AP Physics 4:Linear Momentum and Energyname ______

AP Physics 4:Linear Momentum and Energyname ______

AP Physics 4:Linear Momentum and EnergyName ______

A.Momentum and Energy formulas

1.impulse changes object's momentum:Ft = mv = p

2.quantity of motion (vector quantity): p = mv (kg•m/s)

3.work changes object's energy: W = F||d(J)

a.measured in joules (1 J = 1 N•m)

a.Only component of F parallel to d does work




1.F|| = FcosW = (Fcos)d

2.include sign W > 0 whenF  d 

3.Fwhen d (orbit)

b.variable force—stretching a spring

1.graph spring force (Fs) vs. position (x)

Fs = kx
slope = k
Area = W / x

2.Fs = kxslope = Fs/x = k

3.W = Fsx area = ½x(kx) = ½kx2 = W

c.power is the rate work is done: P = W/t(W)

1.measured in watts (1 W = 1 J/s)

2.P = W/t = F(d/t) = Fvav (v is average)


a.P = W/tslope of W vs. t graph

b.P= Fvavarea under F vs.v graph

4.kilowatt-hour, 1KWh = 3.6 x 106 J

4.mechanical energy

a.work-energy theorem: work done to an object increases mechanical energy; work done byan object decreases mechanical energy

b.scalar quantity, like work

c.kinetic energy—energy of motion

1.positive only

2.K = ½mv2= p2/2m

Steps / Algebra
start with
assume Ko = 0 vo = 0
solve for ad / v2 = v02 + 2ad
v2 = 2ad
ad = ½v2
start with
substitute ma for F
substitute ½v2 for ad
rearrange / K = W = Fd
K = (ma)d = m(ad)
K = m(½v2)
K = ½mv2
start with
square both sides
divide both sides by 2m
substituteK for ½mv2 / mv = p
m2v2 = p2
m2v2/2m = p2/2m
K = p2/2m

d.potential energy—energy of relative position

1.gravitational potential energy

a.based on arbitrary zero

(usually closest or farthest apart)

b.Ug = mgh (near the Earth's surface)

Steps / Algebra
start with
substitute mg for F
substitute h for d / Ug = W = Fd
Ug = (mg)d
Ug = mgh

c.Ug = -GMm/r(orbiting system)

1.G = 6.67 x 10-11 N•m2/kg2

2.r = distance from center to center

3.Ug = 0 when r is Ug < 0 for all values of r because positive work is needed reach Ug = 0

2.spring (elastic) potential energy, Us = ½kx2

a.Us = W to stretch the spring

b.see work by a variable force above

B.Solving Work-Energy Problems

1.work done on object A by a "nonconservative" force (push or pull, friction) results in thechange in amount of mechanical energy

2.work done on object A by a "conservative" force (gravity, spring) results in the change in form of mechanical energy (U  K) for object A, but no loss in energy

a.conservative forces (Fg and Fs)

1.Fg d : Ug K, Fg d : K  Ug

2.Fs d : Us K, Fs d : K  Us

b.process isn't 100 % efficient

1.friction (W = Ffd) reduces mechanical energy

2.mechanical energy is converted into random kinetic energy of the object's atoms and the temperature increases = heat energy—Q

3.total energy is still conserved


Process / Energy
1 / Work done to / pull a pendulum bob off center / WUg
Transformation / release pendulum / UgK
Work done by / hit a stationary object / KW
2 / Work done to / throw a ball into the air / WK
Transformation / ball rises and falls / K Ug
Work done by / falling ball dents ground / KW
3 / Work done to / load a projectile in spring-gun / WUs
Transformation / release projectile / UsK
Work done by / projectile penetrates target / KW

4.general solution for work-energy problems

  • determine initial energy of the object, Eo
  • if elevated h distance: Ug = mgh
  • if accelerated to v velocity: K = ½mv2
  • if spring compressed x distance: Us = ½kx2
  • determine energy added/subtracted due to an external push or pull: Wp = ±F||d
  • determine energy removed from the object by friction: Wf = Ffd = (mgcosd
  • d is the distance traveled
  •  is the angle of incline (0o for horizontal)
  • determine resulting energy, E'= Eo ± Wp – Wf
  • determine d, h, x or v
  • if slides a distanced:0 = Eo ± Wp – mgcosd'
  • if elevates a heighth: E' = mgh'
  • If compresses a springx: Us: E' = ½kx'2
  • if accelerated to velocityv: E' = ½mv'2
  • general equation (not all terms apply for each problem)
K + Ug + Us ± Wp – Wf = K' + Ug'+ Us'
½mv2 + mgh + ½kx2±Fpd – Ffd = ½mv'2 + mgh' + ½kx'2

C.Solving Collision Problems

1.collision between particles doesn't change the total amount of momentum because the impulse on A equals the impulse on B, but in the opposite directions

a.mAvA + mBvB = mAvA’ + mBvB’

Steps / Algebra
start with Newton'sLaw
multiply both side by t
substitute mvfor Ft
substitute v' – v for v
collect like v terms
multiply by -1 / FA = -FB
FAt = -FBt
mAvA = -mBvB
mA(vA’ – vA) = -mB(vB’ – vB)
-mAvA – mBvB = -mAvA’ – mBvB’
mAvA + mBvB = mAvA’ + mBvB’

b.two particles collide and stick together

1.inelastic collisions (vA' = vB')

2.mAvA + mBvB = (mA + mB)v’

Steps / Algebra
start with
substitute v' for vA' and vB'
simplify / mAvA + mBvB = mAvA’ + mBvB’
mAvA + mBvB = mAv’ + mBv’
mAvA + mBvB = (mA + mB)v’

c.two particles collide and bounce off

1.elastic collisions (vA'  vB')

2.both energy and momentum are conserved

a.two unknowns  need two equations

b.mAvA + mBvB = mAvA’ + mBvB’

c.½mAvA2 + ½mBvB2 = ½mAvA’2 + ½mBvB’2

d.simpler equation: vA+ vA' = vB+ vB’

mAvA + mBvB =
mAvA’ + mBvB’
mA(vA – vA') = mB(vB' – vB) / ½mAvA2 + ½mBvB2 =
½mAvA’2 + ½mBvB’2
mA(vA2 – vA'2) = mB(vB'2 – vB2)
mA(vA2 –vA'2) = mB(vB'2 – vB2)
mA(vA – vA') = mB(vB' – vB)
(vA – vA')(vA + vA') = (vB – vB')(vB + vB')
(vA – vA') = (vB' – vB)

3.solving two equations and two unknowns

  • fill in vA and vB into vA+vA'=vB+vB'
  • write expression for vA’ in terms of vB'
  • substitute vA' expression in equation:
mAvA +mBvB =mAvA’ +mBvB’
  • solve for vB’
  • solve for vA’ using the expression for vA' above

4.collisions in two dimensions

a.px is conserved independently of py

b.elastic collision



3.solve two equations & two unknowns

c.inelastic collision



3.solve two equations & two unknowns

d.object explodes into two pieces mA and mB

1.(mA + mB)v = mAvA’ + mBvB’

2.opposite inelastic collision equation

2.solve ballistics problems

m vm
  • bullet collidesinelastically with block: mvm = (M + m)v'
  • block swings or slides (conservation of energy)
  • block swings like a pendulum to height h
  • K = Ug½(M + m)v'2 = (M + m)ghh = v'2/2g
  • block slides a distance d along a rough surface
  • K = Wf½(M + m)v'2 =(M + m)gdd = v'2/2g

A.Momentum and Energy Formulas

Human Power Lab

Run up a flight of stairs at the football stadium and calculate the power that you generated.

a.Collect the following data.

weight Fg-lbs / Number of steps N
time t / height of step y
width of step x

b.Calculate the following from the data.

Formula / Calculation
dy / dy =Ny
dx / dx = Nx
d / d =(dx2 + dy2)½
v / v =d/t
Fg / Fg = Fg-lbs x 4.45
m / Fg = mg
K / K =½mv2
Ug / Ug = Fgdy
P / P = (Ug + K)/t

c.Calculate the power in horse power (1 hp = 746 W).

Questions 1-24 Briefly explain your answer.

1.An open cart rolls along a frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it?

(A)increase(B) the same(C)decrease

C—rain increases mass, but momentum is conserved and p = mv  increase mass = decrease v

Questions2-3 Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each for 1 s.

2.Which box has more momentum after the force acts?


C—impulses are equal  change in momentum is equal

3.Which box has the greater velocity after the force acts?


A—equal change in momentum: p = mv), but smaller mass would produce greater change in v

Questions 4-6 Use the following options

(A)W > 0(B)W = 0(C)W < 0

4.Work done by a person holding a 10-kg box.

B—without displacement, work = 0 (W = Fd)

5.Work done by a person pulling a 10-kg box across a rough floor at constant speed.

A—force and distance are in opposite directions

6.Work done by friction when a 10-kg box is pulled across a rough floor at constant speed.

C—force and distance are in the opposite directions

7.A box is being pulled up a rough incline by a rope. How many forces are doing work on the box?


C—friction, gravity and tension

Questions 8-9 Alice applied 10 N of force over 3 m in 10 s. Bill applies the same force over the same distance in 20 s.

8.Who did more work?

(A)Alice(B)Bill (C)They did the same

C—W = fd, which is the same for Alice and Bill

9.Who produced the greater power?

(A)Alice(B)Bill (C)They did the same

A—P = W/t, since Alice did the work in less time she produced greater power

10.Car A has twice the mass of car B, but they both have the same kinetic energy. How do their speeds compare?

(A)2vA = vB (B)2vA = vB(C)4vA = vB

B—½mAvA2 = ½mBvB2
2mBvA2 = mBvB2 2vA2 = vB22vA = vB

11.A system of particles has total kinetic energy of zero. What can you say about the total momentum of the system?

(A)p > 0 (B)p = 0(C)p < 0

B—kinetic energy is zero only when v = 0  p = mv = 0

12.A system of particles has total momentum of zero. Does it necessarily follow that the total kinetic energy is zero?

(A)yes (B)no

B—momentum is a vector quantity, a system of particles could be moving, but in opposite directions

13.Two objects have the same momentum. Do these two objects also have the same kinetic energy?

(A)yes (B)no

B—K = ½mv2, where p = mv; since v is squared in K, but not p, then differences in v will not have the same effect

Questions14-15 Stone A has twice the mass as stone B. They are dropped from a cliff and reach a point just above ground.

14.What is the speed of stone A compared to stone B?

(A)vA = vB(B)vA = ½vB

(C)vA = 2vB(D)vA = 4vB

A—acceleration due to gravity is equal and they fall the same distance  velocities must be the same

15.What is the kinetic energy of stone A compared to stone B?

(A)KA = KB(B)KA = ½KB

(C)KA = 2KB(D)KA = 4KB

C—stone A has twice the mass as B  it has twice the K

Questions 16-17Car A and Car B are identical, but car A has twice the velocity of car B.

16.What is the kinetic energy of car A compared to car B?

(A)KA = KB(B)KA = ½KB

(C)KA = 2KB(D)KA = 4KB

D—K = ½mv2 car A has 4 times the kinetic energy

17.Car A takes d meters to stop. What is the stopping distance for car B?


D—K = Wf = mgd; car B has ¼ the kinetic energy  it takes ¼ the distance

18.The work Wo accelerates a car for 0 to v. How much work is needed to accelerate the car from v to 3v?


C—W = K: K 0  v  v2 - 02 = v2
K from v to 3v  (3v)2 – v2 = 8v2

19.A golfer making a putt gives the ball an initial velocity of vo, but he has misjudged the putt, and the ball only travels one-quarter of the distance to the hole. What speed should he have given the ball?


A—he needs 4 x the kinetic energy, which requires 2 x v (K = ½mv2)

20.Which can never be negative?


B—mass can't be negative and v is squared  K = ½mv2 is always positive

21.You and your friend both solve a problem involving a skier going down a slope, starting from rest. The two of you have chosen different levels for h = 0 in this problem. Which of the quantities will you and your friend agree on?


(A)I only(B)II only(C)III only(D)II and III

D—your absolute value of Ug is different, but the change is the same and K depends on v not h

22.Two paths lead to the top of a big hill. A is steep and direct, while the B is twice as long but less steep. The change in potential energy on path A compared to path B is

(A)UA < UB(B)UA = UB(C)UA > UB

B—the beginning and end points are the same h is the same

23.How does the work required to stretch a spring 2 cm compare with the work required to stretch it 1 cm?

(A)W2 = W1(B)W2 = 2W1

(C)W2 = 4W1(D)W2 = 8W1

C—W = Us = ½k(x)2, since x is 2 x then W = 22 x

24.A mass attached to a vertical spring causes the spring to stretch and the mass to move downward. Which is true about the sign for Us and Ug?

(A)+Us, +Ug(B)+Us, –Ug

(C)–Us, +Ug(D) –Us, –Ug

B—Us = ½k(x)2: since x is positive Us is positive
Ug = mgh: since h is negative Ug is negative

25.What is the momentum of a 0.5-kg ball traveling at 18 m/s?

p = mv = (0.5 kg)(18 m/s) p = 9.0 kg•m/s

26.What force is generated by a racket, which strikes a 0.06-kg tennis ball that reaches a speed of 65 m/s in 0.03 s?

Ft = mv
F(0.03 s) = (0.06 kg)(65 m/s)F = 130 N

27.How much work is done to move a 50-kg crate horizontally 10 m against a 150-N force of friction?

W = F||d = (150 N)(10 m) = 1500 J

28.How much work is done to pull a 100-kg crate horizontally 10 m using a force of 100 N at 30o?

W = F||d= (100 N)(cos30)(10 m) = 866 J

29.How much work is done to carry a 100-kg crate 10 m up a 30o ramp?

W = F||d=(1000 N)(sin30)(10 m) = 5000 J

30. Why is work not needed to keep the earth orbiting the sun?

Work is not needed to keep the Earth orbiting the sun because there is no force in the direction of motion.

31.How much power is needed to change the speed of a 1500-kg car from 10 m/s to 20 m/s in 5 s?

P = W/t = ½m(v2 – vo2)/t
P = ½(1500)(202 – 102)/5 = 45,000 W

32.How much power does a 75-kg person generate when climbing 50 steps (rise of 25 cm per step) in 12 s?

a.in Watts

P = W/t = U/t = mgh/t
P = (75 kg)(10 m/s2)(50 x 0.25 m)/12 s = 780 W

b.in horse power (1hp = 746 W)

780 W x 1 hp/746 W = 1.05 hp

33.How much power is needed to maintain a speed of 25 m/s against a total friction force of 200 N?

P = Fvav= (200 N)(25 m/s) = 5000 W

34.How long will it take a 1750-W motor to lift a 285-kg piano to a sixth-story window 16 m above?

P = W/t = mgh/t
1750 W = (285 kg)(10 m/s2)(16 m)/t  t = 26 s

35.A 1000-kg car travels at 30 m/s against 600-N of friction.

a.How much power does the car engine deliver?

P = Fvav= (600 N)(30 m/s) = 18,000 W

b.The car accelerates at 2 m/s2. How much power does the car engine deliver now?

P = Fvav = (ma + Ff)vav
P = [(1000 kg)(2 m/s2) + 600 N](30 m/s) = 78,000 W

c.The car goes up a 10o incline at 30 m/s. How much power does the car engine deliver now?

P = Fvav = (mgsin + Ff)vav
P = [(1000 kg)(10 m/s2)(sin10)+600 N](30 m/s)=70,000 W

36.What is the kinetic energy of a 2-kg block moving at 9 m/s?

K = ½mv2= ½(2 kg)(9 m/s)2 = 81 J

37.What is the gravitational potential energy of a 2-kg block that is 6 m above zero potential energy?

Ug = mgh= (2 kg)(10 m/s2)(6 m) = 120 J

38.What is the gravitational potential energy of the earth-moon system? (MEarth = 5.97 x 1024 kg, Mmoon = 7.35 x 1022 kg, distance between Earth and moon, r = 3.84 x 108 m)

Ug = -GMm/r
Ug= -(6.67x10-11)(5.97x1024)(7.35x1022)/3.84x108
Ug = -7.6 x 1028 J

39.Consider the following spring (k = 100 N/m).

a.Calculate the force (F = kx) needed to stretch a spring from 0.0 m to 0.5 m.

x (m) / 0.0 / 0.1 / 0.2 / 0.3 / 0.4 / 0.5
F (N) / 0 N / 10 N / 20 N / 30 N / 40 N / 50 N

b.Graph the data

40 N
20 N
0 N
0.2 m / 0.4 m

c.Calculate the area under the graph.

A = ½BH = ½(0.50)(50) = 12.5 N•m

d.How much potential energy is stored in the stretched spring?

Us = ½kx2= ½(100)(0.50)2= 12.5 J

B.Solving Work-Energy Problems

Questions 40-47 Briefly explain your answer.

40.Three balls of equal mass start from rest and roll down different ramps. All ramps have the same height. Which ball has the greatest speed at the bottom of the ramp?

 


(A)A(B)B(C)C(D)All the same

D—the same amount of Ug is converted to K  same speed

41.A stationary block slides down a frictionless ramp and attains a speed of 2 m/s. To achieve a speed of 4 m/s, how many times higher must the block start from?

(A)2 times(B)4 times(C)8 times

B—a speed of 4 m/s requires 4 x as much K  the ball must lose 4 x Ug = 4h

42.A box sliding on a frictionless flat surface runs into a fixed spring, which compresses a distance x to stop the box. If the initial speed of the box is doubled, how much would the spring compress?


C—Us = K  ½kx2 = ½mv2 x  v (double v = double x)

Questions 43-44 Alice and Bill start from rest at the same elevation on frictionless water slides with different shapes.

 Bill

 Alice

43.At the bottom whose velocity is greater?


C—Ug = K  mgh = ½mv2 v = (2gh)½, since h is the same v is the same

44.Who makes it to the bottom in the least amount of time?

(A)Alice(B)Bill(C)both are the same

A—Alice is faster at all times except the very end (she has fallen a greater h at each instance)

45.A cart starting from rest rolls down a hill and at the bottom has a speed of 4 m/s. If the cart were given a push, so its initial speed at the top of the hill was 3 m/s, what would be its speed at the bottom?

(A)4 m/s(B)5 m/s(C)6 m/s(D)7 m/s

B—K' = Ko + Ug
½mv2 = ½m(3)2 + ½m(4)2 v2 = 32 + 42 = 25  v = 5 m/s

46.You see a leaf falling to the ground with constant speed. When you first notice it, the leaf has initial total mechanical energy Ei. You watch the leaf until just before it hits the ground, at which point it has final total mechanical energy Ef. How do these total energies compare?

(A)Ei < Ef(B)Ei = Ef(C)Ei > Ef

C—energy is lost due to air resistance

47.You throw a ball straight up into the air. In addition to gravity, the ball feels a force due to air resistance. Compared to the time it takes the ball to go up, the time to come back down is


C—at each elevation, the ball is traveling faster on the way up compared to the way down

48.A rock is dropped from 20 m. What is the final velocity?

a.Use kinematics to solve this problem.

v2 = vo2 + 2ad
v2 = 2(-10 m/s2)(-20 m) = ±20 m/s

b.Use energy to solve this problem.

Ug-0 + K0 = Ug-t + Ktmgh = ½mv2
(10 m/s2)(20 m) = ½v2v = ±20 m/s

49.A pendulum bob reaches a maximum height of 0.6 m above the lowest point in the swing, what is its fastest speed?

Ug-0 + K0 = Ug-t + Ktmgh = ½mv2
(10 m/s2)(0.6 m) = ½v2v = 3.5 m/s

50.How far must a 1 kg ball fall in order to compress a spring 0.1 m?(k = 1000 N/m)

Ug-0 + Us-0 = Ug-0 + Us-0mgh = ½kx2
(1)(10 m/s2)h = ½(1000 kg)(0.1 m/s)2h = 0.50 m

51.A 10-kg box is initially at the top of a 5-m long ramp set at 53o. The box slides down to the bottom of the ramp. The force of friction is 31 N. Determine the

a.potential energy at the top of the ramp.

Ug = mgh = mgLsin
Ug= (10 kg)(10 m/s2)(5 m)sin53 = 400 J

b.work done by friction during the slide.

Wf = Ffd= (-31 N)(5 m) = -155 J

c.velocity of the box at the bottom of the ramp.

Ug+ Wf = Kt = ½mv2
400 J – 155 J = ½(10 kg)v2v = 7 m/s

52.A spring (k = 500 N/m) is attached to the wall. A 5-kg block on a horizontal surface ( = 0.25) is pushed against the spring so that the spring is compressed 0.2 m. The block is released and propelled across the surface.

a.Determine the potential energy of the spring.

Us = ½kx2= ½(500 N/m)(0.2 m)2 = 10 J

b.Determine the distance that the block travels.

Us = Wf = Ffd = mgd
10 J = (0.25)(5 kg)(10 m/s2)dd = 0.8 m

C.Solving Collision Problems

Collision Lab

Observe an elastic collision between a swinging 200-g weight and a golf ball, and compare the actual post-collision velocities with the theoretical velocities.

a.Collect the following data.

200-g Weight (A) / Golf Ball (B)
string length LA / mass mB
initial angle A / 25o / table height dyB
final angle A' / distance dxB'

b.Calculate the following from the data.

Formula / Calculation
dyA / dyA = LA(1 – cosA)
vA / vA2 = 2gdyA
dyA' / dyA' = LA(1 – cosA')
vA' / vA'2 = 2gdyA'
tB' / dyB = ½gtB'2
vB' / vB' = dxB'/tB'

c.Calculate the following theoretical values.

Formula / Calculation
vA' / vA = -(vA' – vB')
mAvA = mAvA' + mBvB'
vB' / vB' = vA + vA'

d.Calculate the percent differences.

Formula / Calculation
vA' / % = 100|v|/vtheory

Questions 53-59 Briefly explain your answer.

53.A small car and a large truck collide head-on and stick together. Which one has larger momentum change?


C—impulses are equal (Newton's third law)  change in momentums are equal

54.A small beanbag and a bouncy rubber ball are dropped from the same height above the floor. The both have the same mass. Which would hurt more if it hit you on the head?

(A)beanbag(B)rubber ball(C)doesn't matter

B—the rubber ball would bounce back which would generate a greater p and impulse force

55.A box slides with initial velocity 10 m/s on a frictionless surface and collides inelastically with a stationary identical box. What is the final velocity of the combined boxes?

(A)0 m/s(B)5 m/s(C)10 m/s(D)20 m/s

B—mAvA + mBvB = (mA + mB)v'
mA(10 m/s) + 0 = 2mAv'  v' = 5 m/s

Questions 56-57 A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light.

56.Which fragment has the greater momentum?

(A)heavier one(B)lighter one(C)tie

C—initial momentum = final momentum = zero  equal momenta in opposite directions

57.Which fragment has the greater speed?

(A)heavier one(B)lighter one(C)tie

B—momentums are equal, p = mv, but with less mass, the lighter one has greater speed

58.Alice (50 kg) and Bill (75 kg) are standing on slippery ice and push off of each other. If Alice slides at 6 m/s, what speed does Bill have?

(A)2 m/s(B)3 m/s(C)4 m/s(D)6 m/s

C—0 = (mAvA' + mBvB'
0 = (50 kg)(6 m/s) + (75 kg )vB'  vB' = -4 m/s

59.A cannon sits on a stationary railroad flatcar with a total mass of 1000 kg. When a 10-kg cannon ball is fired at a speed of 50 m/s, what is the recoil speed of the flatcar?

(A)0 m/s(B)0.5 m/s(C)10 m/s(D)50 m/s

B—0 = (mA + mB)v'
0 = (10)(50 m/s) + (1000)vB'  vB' = -0.5 m/s

60.A 25-kg child in a stationary 55-kg boat with a 5-kg package throws the package out horizontally at 8 m/s. What is the boat and child'sresultant velocity?

0 = mAvA' + mBvB'
0 = (25 kg + 55 kg)v1' + (5 kg)(8 m/s)v = -0.5 m/s

61.An 85-kg safety running at 5 m/s tackles a 95-kg fullback traveling at 4 m/s from behind. What is their mutual speed just after the tackle?

mAvA + mBvB = (mA + mB)v'
(85)(5) + (95)(4) = (85+ 95)v'v' = 4.47 m/s

62.A 0.45-kg ice puck, moving east with a speed of 3.0 m/s, has a head-on elastic collision with a 0.9-kg puck initially at rest. What are the resulting speeds and directions?

vA + vA' = vB + vB'
3.0 + vA' = 0 + vB'  vB' = vA' + 3.0
mAvA + mBvB = mAvA' + mBvB'
(.45 kg)(3.0 m/s) = (45 kg)(vA') + (.9 kg)(vA' + 3.0)
vA' = -1 m/s (west)
vB' = vA' + 3.0 = -1 m/s + 3 = 2 m/s (east)

63.A 1-kg block traveling at 5 m/s in the direction of 30o south of east collides and sticks with a 2-kg block traveling north at 3 m/s. Determine

a.The x-component of the resulting velocity, vx'.

px: mAvAcosA + mBvBcosB = (mA + mA)vx'
(1)(5)cos-30+0 = (1+2)vx'vx' = 1.4 m/s

b.The y-component of the resulting velocity, vy'.

py: mAvAsinA + mBvBsinB = (mA + mA)vy'
(1.0)(5)sin-30 + (2)(3)sin90 = (1 + 2)vy'vy' = 1.2 m/s

c.The resultant speed.

v' = (vx'2 + vy'2)½
v' = [(1.4 m/s)2 + (1.2 m/s)2]½ = 1.9 m/s

d.The resultant direction.

tan = vy'/vx'
tan = 1.2 m/s/1.4 m/s = 0.81   = 54o

64.A 0.50-kg softball is traveling at 40 m/s. A bat makes contact with the ball for 0.025 s, after which, the ball's velocity is 35 m/s in the opposite direction (v = -75 m/s).