Download presentation

Presentation is loading. Please wait.

1
**Linear Momentum of a Particle**

Momentum is defined as the product of mass and velocity: p = m• v Units: kg • m/s Scalar times a vector that produces a vector answer! A net force acting upon a particle will change its net momentum in the direction of that net force: ∑F = dp dt

2
**A 4. 88 kg object with a speed of 31**

A 4.88 kg object with a speed of 31.4 m/s strikes a steel plate at an angle of 42.0˚ with the horizontal and rebounds at the same speed and angle: What is the change (magnitude and direction) of the linear momentum of the object? v0 = ˚ v = 31.4 m/s @ 42.0˚ m = 4.88 kg

3
**p0 = mv0cosø1 + mv0sinø1 = (114i – 103j) kg-m/s**

∆p = p – p0 = (0i + 206j) kg-m/s ∆p = px2 + py2 = kg-m/s Ø = tan-1 (py/px) = ˚ p ∆p po 42.0˚

4
Impulse and Momentum A collision occurs when a relatively large force acts over a relatively short period of time. A net force will produce acceleration, and therefore a change in momentum: F = dp dt Fdt = dp 8.2 The net force acting over time will produce a change in momentum-- this is called impulse (J)!

5
The impulse-momentum theorem states that the net force acting on a particle during any given time interval is equal to the change in momentum of the particle during that time period: 8.3

6
**For a constant force: J = ∆p = pf - pi**

For a varying force: J = ∫ F(t)dt = ∆p Impulse-Momentum Lab 8.4

7
A baseball (m = .14 kg) is thrown horizontally with a speed of 42 m/s is struck by a bat. The ball leaves the bat with a velocity of 50.0 m/s at an angle of 35˚ with the horizontal. A) What is the impulse of the force exerted on the ball? B) Assuming the collision lasts for 1.5 ms, what average force acts upon the ball? C) Find the change in momentum of the bat. A) J = ? J = p – p0 J p α ø po

8
**Solve using components (or trig):**

pxo = mvxo = (.14 kg)(- 42 m/s) = kg• m/s pyo = 0 px = mvxcosø = (.14 kg)(50 m/s)(cos 35˚) = 5.7 kg • m/s py = mvysinø = (.14 kg)(50 m/s)(sin 35˚) = 4.0 kg • m/s Jx = px - pxo = (5.7) - (- 5.9) = 11.6 kg • m/s Jy = py - pyo = (4.0) - 0 = 4.0 kg • m/s J = Jx2 + Jy2 = kg • m/s

9
**acting in the same direction as the impulse! **

α = tan-1 (Jy / Jx ) = 19˚ J = 12.3 kg • m/s at 19˚ B) J = Fav • ∆t Fav = J / ∆t = kg • m/s / s = 8200 N about 1840 lbs of force! acting in the same direction as the impulse! C) the ∆p of bat must follow ∆p = ∆p1 + ∆p2 = 0 Therefore, ∆p for bat must equal - ∆p for ball Online HW assiognment ∆pbat = (- 11.6i j ) kg• m/s or: = 12.3 kg• m/s at 199˚

10
**Conservation of momentum in collisions**

If there are no external forces acting upon a system of particles, then there can be no net change in total momentum of the system: ∆P = ∆p1 + ∆p2 + … = 0 8.7 to 8.10 Including Derivation of Cons. Of Momentum Law

11
**Conservation of momentum in collisions**

In any type of collision, linear momentum is always conserved! Total energy is also always conserved, but often times mechanical energy will be converted to internal, rotational, radiant, etc., forms of energy. If the mechanical energy is conserved, then the collision is known as an elastic collision. If Ki Kf , then it is an inelastic collision.

12
Elastic Collisions: p is always conserved: m1v1i + m2 v2i = m1v1f + m2 v2f K is also conserved: .5m1v1i2 + .5m2v2i2 = .5m1v1f2 + m2v2f2 P eq. can be expressed: m1(v1i - v1f ) = m2(v2f - v2i ) K eq. can be: m1(v1i2 - v1f2 ) = m2(v2f2 - v2i2 ) Dividing these equations: v1i + v1f = v2f + v2i The following can then be derived:

13
**If the target particle is at rest:**

v1f = (m1 – m2)(v1i) m2 (v2i) m1 + m m1 + m2 v2f = m1 (v1i) (m2 – m1) (v2i) m1 + m m1 + m2 If the target particle is at rest: v2f = m1 (v1i) m1 + m2 v1f = (m1 – m2)(v1i) m1 + m2 8.15

14
**Inelastic Collisions:**

Momentum is conserved (as always), but K is not, so we examine one special case: In a completely inelastic collision, the two particles stick together and have a common velocity afterward: 8.20 v1f = v2f = vf

15
**Two - Dimensional Collisions**

Momentum is always conserved in collisions: v = 0 ø2 ø1 pi = pf 8.17,8.18 m1v1ix = m1v1f cosø1 + m2v2f cosø2 m1v1iy = m1v1f sinø1 + m2v2f sinø2 Remember: if the collision is elastic Ki = Kf

16
**A 55 kg skater head north at 8**

A 55 kg skater head north at 8.8 km/h and collides with an 83 kg skater heading east at 6.4 km/h. They embrace in a completely inelastic collision. A) What is the magnitude and direction of the velocity of the couple after the collision? B) What is the fractional change in kinetic energy of the skaters due to the collision? V ø A B

17
**Momentum is conserved in the x and y direction:**

mAvA = (mA + mB)(Vcosø) (1) mBvB = (mA + mB)(Vsinø) (2) Eq. (2) (1) : tanø = mBvB mAvA = . 911 ø = tan = 42.3˚ V = mAvA (mA + mB)(cosø) = 5.21 km/h

18
**51% of K is dissipated in some form or other**

Loss of K = Kf - Ki Ki Ki = .5mAvA2 + .5mBvB2 = 3830 Kf = .5(mA + mB)V2 = 1870 Loss of K = 3830 = 51% of K is dissipated in some form or other

19
**used to measure projectile speeds before electronic timing!**

Ballistic Pendulum used to measure projectile speeds before electronic timing! ∆h 8.21 A bullet (m = 9.5 g) is fired into the block (M = 5.4 kg) and they rise to a height of 6.3 cm. What was the initial speed of the bullet?

20
**Momentum is conserved:**

mv = (M + m)V Total energy of the swinging bullet + block pendulum is conserved: .5(m +M)V2 = (m + M)g∆h V = 2 g∆h v = (M + m) 2g∆h m = 630 m/s

21
**Show how this is obviously an inelastic collision:**

Ki = Kb = .5mv2 = 1900 J Kf = K b + B = .5(m + M)V2 = U b + B Ki ≠ Kf U b + B = (m + M)g∆h = 3.3 J Only 3.3 J out of 1900 J, or 0.2%, is transferred to the mechanical energy of the pendulum! The rest would be transferred as internal energy-- heat, sound, etc.

22
**A steel ball of mass. 514 kg is fastened to a cord 68**

A steel ball of mass .514 kg is fastened to a cord 68.7 cm long and released when the cord is horizontal. At the bottom of its path, the ball strikes a 2.63 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find A) the speed of the ball and B) the speed of the block, both just after the collision. C) Suppose that on collision, one-half the mechanical kinetic energy is converted to internal energy and sound energy. Find the final speeds.

23
**Two pendulums of equal mass and length are situated as shown**

Two pendulums of equal mass and length are situated as shown. The first pendulum is released from height d and strikes the second. L L m1 d m2 Assume the collision between the pendulums is completely inelastic and neglect the mass of the strings and any frictional effects. How high will the center of mass rise after the collision?

24
**Particle Systems and Center of Mass**

8.22

25
In this case, the motion of the two particles is complicated, but the overall motion of the system is relatively simple. The center of mass (cm)– the “average” location of the masses follows a relatively simple motion of a single particle. The location of the center of mass of the system is proportional to the each mass’ location: 8.22 1 M xcm = (m1x1 + m2x2) M = m1 + m2

26
**For a many particle system:**

xcm = 1/M(m1x1 + m2x2 + …. mnxn) M = m1 + m2 + …. mn = ∑mn xcm = (1/M)∑mnxn ycm = (1/M)∑mnyn zcm = (1/M)∑mnzn 8.23 rcm = (1/M)∑mnrn vcm = (1/M)∑mnvn acm = (1/M)∑mnan

27
**Three particles are arranged as follows:**

1 2 3 What is the magnitude and direction from the origin of the cm of the system? m1 = 4.1 (-2 cm, 3 cm) m2 = 8.2 ( 4 cm, 2 cm) m3 = 4.1 ( 1 cm, -2 cm)

28
xcm = (1/M)(m1x1 + m2x2 + m3x3) xcm = (1/16.4 kg)[(4.1 kg)(-2 cm) + (8.2 kg)(4 cm) + (4.1 kg)(1 cm)] = 1.8 cm ycm = (1/M)(m1y1 + m2y2 + m3y3) ycm = (1/16.4 kg)[(4.1 kg)(3 cm) + (8.2 kg)(2 cm) + (4.1 kg)(-2 cm)] = 1.3 cm rcm = (1.8 cm)2 + (1.3 cm)2 = 2.22 cm ø = tan-1 (y/x) = tan-1 (1.3/1.8) = 36˚

29
**If each side measures L, where is the cm located?**

30
**Newton’s Laws then hold for each individual parts as well as the center of mass for the system.**

Therefore, if the net external forces acting on the parts of the system are not balanced, there will be a net force acting on the center of mass of the system: ∑Fext = macm If the net external force acting upon the parts of the system is balanced, then the center of mass of the system will move at constant speed: ∑Fext = 0

31
**Three particles are arranged as follows: 6 N**

1 12 45˚ 2 What is the magnitude and direction from the origin of the acceleration of the system? 3 14 N m1 = 4.1 (-2 cm, 3 cm) m2 = 8.2 ( 4 cm, 2 cm) m3 = 4.1 ( 1 cm, -2 cm)

32
Fx = F1x + F2x + F3x = - 6 N + cos45˚(12 N) + 14 N = 16.5 N Fy = F1y + F2y + F3y = 0 + sin45˚(12 N) + 0 = 8.5 N a = F M Fext = Fx2 + Fy2 = N ø = tan-1(Fy /Fx)= 27˚ 18.6 N 16.4 kg = 1.1 m/s2

33
**Two masses, one of 1. 10 kg and the other of 1**

Two masses, one of 1.10 kg and the other of 1.40 kg, are connected by a light string and passed over a frictionless pulley of radius 25.0 cm (Atwood machine). The lighter mass is held in place on the left level with the heavier mass. A) Where is the center of mass of the two masses at this time? The lighter mass is then released. B) What is the acceleration of the center of mass of the two masses?

34
**A shell is fired with a velocity of 466 m/s at an angle of 57**

A shell is fired with a velocity of 466 m/s at an angle of 57.4˚ with the horizontal. At the top of its trajectory the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls straight down. How far from the cannon will the other shell land, assuming it lands at the same elevation as which it was fired from?

35
**vcmx = vcosø = 251 m/s ∆t = vy - vyo g vcmy = vsinø = 393 m/s**

vcmx = (1/M)(m1v1 + m2v2) = (-393) 9.80 m1 = m2 = m = 80.2s vcmx = mv2 2m = v2 2 v2 = 502 m/s 8.24 251 m/s = .5v2 x2 = vcmx(.5∆t) + v2 (.5∆t) = 30, 200 m

36
**Neglecting any external forces, the cm of the system will not change!**

mr = 93 kg mb = 52 kg xro = 1.5 m xbo = 3.0 m xr = ? xb = xr – 1.5 m Neglecting any external forces, the cm of the system will not change! 8.26 xcm = mrxro + mbxbo mr + mb = 2.0 m

37
**A dog weighing 10. 8 lb is standing on a flatboat so that he is 21**

A dog weighing 10.8 lb is standing on a flatboat so that he is 21.4 ft from the shore. He walks the 8.50 ft length of the flatboat and then halts. The boat weighs 46.4 lb, and you can assume the water to be frictionless. How far is the dog from the shore after walking? (Assume the boat is perpendicular to the shore and realize that you have two possible answers depending upon whether the dog walks toward the shore or away).

38
Richard, mass 78.4 kg, and Judy, who is less massive, sit at opposite ends of a canoe, 2.93 m apart, on placid water. The canoe is 31.6 kg. The two changes seats and Richard notices that the canoe has moved 41.2 cm relative to a sunken log. Richard then decides to calculate Judy’s mass. (Richard is a loser who will die alone playing a video game in his mother’s basement which he will call his mancave.) What is Judy’s mass? ∆xc = .412 m R J 55.2 kg

39
An 84.4 kg man is standing at the rear of a 425 kg iceboat that is moving at 4.16 m/s across the ice that is frictionless. He decides to walk to the front of the 18.2 m boat and does so at a speed of 2.08 m/s relative to the boat. How far does the boat move across the ice while he is walking?

Similar presentations

OK

Chapter 6 Table of Contents Section 1 Momentum and Impulse

Chapter 6 Table of Contents Section 1 Momentum and Impulse

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on standing order act waivers Ppt on types of forests in india Download ppt on turbo generator manufacturers Ppt on network theory and analysis Ppt on good manners for kindergarten Ppt on javascript events on change Ppt on water pollution in hindi Ppt on preservation of public property auction Ppt on textile industry in mumbai Ppt on resistance spot welding