Impulse, Momentum, Work & Energy

Impulse, Momentum, Work & Energy

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Impulse Force acting for a period of time J = F∆t Unit of measurement: N·s or

kg·m/s Describes how hard and how long

we push to change a motion

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Momentum Impulse causes a change in

momentum.

Impulse-Momentum Theorem F∆t = ∆p Linear momentum is defined as

the product of mass x velocity p = m∆v

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Unit of Measurement for ‘p’ MKS N·s or kg·m/s

CGS dyn·s or g·cm/s

FPS lb·s

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Vector Quantities

Both impulse and momentum are vector quantities.

They have both magnitude and direction.

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Newton’s Second Law F∆t = m∆v F/m = ∆v/t a = a F = m ∆v/t F = ma

Force is directly proportional to the change in velocity if the mass is held constant.

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Exerting a given impulsea) A big force for a short time (F∆t)

b) A smaller force for a longer time (F∆t)

c) A force that changes while it acts (F∆t)

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Direction of Impulse Impulse points in the same

direction as the force and the change in momentum.

Impulse and Momentum

Sample Problems

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1.A baseball of mass 0.14 kg is moving at

+35 m/s. a) Find the momentum of the baseball.

m = 0.14 kg p = m∆vv = +35 m/s = (0.14 kg)(+35 m/s)p = ? p = +4.9 kg·m/s

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b.Find the velocity at which a bowling ball,

mass 7.26 kg, would have the same momentum as the baseball.

p = +4.9 kg·m/s p = mvm = 7.26 kg ∆v = p/m∆V = ? = +4.9 kg·m/s 7.26 kg ∆v = +0.67 m/s

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c.Find the average acceleration of the ball

during its contact with the bat if the average force is -1.4 x 104 N and

m = 0.144 kg.

F = -1.4 x 104 N F = mam = 0.144 kg a = F/m a = ? = -1.4 x 104N 0.144 kg a = -9.7 x 104 m/s2

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2.A 54 N·s impulse if given to a 6.0 kg

object. What is the change of momentum for the object?

J = 54 N·s p = F∆tm = 6.0 kg p = Jp = ? P = 54 N·s

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3. Which quantities do not always

occur in equal and opposite pairs when an interaction takes place within a system?

a. Impulsesb. Accelerationc. Forcesd. Momenta changes

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Answer to 3

accelerations

Conservation of Momentum

Sum of Momenta before =Sum of Momenta after

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Any gain in momentum by an object occurs only by the lost of a corresponding amount of momentum by another object.

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In a system consisting of objects upon which no external force is acting, the momentum of the system is conserved.

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Two objects of equal mass approach each other head-on with the same speed. The total momentum of the system before the collision takes place equal zero.

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Scenario 1Before collision After

Collision -----> at rest ----> ------>m1v1 + m2v2 = m1v1’ + m2v2’

p1 + p2 = p1’ + p2’

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Scenario 2Before collision After

Collision -----> < ---- <---- ------>m1v1 - m2v2 = - m1v1’ + m2v2’

p1 - p2 = -p1’ + p2’

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Scenario 3Before collision After

Collision at rest <---- <----- < ----m1v1 + m2v2 = m1v1’ + m2v2’

p1 + p2 = - p1’ + - p2’

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Scenario 4Before collision After

Collision -----> at rest --- >------>m1v1 + m2v2 = (m1 + m2) v’

p1 + p2 = p’

Examples

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1. Object A has a momentum of 60 N·s.

Object B, which has the same mass, is standing motionless. Object A strikes object B and stops. What is the velocity of object B after the collision if the mass of object B is 6 kg?

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Before collision After Collision

-----> at rest stops ?m1v1 + m2v2 = m1v1’ + m2v2’ p1 + p2 = p1’ + p2’60 N·s + 0 N·s = 0 N·s + (6.0 kg)v2’ v2’ = 10 m/s, to the right

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2.Ball A of mass 0.355 kg moves along a

frictionless surface with a velocity of +0.095 m/s. It collides with ball B of mass 0.710 kg moving in the same direction at a speed of +0.045 m/s. After the collision, ball A continues in the same direction with a velocity of +0.035 m/s. What is the velocity and direction of ball B after the collision?

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2.Before collision After Collision -----> ----- > ------ > ?m1v1 + m2v2 = m1v1’ + m2v2’

(0.355 kg)(0.095 m/s) + (0.710 kg)(0.045 m/s) = (0.355 kg)(0.035 m/s) + (0.710 kg)V2’

v2’ = 0.075 m/s, to the right

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3.A 0.105 kg hockey puck moving at

+48 m/s is caught by a 75 kg goalie at rest. With what speed does the goalie slide on the ice?

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3.Before collision After

Collision -----> at rest ?m1v1 + m2v2 = (m1 + m2)v’

(0.105 kg)(48 m/s) + 0 N·s = (0.105 kg + 75 kg)v’

v’ = 0.067 m/s

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4.A 0.50 kg ball traveling at +6.0 m/s

collides head on with a 1.00 kg ball moving in the opposite direction at a velocity of -12.0 m/s. The 0.50 kg ball moves away at -14 m/s in the opposite direction after the collision. Find v2’.

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4.Before collision After

Collision -----> < ---- < ---- ?m1v1 + m2v2 = m1v1’ + m2v2’

(0.50 kg)(+6.0 m/s) + (1.00 kg)(-12.0 m/s) = (0.50 kg)(-14 m/s) + (1.00 kg)v2’

v2’ = -2.0 m/s

Momentum at an Angle

Right AngleOther Angles

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1.

A 1325 kg car is moving due east at 27 m/s. It collides with a 2165 kg car moving due north at 17.0 m/s.

a. Calculate the momentum of each car before the collision.

b. The two cars stick together and move off at an angle. Calculate the resultant momentum, speed and angle of the cars.

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a.PN = mv =(1325 kg)(27 m/s)PN =3.6 x 104 N·s

PE= mv = (2165 kg)(17.0 m/s)PE = 3.7 x 104 N·s

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b.p’ 2 = (3.6 x 104 N·s)(3.7 x 104 N·s)p’ = 5.2 x 104 N·s

v’ = p’ / mt

= 5.2 x 104 N·s / (1325 kg + 2165 kg)v’ = 14.9 m/s

tan Ө = PN / PE = 3.6 x 104N·s / 3.7 x 104 N·s Ө = 44o N of E

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2. A 6.0 kg ball (A) is moving due east at 3.0 m/s.

It collides with another 6.0 kg ball (B) that is at rest. Ball A moves off at an angle of 40.o north of east. Ball B moves off at an angle of 50.o south of east.

a. Calculate the momentum of ball A before the collision.

b. Calculate the x-component of ball A.c. Calculate the y-component of ball B.d. Calculate the velocity of ball A.e. Calculate the velocity of ball B.

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a) PA = mv= (6.0 kg)(3.0 m/s)PA = 18 kg·m/s

b) PA’ = PA cos 40.o

= 18 kg·m/s (cos 40.o) PA’ = 14 kg·m/s

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c) PB’ = PA sin 40.o

= 18 kg·m/s (sin 40.o) PB’ = 12 kg·m/s

d) VA’ = PA’ / mA

= 14 kg·m/s / 6.0 kg vA = 2.3 m/s

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e) vb’ = PB / mB

= 12 kg·m/s / 6.0 kg vb

’ = 2.0 m/s

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3.Two objects of masses M1 = 1 kg and M2=

4 kg are free to slide on a horizontal frictionless surface. M1 is moving due east at 16 m/s. M2 is at rest. The objects collide and the magnitudes and directions of the velocities of the two objects before and after the collision are M2 moves 37o N of E at 5 m/s; M1 movess 90o due S at 12 m/s.

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a. Calculate the x and y components of the momenta of the balls before and after the collision.

b. Show, using the calculations that momentum is conserved.

c. Calculate the kinetic energy of the two-object system before and after the collision.

d. Is kinetic energy conserved in the collision?

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Before CollisionPx = M1v1 Py = M1v1

= (1 kg)(16 m/s) = (1 kg)(0 m/s) Px = 16 kg·m/s py = 0 kg·m/s

Px = M2v2 py = M2v2

=(4 kg)(0 m/s) = (4 kg)(0 m/s)Px = 0 kg·m/s py = 0 kg·m/s

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After Collision

P’1x = M1v1cos 900 p’1y = M1v1 sin 90o

=(1kg)(12m/s)(0) =(1kg)(12 m/s)(1) P’1x = 0 kg·m/s p’1y = 12 kg·m/s

P’2x = M2v2 cos 37o P’2x = - M2v2 sin 37o

= (4 kg)(5 m/s)(0.80) = (4 kg)(5 m/s)(0.60)P’2x = 16 kg·m/s P’2x = - 12 kg·m/s

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b.Sum of Px before: (16 + 0) kg·m/s = 16 kg·m/sSum of Py before: (0 + 0) kg·m/s = 0 kg·m/s Total = 16 kg·m/s

Sum of Px after = (0 + 16 kg·m/s) = 16 kg·m/sSum of Py after = [(+12) + (-12)] = 0 kg·m/s Total = 16 kg·m/sPx & py are conserved.

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c. K.E. = ½ mv2

Before: K.E. = ½ (1 kg)(16 m/s)2

= 128 JAfter: K.E. = ½ [(1 kg)(12 m/s)2 + (4 kg)(5m/s)2

K.E. = 122 J

d. K.E. isn’t conserved.

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4.A fireworks rocket is moving at a speed

of 50.0 m/s. The rocket suddenly breaks into two pieces of equal mass, and they fly off with velocities v1 and v2. v1 is acting at an angle of 30.0o with the original direction and v2 is acting at an angle of 60.0o with the original direction. What are the magnitudes of v1 and v2?

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X - components2mv1 = m1v1’ + m2v2’2mv1 = m1v1’ cos 30.0o + m1v1’ cos 60.0o

2(50.0 m/s) = 0.866v1’ + 0.500 v2’ v1 = 1.73 v2

Y-components2mv1 = m1v1’ sin 30.0o + m1v1’ sin 60.0o

0 = 0.500 v1 + 0.866v2

=(0.500)(1.73 v2) + 0.866 v2 V2 = 50.0 m/s

V1 = (1.73)(50.0 m/s)V1 = 86.5 m/s

Work

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Work is done only when a force moves an object.

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The distance the object moves must be in the same direction as the force applied to the object.

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Work = force x distance

when both the force and displacement are in the same direction

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Work = force (cos Ө) x distance

When both the force and displacement are not in the same direction

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If W = F cos Ө d, then the units of measurements are

MKS ---- > N·m or J (Joule)CGS ---- > dyn·cm or ergFPs ---- > ft·lb

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Example

A force of 10,000 N is applied to a stationary wall. How much work is performed?

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0. The wall didn’t move.

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Example 2

A 950 N skydiver jumps from an altitude of 3000 m. What is the total work performed on the skydiver?

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Answer:

W = F cos Ө d = (950 N) (3000 m)W = 2.9 x 106 N·m

Work is done by gravity.

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Example 3A 200 N sled is pulled a distance of

20. m with a rope held at an angle of 30o. Calculate the force exerted by the rope if W = 2.0 x 104J.

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Answer:

W = F cos Ө d2.0 x 104 J = F (cos 30o) (20. m)F = 1.2 x 103 N

Mechanical Energy

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Mechanical energy is what an object has due to its motion or position.

Potential energyKinetic energy

Kinetic Energy

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K.E.Energy due to the motion of an object

Work can increase or decrease K.E.

Equation: K.E. = ½ mv2

Units of measurements:MKS: N·m or J (J = Joule)CGS: dyn·cm or ergFPS: ft·lb

Potential Energy

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P.E. Stored in an object because of its state or

position Gravitational P.E. depends on an object’s

position above the earth surface. (base level)

Equation: P.E. = mgh Units of measurements MKS: N·m or J (J = Joule) CGS: dyn·cm or erg FPS: ft·lb

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Gravitational Potential EnergyNewton’s Law of Univ. Grav. P.E. Fg

= mgh = Gm1m2

r2

= m x Gm x h r2

= Gm1m2 r

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Blue Book Reviewing ConceptsP. 236Type or form of energy2. wound-up watch spring = P.E. work was done to wind up watch running mechanical watch use = P.E. converted to K.E. watch runs down = energy converted to heat and sound

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3. Types of mechanical energy in the Earth-Sun system.

Both P. E. and K.E. Both changing. K.E. is greatest when earth is closest to sun.

P. E. is greatest when earth is farthest from the sun.

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B.B. = Applying Concepts5. Can the kinetic energy of a baseball

ever have a negative value?

No. Mass is positive. The square of the velocity is positive.

The change in K.E. can be negative.

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B.B. P. 257 Applying Concepts1. Which has more kinetic energy Compact car or b. semi-truck. Both

are traveling at the same velocity.

Answer:Semi-truck. It has more mass. K.E. is

directly proportional to the mass.

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4. Is it possible to exert a force and yet not cause a change in kinetic energy?

Answer:Yes. The force cannot be applied through

a distance. No work is done and there is no change in kinetic energy

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Example 11. A 1600-kg car travels at a speed of

12.5 m/s. What is its kinetic energy?

Answer:K.E. = ½ mv2

= ½ (1600 kg)(12.5 m/s)2

K.E. = 1.3 x 105 J

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Example Problem 212. How much potential energy does Tim,

mass 60.0 kg gain when he climbs a gymnasium rope a distance of 3.5 m?

Answer:P.E. = mgh = (60.0 kg)(9.80 m/s2)(3.5 m)P.E. = 2.1 x 103 J

Applications of Work and Energy Equations

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Work-Energy TheoremWork causes a change in either

potential or kinetic energy. W = ∆P.E. W = ∆K.E. = mg(h1 – h2) = ½ m (v2

2 –v12)

or or Fd = mg(h1 – h2) Fd = ½ m (v2

2 –v12)

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Conservation of Energy ∆P.E. = ∆K.E.

There is no lost or gain of energy. P.E. is converted to K.E. or K.E. is converted to P.E.

Energy can also be converted to heat, sound or other forms of energy.

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Conservation of Energyhttp://www.glenbrook.k12.il.us/gbssci/phys/mmedia/energy/ie.html

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Nonconservative (external) Conservative (internal) Forces

Internal Forces External Forces

Fgrav Fapplied

Fspring Ffriction

Fair

Ftension

Fnormal

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Nonconservative ForcesTotal mechanical energy of an object is

changedTotal mechanical energy = K.E. + P.E.

Work is positive - object will gain energy

Work is negative - object will lose energy.

The gain or loss in energy can be in the form of P. E., K.E. or both.

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Conservative ForcesInternal forces that change the form of

energy without changing the total amount of mechanical energy.

Energy changes forms from kinetic to potential (or vice versa)

The total amount of mechanical is conserved.

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Example1. A 98-N sack of grain is hoisted to a

storage room 50 m above the ground floor or a grain elevator.

a) W = Fd cosӨ = (98 N)(50 m)(cos Oo) W = 5 x 103 Jb) P.E. = mg∆h or ∆P.E. = W =(98 N)(50 m) = 5 x 103 J P.E. = 5 x 103 J

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1 c) ∆P.E. = ∆K.E. = 5 x 103 J

Using free-fall equationV2

2= V12 + 2 g d

= (0 m/s)2 + 2(9.80 m/s2)(50 m) V2 = 31 m/s

m = W/g = 98 N / 9.8 m/s2 = 10 kg

K.E. = ½ (10 kg)(31 m/s)2 = 5 x 103 J

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Example 23. A 2.0 kg rock initially at rest loses 400 J of potential

energy while falling to the ground.a. Calculate the kinetic energy that the rock gains while

falling.

∆K.E. = ∆P.E. ∆K.E. = 400 J

b. V2 = 2(∆K.E.) / m = (2)(400 J) / 2.0 kg V = 20 m/s2

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Example: Roller Coaster

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Starting from resta) When is the P.E. at its maximum?b) When is K.E. at its maximum?c) What stops the roller coaster?d) Did the wall do work on the roller

coaster?e) Explain how you can determine

the velocity at each point?

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1. At the top. Greatest height2. At the bottom. Greatest change in

height from beginning.3. the wall4. Yes. The wall applied a force through

a distance.5. Use conservation of energy equation. ∆K.E. = ∆P.E. in the forward direction. W = ∆K.E. + ∆P.E.

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Example. The Pendulum and Conservation of Energy

animation: http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/energy/pe.html

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1. Maximum P.E.2. Maximum K.E.3. Equation used to calculate the

maximum speed.4. Do you need to know the mass of the

pendulum bob? Why?5. Equation used to calculate the speed

between the maximum height and bottom of the swing.

animation: http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/energy/pe.html

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1. At beginning and end of swing

2. At the bottom of the swing

3. ∆K.E. = ∆P.E. m(v2

2 – v12) = m(h1 – h2)

h = L – L cosӨ4. No. Mass cancels out.

5. V2 = (L)(g) tan Ө or T sin Ө = mv2/L animation: http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/energy/pe.html

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PowerRate at which work is done by the net force

Power = work / time

Units in MKS system = J/s or Watt (W)

Unit in FPS system is the horsepower (hp)

1 horsepower = 550 ft·lb/sec = 745 watts

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Since it is measured in J/s and Joule is a unit of measurement for energy

Power can also be defined as

P = change in energy/ time or P = ∆E/t

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Deriving another equation for Power work = force x displacement

W = Fd t tOr P = Fv v = d/t

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60.a) P = Fv = (2.00 x 102N)(20.0 m/s) P = 4.00 x 103 Wb) F = Fa + mg sin 37.0o

P = Fv = (Fa + mg sin 37.0o)v=[2.00 x 102 N + (2.0 x 102 kg)(9.80 m/s2)sin 37.0o](20.0 m/s)P = 3.35 x 104 W

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61. P1 = P2 Fwater = 7.50 x 104 W = 6250 N Fwater Fwater + T 2 m/s

T = Fwater [(P2 /P1)] – 1]

= 6250 N[(8.30 x 104 W / 7.50 x 104 W) – 1]

T = 6.7 x 102 N

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61.Constant speed: FBoat(1) = FWater

Pulling a skier: FBoat(2) = Fwater + Trope

P = FvV is the same in both cases P1 = P2

F1(boat) F2(boat)

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64. a. W01 = ½ (+6.0 N)(1.0 m – 0 m) = 3.0 J

b. No area. Work12 = 0 J

c. W23 = ½ (-6.0 N)(3.0 m – 2.0 m) = -3.0 J W34 = (-6.0 N)(4.0 m – 3.0 m) = - 6.0 J

Total Work24 = (-3.0 J) + (-6.0 J) = - 9.0 JWork is negative because force is negative.

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65.a) W03 = (31 N)(3.0 m – 0 m) = 93 J

b. W36 = 0 J No area under curve

c. W = ½ m (v22 - v1

2) 93 J = ½ (65 kg) [(v2)

2 - (1.5 m/s)2] v2 = 2.3 m/s

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67.W0,10= ½ (10.0 N)(10.0 m – 0 m) =5.00 x 101 J

W10,20=(10.0 m)(20.0 m–10.0 m) = 1.00 x 102 J

Total Work = 1.50 x 102 J

b. W = ½ m (v22 - v1

2) 1.50 x 102 J = ½ (6.00 kg) [(v2)

2 - ( 0 m/s)2] v2 = 7.07 m/s

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