LECTURE 13- PROBLEMS. Chapter 1-9,13 Professor Noronha-Hostler Professor Montalvo

LECTURE 13- PROBLEMS Chapter 1-9,13 Professor Noronha-Hostler Professor Montalvo

FARADAY LECTURES! Physics Lecture Hall Friday Dec. 7 Demos: 6pm Show: 7-8:30pm Saturday Dec. 8 Demos: 2pm Show: 3-4:30pm Sunday Dec. 9 Demos: 12pm Show: 1-2:30pm

FINAL EXAM REMEMBER THAT NOTES ARE ALLOWED!!! No formulas will be provided, you do this yourself

PROBLEMS COVERED TODAY Ball hitting a spring Momentum vs. Energy conservation, hooks law Dog sled Friction, work, forces 2d Billiard balls Momentum vs. Energy conservation, vectors Center of mass Center of mass, area Projectile motion Projectile motion, energy conservation, equations of motion

#1 BALL HITTING A SPRING Spring constant k v 0 m 1 m 2 Block at rest An incoming ball traveling on a frictionless surface hits a block of wood and they stick together. What is the maximal compression of the spring?

READING CLUES An incoming ball traveling on a frictionless surface hits a block of wood and they stick together. What is the maximal compression of the spring? f k = 0 Stick total inelastic collisions Maximal compression v f = 0 K f = 0 U = 1 2 kx2 > 0

#1 BALL HITTING A SPRING Spring constant k v 0 m 1 m 2 Block at rest Step 1 Ball travels an unknown distance until it reaches block Step 2 Ball collides with block (totally inelastic) Step 3 Ball+block compress spring until v f = 0

#1 BALL HITTING A SPRING Spring constant k v 0 m 1 m 2 Constant velocity, Block at rest Step 1 Ball travels an unknown distance until it reaches block Step 2 Ball collides with block (totally inelastic) Step 3 Ball+block compress spring until v f = 0 no forces acting here

BALL+BLOCK COMPRESSES SPRING v 0 v 12 m 1 m 2 m 1 m 2 Before collision After collision Conservation of momentum P i = P f initially at rest m 1 v 1 = 0 m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v 12 m 2 v 2 = (m 1 + m 2 )v 12

BALL COLLIDES WITH BLOCK v 0 v 12 m 1 m 2 m 1 m 2 Before collision After collision Conservation of momentum m 2 v 2 = (m 1 + m 2 )v 12 v 12 = m 2 m 1 + m 2 v 2

#1 BALL HITTING A SPRING v 12 v f = 0 m 1 m 2 m 1 m 2 At collision Conservation of Mechanical Energy K i + U i = K f + U f x Maximal compression K i = 1 2 ( m 1 + m 2 ) v 2 12 U i = 0 K f = 0 U = 1 2 kx2

#1 BALL HITTING A SPRING v 12 v f = 0 m 1 m 2 m 1 m 2 At collision x Maximal compression K i + U i = K f + U f 1 2 ( m 1 + m 2 ) v12 2 + 0 = 0 + 1 2 kx2 (m 1 + m 2 ) v12 2 = kx2 x = (m 1 + m 2 ) k v 12

COLLECTING EQUATIONS Conservation of Momentum Conservation of Mechanical Energy v 12 = m 2 m 1 + m 2 v 2 x = (m 1 + m 2 ) k v 12 Substitute in v 12 x = (m 1 + m 2 ) k m 2 m 1 + m 2 v 2 x = m 2 v 2 k (m 1 + m 2)

IF THE TABLE HAD A FRICTIONAL FORCE, WHEN WOULD YOU CONSIDER IT IN THE PREVIOUS PROBLEM? A.) You wouldn t because it would cancel out. B.) Only at the point of the collision. C.) Only after the collision because the frictional force would counteract the force of the incoming block+ball. D.) Before the collision since the ball would slow down and after the collision because the spring would compress more. E.) Before the collision since the ball would slow down and after the collision because the spring would compress less.

FINAL EXAM REMEMBER THAT NOTES ARE ALLOWED!!! No formulas will be provided, you do this yourself

DOG SLED How much work done by the dogs pulling the sled? Assume the snow has a coefficient of friction of μk, the sled has a normal force of FN and travels a distance d.

DOG SLED- FREE BODY DIAGRAM F N v 0 a f k F p d F g

DOG SLED- FREE BODY DIAGRAM F N v 0 a f k F p y direction F N = F g d F g x direction F net = F p f k = ma Force from the dogs, what we need!

DOG SLED- NEWTON S 2ND LAW f k F N v 0 a F g d F p F net = F p f k = ma F p = ma + f k Recall f k = μ k F N F p = ma + μ k F N

DOG SLED- WORK f k F N v 0 a F p F p = ma + μ k F N F g d W = F d Force and displacement in the SAME direction W = + F p d = d(ma + μ k F N )

BILLIARD BALLS

2-DIMENSIONS BILLIARD BALLS m 1 A cue ball ( ) traveling with a velocity of v 1 = 4 m/s hits another ball ( m 2 ) and each travel away at an angle. If the final velocity of m is and θ 1 = 60 1 v f1 = 2 m/s then what is θ 2? Assume mechanical energy is conserved. Assume that they are equal in mass.

WHAT TYPE OF COLLISION IS THIS? A.) Elastic because the potential energy doesn t change, which means that kinetic energy is conserved. B.) Totally inelastic. C.) Inelastic because the balls go off at angles. D.) Elastic because billiard balls always act like elastic collisions. E.) Inelastic because mechanical energy is conserved.

BILLIARD BALLS Elastic Collisions Conservation of Momentum P i = P f Conservation of Kinetic Energy K i = K f Momentum m 1 v 1 = m 1 v f1 + m 2 v f2 Kinetic Energy 1 2 m 2 v2 1 = 1 2 m 2 v2 f1 + 1 2 m 2 v2 f2

CONSERVATION OF KINETIC ENERGY 1 2 m 2 v2 1 = 1 2 m 2 v2 f1 + 1 2 m 2 v2 f2 Recall m 1 = m 2 v 2 1 = v2 f1 + v2 f2 Recall v 1 = 4 m/s v f1 = 2 m/s v 2 f2 = v2 1 v2 f1 v 2 f2 = 42 2 2 v f2 = 3.5 m/s

VELOCITY DECOMPOSITION v f1 sin θ 1 θ m 1 m 1 2 v f1 cos θ 1 θ 2 v f2 cos θ 2 v f2 sin θ 2 v f1 = v f1 cos θ 1 i + v f1 sin θ 1 j v f2 = v f2 cos θ 2 i v f2 sin θ 2 j

MOMENTUM CONSERVATION m 1 v 1 = m 1 v f1 + m 2 v f2 x-direction m 1 v 1 = m 1 v f1 cos θ 1 + m 2 v f2 cos θ 2 m 2 v f2 cos θ 2 = m 1 ( v 1 v f1 cos θ 1) cos θ 2 = m 1 m 2 v f2 (v 1 v f1 cos θ 1) y-direction 0 = m 1 v f1 sin θ 1 m 2 v f2 sin θ 2 sin θ 2 = m 1 v f1 m 2 v f2 sin θ 1

MOMENTUM CONSERVATION Recall m 1 = m 2 x-direction cos θ 2 = m 1 m 2 v f2 (v 1 v f1 cos θ 1) cos θ 2 = 1 v f2 (v 1 v f1 cos θ 1) y-direction m 1 v f1 sin θ 2 = m 2 v f2 sin θ 1 v f1 sin θ 2 = v f2 sin θ 1

WHICH EQUATION SHOULD WE USE? Equation A Equation B cos θ 2 = 1 v f2 (v 1 v f1 cos θ 1) sin θ 2 = v f1 v f2 sin θ 1 v 1 = 4 m/s v f1 = 2.0 m/s v f2 = 3.5 m/s Which equation should we use?

WHICH EQUATION SHOULD WE USE? Equation A sin θ 2 = v f1 v f2 sin θ 1 v 1 = 4 m/s v f1 = 2.0 m/s v f2 = 3.5 m/s sin θ 2 = 2 m/s 3.5 m/s sin 60 θ 2 = 30

WHICH EQUATION SHOULD WE USE? Equation B cos θ 2 = 1 v f2 (v 1 v f1 cos θ 1 ) v 1 = 4 m/s v f1 = 2.0 m/s v f2 = 3.5 m/s cos θ 2 = 1 3.5 m/s ( 4 m/s 2 m/s cos 60 ) θ 2 = 30

Which one of the following statements concerning a planet orbiting the Sun is true? a) The gravitational force between the planet and Sun must be larger than the centripetal force needed to keep the planet in its orbit. b) The centripetal force that keeps the planet moving in a stable orbit is provided by the gravitational attraction between the planet and Sun. c) If the gravitational force between the planet and Sun is less than the centripetal force needed to keep the planet in its orbit the planet would spiral in until the planet reaches a stable orbit. d) The gravitational force of the planet that acts on the Sun is always smaller than the gravitational force of the Sun that acts on the planet. e) None of the above statements are true.

13.7.5. The mean distance between the Earth and the Sun is 1.5 10 11 m. Using this fact and your knowledge of the period of the Earth s orbit around the Sun, determine the approximate mass of the Sun. a) 2 10 14 kg b) 2 10 18 kg c) 2 10 24 kg d) 2 10 27 kg e) 2 10 30 kg

FIND THE CENTER OF MASS r/2 r Find the center of mass of the following object: There is a a circle of radius r that has a circle of radius r/2 cut out of it. The surface density of the circle is ρ Find the Area of the big and little circle Find the Mass of the big and little circle Calculate the center of mass subtracting out the little circle

STEP #1 FIND THE AREA r/2 r Large circle Small Circle A l = πr 2 r A s = π ( 2 ) 2 = 1 4 πr2

STEP #2 FIND THE MASS r In 2D Mass=density*Area r/2 Large circle M l = ρπr 2 Small Circle M s = ρ 1 4 πr2

STEP #2 FIND THE CENTER OF MASS r/2 r How do we find the Center of Mass when we need to remove part of an object? r com = M l r l M s r s M l M s

STEP #2 FIND THE CENTER OF MASS The center mass is equally distributed in the x -direction. We y x-direction see this by drawing the y axis and comparing both halves (A and B). Since A=B then the center of mass in the x-direction is at the origin. A B x x com = 0 r com = M lr l M s r s M l M s

STEP #2 FIND THE CENTER OF MASS y r/2 M l = ρπr 2 M s = 1 4 ρπr2 r y-direction x Where On the positive and negative half in the y-direction, they are NOT symmetric. Thus, we calculate: y com = M l y l M s y s M l M s y l = r y s = r 2 This is at the com of each individual circle!

STEP #2 FIND THE CENTER OF MASS y r/2 r y-direction On the positive and negative half in the y-direction, they are NOT x symmetric. Thus, we calculate: ρπr 2 0 1 4 ρπr2 ( r 2 ) y com = ρπr 2 1 4 ρπr2 y com = 1 8 ρπr3 3 = 4 ρπr2 r 6

STEP #2 FIND THE CENTER OF MASS y r r/2 x r com = 0i + r 6 j

PROJECTILE MOTION v 0 = 50 m/s m θ = 53 Q1 What is the minimum kinetic energy of the ball?

WHERE IS THE MINIMUM KINETIC ENERGY? II v 0 = 50 m/s I m θ = 53 III A.) I B.) II C.) III D.) I and III E.) None of the above

WHERE IS THE MINIMUM KINETIC ENERGY? II v 0 = 50 m/s I m θ = 53 III At the top, is the Kinetic Energy zero?

WHERE IS THE MINIMUM KINETIC ENERGY? II v 0 = 50 m/s m θ = 53 At the top, is the Kinetic Energy zero? No, the horizontal velocity is constant v min = v 0 cos θi + 0j v min = v 0 cos θ K min = 1 2 mv 0 cos θ

PROJECTILE MOTION v 0 = 50 m/s m θ = 53 Q2 What is the height of the maximum potential energy of the ball?

HEIGHT AT MAXIMUM POTENTIAL ENERGY II v 0 = 50 m/s I m θ = 53 So far, conservation of mechanical energy K f K i = U f U i K i = 1 2 mv2 0 K f = K min = 1 2 mv 0 cos θ U i = 0 U f = mgh

HEIGHT AT MAX. POTENTIAL E K f K i = U f U i K i = 1 2 mv2 0 K f = K min = 1 2 mv2 0 cos2 θ U i = 0 U f = mgh Identity 1 = cos 2 θ + sin 2 θ K f K i = 1 2 mv2 0 cos2 θ 1 2 mv2 0 = 1 2 mv2 0 (cos 2 θ 1) = 1 2 mv2 0 sin2 θ

HEIGHT AT MAX. POTENTIAL E K f K i = U f U i U i = 0 K f K i = 1 2 mv2 0 sin2 θ 1 2 mv2 0 sin2 θ = mgh U f = mgh h = 1 2g v2 0 sin2 θ h = 1 2g (50 m/s)2 sin 2 53 = 2.0 m

LANDING v 0 = 50 m/s m θ = 53 Q3 if at the top of flight path the ball explodes in half and one piece lands at the origin, where does the other half land?

LANDING Where does the second piece land? m Remember, in the case of projectile motion, the center of mass of the system always follows the path of the projectile (as if it stayed together)

RECALL THE RANGE

AT THE CENTER OF MASS m 1 m 2 The center of mass lands at m Mass 1 lands at x 1 = 0 Thus, x com = m 1 x 1 + m 2 x 2 m 1 + m 2 x com = v2 0 sin 2θ g Solve for x2 v 2 0 g sin 2θ = M/2 0 + M/2 x 2 M = 2x 2 x 2 = 2 * x com

Q4 is energy and momentum conserved before to after the explosion? A.) Energy and momentum are always conserved, but mechanical energy is not conserved. B.) Only momentum is conserved due to heat loss of the explosion. C.) Only energy is conserved because of the change in velocity from the explosion. D.) None of the above. E.) Only mechanical energy is conserved.

FINAL EXAM REMEMBER THAT NOTES ARE ALLOWED!!! No formulas will be provided, you do this yourself

GOOD LUCK!