…

17. Cue Ball (A) m A = 0.40-kg V Ao = 1.80 m/s V Af = 0.20 m/s Eight Ball (B) m B = 0.50-kg V Bo = 0 m/s V Bf = 1.28 m/s ½ m A v Ao 2 + ½ m B v Bo 2 = ½ m A v Af 2 + ½ m B v Bf 2 ½ m A v Ao 2 = ½ m A v Af 2 + ½ m B v Bf 2 0.648 J = 0.418 J 0 These are NOT equal….kinetic energy was lost. Therefore, this collision is inelastic. If the collision is elastic then KE o = KE f ….check it….

18. A Ford F-150 (4690 – Lbs) is stopped at a red light when it is rear-ended by a Honda Civic (2600-Lbs) traveling north at 20.0 m/s. The Honda becomes wedged beneath the rear bumper of the Ford, causing the two vehicles to travel together after the crash. A) What is the resulting velocity of the wreckage? B) Verify (mathematically) that the collision is inelastic.

19. Honda Civic (A) m A = 1181.82-kg V Ao = 20 m/s Ford F150 (B) m B = 2131.82-kg V Bo = 0 m/s m A v Ao + m B v Bo = m A v Af + m B v B f The vehicles get STUCK TOGETHER, so they will have the SAME FINAL VELOCITY. m A v Ao = (m A + m B ) v Bf 7.13 m/s= v f V f = ?