Lecture 11. Linear Momentum and Impulse. Collisions.

Transcription

1 Lecture 11 Linear Momentum and Impulse. Collisions.

2 Momentum and Newton s Second Law F net = m a= m Δ v Δ t = Δ (m v ) Δ t = Δ p Δ t Linear momentum p = m v Newton s second law in terms of linear momentum: F net = Δ p Δ t Impule gained by an object is: Δ p = F net Δ t

3 Impulse The change in linear momentum for a certain time interval is called the impulse J (something like how much of an effect did the force have by being applied during a given time interval ). F J = area t 0 t f t

4 Impulse The change in linear momentum for a certain time interval is called the impulse J (something like how much of an effect did the force have by being applied during a given time interval ). F J =Δ p = F average Δ t J = area Equal area J F average = Δ t = Δ p Δ t F average t 0 t f t

5 F Short-duration collision, large maximum force. eg. Hitting a hard ball with a bat Same impulse Long-duration collision, small maximum force. eg. Hitting a soft ball with a bat t Same change in momentum

6 Units of Momentum and Impulse SI units p =m v kg m/s It s the same! J = Δ p = F net Δt N s

7 ACT: Egg tossing In an egg-tossing contest, two people toss a raw egg back and forth. After each successful toss, each person takes a step back. Catching the egg without breaking it becomes harder and harder. Usually the trick is moving your hand down with the egg when you receive it. This works better because: A. It decreases the change in momentum. B. It decreases the impulse. C. It decreases the force on the egg.

8 ACT: Egg tossing In an egg-tossing contest, two people toss a raw egg back and forth. After each successful toss, each person takes a step back. Catching the egg without breaking it becomes harder and harder. Usually the trick is moving your hand down with the egg when you receive it. This works better because: A. It decreases the change in momentum. B. It decreases the impulse. C. It decreases the force on the egg.

9 If the flying egg has speed v, the change in momentum is: Δp = 0 mv = - mv (independent of how you catch it) The impulse is just the same! J = Δp = - mv But by moving your hand with the egg, you are increasing the time interval over which this Δp must take place. So the average force on the egg F ave = Δ p Δ t decreases. By the way: Catching the egg is harder and harder because its speed becomes larger (and the required change in momentum, too), so exerting a small force becomes harder as well.

10 Example: Do you git much work nowadays, Jerry? Sure! LOOK OUT!! I m p u Tush tush! Don t make so much noise! Ooooh! Did you see the size of that brick? I know, but it wuz only on your head a second. l s e

11 Jiggs is completely missing the point If the brick has a mass of 1 kg and has been dropped from 10 m, what is the average force on Jerry s head during the 10 ms it is in contact with his head? Right before hitting Jerry s head, the speed of the brick is: 1 mgh mv v gh (9.8 m/s )(10 m) 14 m/s If the brick is completely stopped by his head, the change in momentum is: Δp = 0 mv = (1kg) (14m/s ) = 14kg m/s Therefore, the average force is: F average = J Δ t = Δp 14kg m/s = Δ t 0.01 s = 1400 N DEMO: Karate boards Equivalent to 315 lb. Ouch!!

12 A: Δ p A Δ t = F A,B + F A,C + F A,ext A F A,C F A,B C F C,B F A,ext F C,A F B,A F B,C B C: Δ p C Δ t = F C,A + F C,B B: Δ p B Δ t = F B,A + F B,C

13 A: B: C: Δ p A Δ t = F A,B + F A,C + F A,ext Δ p B Δ t = F B,A + F B,C Δ p C Δ t = F C,A + F C,B Δ p A Δ t + Δ p B Δ t + Δ p C Δ t = F A,B + F A,C + F A,ext + F B,A + F B,C + F C,A + F C,B Δ p total Δ t = F A,ext

14 Define total linear momentum for the system: p total = i p i = i m i v i Δ p total Δ t = F ext

15 Conservation of momentum The really important case: if F =0, ext Δ p total Δ t = 0 p i = p f

16 EXAMPLE: Tug of war on ice Two guys of masses m 1 = 75 kg and m = 90 kg pull on both ends of a rope on an ice rink. After a couple of seconds, the thin one is moving at 0. m/s. What is the speed of the big one? Grrr Grrr m 1 m

17 Two guys of masses m 1 =75 kg and m =90 kg pull on both ends of a rope on an ice rink. After a couple of seconds, the thin one is moving at 0. m/s. What is the speed of the big one? No friction No net vertical force

18 Two guys of masses m 1 =75 kg and m =90 kg pull on both ends of a rope on an ice rink. After a couple of seconds, the thin one is moving at 0. m/s. What is the speed of the big one? No friction No net vertical force 1, f 1, f m F ext = 0 p total is conserved p1, i p, i p p 1, f, f 0 0 mv 1 mv 1, f, f m 75 kg v v 0. m/s 0.17 m/s 90 kg Changes in linear momentum: Δp 1 = m 1 (v 1, f v 1,i ) = (75kg ) (0.m/s 0 ) = 15 kg m/s Δp = m (v,f v, i ) = (90kg ) ( 0.17m/s 0 ) = 15kg m/s The force produced a momentum tranfer.

19 Conservation of momentum in a - particle system F BA F AB F BA = F AB Δ p A Δ t = Δ p B Δ t The interaction means there is an exchange of linear momentum between two objects.

20 This is general for any isolated system (a system which is not subject to a net external force). A F A,C F A,B C F C,B F C,A F B,C F B,A B DEMO: Carts with springs Each interactions within the system represents momentum flows between the particles but the total momentum of the system remains constant.

21 ACT: Big block, small block Consider the following two collisions between two blocks of masses m and M (> m). In both cases, one of the blocks is initially moving with speed v and the other is at rest. After the collision, they move together. The final speed of the two objects is larger when: A. The big block is initially at rest. B. The small block is initially at rest. C. The speed is the same in both cases.

22 ACT: Big block, small block Consider the following two collisions between two blocks of masses m and M (> m). In both cases, one of the blocks is initially moving with speed v and the other is at rest. After the collision, they move together. The final speed of the two objects is larger when: A. The big block is initially at rest. B. The small block is initially at rest. C. The speed is the same in both cases.

23 What happens to the total kinetic energy in collisions? 1. Constant KE. KE decreases Elastic collisions When internal forces are conservative or objects are hard. Examples: Elementary particles collisions, billiard balls Inelastic collisions Whenever a deformation is involved. Example: Most macroscopic collisions Special case: Perfectly inelastic collisions, when the objects stick together. Example: Pin and putty Explosions 3. KE increases Superelastic collisions One body breaks into a number of parts. The explosion mechanism provides the extra kinetic energy. Some internal energy is transformed into KE because of a collision. Example: An excited atom hits another atom and drops to a lower state without radiation.

24 1D Inelastic: Shooting at a block A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with speed v. The bullet lodges in the block. Determine the final speed of the block (with the bullet). At rest m v M Before

25 1D Inelastic: Shooting at a block A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with speed v. The bullet lodges in the block. Determine the final speed of the block (with the bullet). At rest m v M Before M +m v After

26 p total, initial = mv + 0 p total, final = (M + m)v mv = (M + m)v v '= m M+ m v 0 as M v as M 0 At rest m v M Before M +m v After

27 Is total kinetic energy conserved? KE total, initial = 1 mv Clearly KE total, final = 1 (M+m) (v ') = 1 (M+m) ( KE KE KE total total, final total, initial 1 mv ( 1) 1 m M m M mv M m < 0 inelastic collision )= m 1 mv( m ) M+m M+m (Work done by friction to stop the bullet)

28 1D Elastic: Two steel balls head-on A steel ball with mass m 1 = 1 kg and initial speed v 0 collides head-on with another ball of mass m = kg that is initially at rest. What are the final speeds of the balls?

29 The hard way to solve this: Head-on: 1D process. v 0 v v 0 1 Hard steel balls: elastic collision (KE total is conserved ) v0 0 v1 v v v v 0 1 v 0 v 1 v

30 v v v 0 1 v v v 1 0 v0 v1 v v v v v 0 0 v v v 1 0 6v 4v v 0 0 Two solutions: v 0 v v v v v v No collision!

31 But there is an easier way to solve elastic collisions. mv mv m v mv mv mv m v0 v1 v m m v0 v1 v m 1 1 m v v 0 v 1 m 1 v v v 0 1 v ( v v ) 0 1 m v v 0 v 1 m 1 Velocity of 1 relative to before the collision Velocity of 1 relative to after the collision

32 So the relative velocity has the same magnitude and opposite sign before and after the collision: v ( v v ) 0 1 But relative velocity is the same independently of the original frame of reference (including one in which both balls are initially moving) The relative velocity has the same magnitude and opposite sign before and after any elastic collision between two bodies: v v ( v v ) 1,i,i 1,f,f

33 DEMO: Basketball and superball Assume all collisions are elastic and M >> m m v V? M v v v v

34 DEMO: Basketball and superball Assume all collisions are elastic and M >> m m v Relative speed is v V? 3v M v v Here too! v v

35 ACT: Newton s craddle Consider a row of adjacent steel-ball pendula. If two balls on the left are pulled to a certain height h and released, what happens? A. One ball rises on the right, but higher than h. B. Two balls rise on the right to height h. C. A ball rises on each side to height h (i.e., one of the initial balls bounces back)

36 A. One ball rises on the right, but higher than h. We could have mv i = mv f,, so v f = v i. But then kinetic energy increases!!: KE i = ½mv i = mv i KE f = ½mv f = mv i B. Two balls rise on the right at height h. mv i = mv f,, so v f = v i. KE i = ½mv i = mv i KE f = ½mv f = mv i Ok! C. A ball will rise on each side to height h (i.e., one of the initial balls bounces back) Same height means v i = v f. But this violates conservation of momentum: p i = mv i p f = mv i mv i = 0 DEMO: Newton s craddle