Classical Mechanics Lecture 11 Today s Examples Center of Mass Today s Concept: Conservation of Momentum Inelastic Collisions Mechanics Lecture 11, Slide 1
Unit 10 Homework Problems Mechanics Lecture 10, Slide 2
Homework Problem i x N i i Total cm x v m M v, 1, 1 = = = = N i M Total m i 1 i y N i i Total cm y v m M v, 1, 1 = = Mechanics Lecture 10, Slide 3
m 5 CM moves toward m 5 cm Mechanics Lecture 10, Slide 4
Man on Boat F a net, external cm = 0 x cm = = 0 x cm,0 x cm = 1 M x = Total x N i= 1 cm, new cm,0 m i x i Let d be the distance the boat moves to the left. The man s new position is L-d. x j = M Total x cm m The boat s new position in L/2-d. ( m + m ) x = M x 1 2 1 Total cm j N i j m x i i x = 1 x cm Mechanics Lecture 10, Slide 5
Let s try two men on a Boat x new = d 90, L xcm = ; xcm = xcm = L boat,70 0;,, 90 2 x cm,0 = m 70 x cm,70 + m m 70 90 x + m cm,90 90 + m + m boat boat x cm, boat 0 + (90)(20) + (400)(10) x cm, 0 = = 10. 357m 70 + 90 + 400 x cm, new d = = m ( m 70 70 x x 70, new = 90, new = d L xboat, new = + d 2 L ( d) + m90( d) + mboat ( + d) 2 m70 + m90 + mboat x cm, new = xcm,0 L + m90 + mboat ) xcm,0 mboat 2 m + m + m 70 90 boat (560)10.357m (400)(10) d = = 3. 2m (560) Mechanics Lecture 10, Slide 6
x cm = 1 M Total N i= 1 m i x i x = x cm, new cm,0 Mechanics Lecture 10, Slide 7
x x cm j = = 1 M M Total Total x N i= 1 cm m j m x i N i j i m x i i Mechanics Lecture 10, Slide 8
x x cm j = = 1 M M Total Total x N i= 1 cm m j m x i N i j i m x i i x x 1 1 1 = x2 = x3 x4 x = M Total xcm m2x2 m3x3 m4x = m1 M Total xcm m2x1 m3x1 m4x1 = m m m + m + m 1 x = M x Total ( + 1 2 3 4 ) 1 cm 4 x = 1 x cm Mechanics Lecture 10, Slide 9
Classical Mechanics Lecture 11 Today s Concept: Conservation of Momentum Inelastic Collisions Mechanics Lecture 11, Slide 10
Unit 11 Lecture Thoughts Mechanics Lecture 11, Slide 11
Main Points Mechanics Lecture 11, Slide 12
Main Points Mechanics Lecture 11, Slide 13
Conservation of Total Momentum Mechanics Lecture 11, Slide 14
Conservation of Total Momentum Mechanics Lecture 11, Slide 15
Total Momentum & Center of Mass Velocity Mechanics Lecture 11, Slide 16
Conservation of Total Momentum Mechanics Lecture 11, Slide 17
Conservation of Momentum http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html Mechanics Lecture 11, Slide 18
Example Mechanics Lecture 11, Slide 19
Example Mechanics Lecture 11, Slide 20
Example Mechanics Lecture 11, Slide 21
Momentum in 2-d More about this later Mechanics Lecture 11, Slide 22
Collisions: Conservation of Total Momentum Mechanics Lecture 11, Slide 23
Collision and Momentum Conservation Mechanics Lecture 11, Slide 24
Collision and Momentum Conservation What about internal forces? F 21 = F12 p 1 p 2 0 0 0 F net, external = 0 P tot = 0 Mechanics Lecture 11, Slide 25
Checkpoint A. B. C. 0% 0% 0% Mechanics Lecture 11, Slide 26
Checkpoint Mechanics Lecture 11, Slide 27
Inelastic Collisions Internal forces do not affect conservation of momentum! Kinetic Energy of Center of Mass motion also unchanged However sum of Kinetic Energy of individual element may be reduced Mechanics Lecture 11, Slide 28
Collisions:Kinetic Energy Mechanics Lecture 11, Slide 29
Collisions: Kinetic Energy Total Kinetic Energy is not conserved Total Kinetic Energy is reduced in inelastic Collisions Mechanics Lecture 11, Slide 30
Inelastic Collision Mechanics Lecture 11, Slide 31
Which Kinetic Energy? Mechanics Lecture 11, Slide 32
Energy loss in collision Totally Inelastic Stick Together Mechanics Lecture 11, Slide 33
Mechanics Lecture 11, Slide 34
Center of Mass Frame Mechanics Lecture 11, Slide 35
Center of Mass Frame Mechanics Lecture 11, Slide 36
Center of Mass Frame Mechanics Lecture 11, Slide 37
Center of Mass Frame Mechanics Lecture 11, Slide 38
Checkpoint A. B. C. 0% 0% 0% Mechanics Lecture 11, Slide 39
Mechanics Lecture 11, Slide 40
What is the different between conservation of Momentum and conservation of energy? How do I know if the Momentum is conserved or energy is conserve? Momentum (Prelecture 11) Energy (Prelecture 8) Mechanics Lecture 11, Slide 41
Clicker Question A wood block rests at rest on a table. A bullet shot into the block stops inside, and the bullet plus block start sliding on the frictionless surface. The momentum of the bullet plus block remains constant A. B. C. D. E. Before After 20% 20% 20% 20% 20% A) Before the collision. B) During the collision C) After the collision D) All of the above E) Only A and C above As long as there are no external forces acting on the system Mechanics Lecture 11, Slide 42
Recap P tot = i m v i i Mechanics Lecture 11, Slide 43
CheckPoint Suppose you are on a cart initially at rest that rides on a frictionless track. If you throw a ball off the cart towards the left, will the cart be put into motion? Left Right A) Yes, and it moves to the right. B) Yes, and it moves to the left. Conservation of momentum means the cart must move to the right since the ball moves to the left. C) No, it remains in place. Mechanics Lecture 11, Slide 44
CheckPoint Suppose you are on a cart which is initially at rest that rides on a frictionless track. You throw a ball at a vertical surface that is firmly attached to the cart. If the ball bounces straight back as shown in the picture, will the cart be put into motion after the ball bounces back from the surface? Left Right A) Yes, and it moves to the right. B) Yes, and it moves to the left. C) No, it remains in place. Mechanics Lecture 11, Slide 45
A. Clicker Question B. C. Suppose you are on a cart that is moving at a constant speed v toward the left on a frictionless track. If you throw a massive ball straight up (relative to the cart), how will the speed of the cart change? Left Right v 33% 33% 33% A) Increase B) Decrease C) Will not change As long as there are no external forces acting on the system, P total is conserved Mechanics Lecture 11, Slide 46
CheckPoint Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck. Which box ends up moving faster? A) Box 1 B) Box 2 C) same 1 2 Mechanics Lecture 11, Slide 47
CheckPoint Which box ends up moving faster? A) Box 1 B) Box 2 C) same 1 2 A) As the ball bounces to the left, cart 1 moves faster to the right to conserve momentum. B) Some of the energy is not transferred to box 1, since the ball bounces back C) The starting velocity and mass is all the same so the momentum will be the same in both cases. Think of a 2-step bounce Mechanics Lecture 11, Slide 48
Clicker Question A. B. C. Two equal-mass balls swing down and hit identical bricks while traveling at identical speeds. Ball A bounces back, but ball B just stops when it hits the brick. Which ball is more likely to knock the brick over? 33% 33% 33% A B A) A B) B C) They both have the same chance. Mechanics Lecture 11, Slide 49
demo A B P A P B P A > P B The change in the momentum of the ball is bigger in A Mechanics Lecture 11, Slide 50
Ballistic Pendulum http://hyperphysics.phy-astr.gsu.edu/hbase/balpen.html Mechanics Lecture 11, Slide 51
Ballistic Pendulum m v M H A projectile of mass m moving horizontally with speed v strikes a stationary mass M suspended by strings of length L. Subsequently, m + M rise to a height of H. Given H, what is the initial speed v of the projectile? Mechanics Lecture 11, Slide 52
Breaking it down into steps before during after m v M V H splat Which quantities are conserved before the collision? A) momentum B) mechanical energy C) both momentum and mechanical energy Mechanics Lecture 11, Slide 53
Breaking it down into steps before during after m v M V H splat Which quantities are conserved during the collision? A) momentum B) mechanical energy C) both momentum and mechanical energy mv = ( m + M ) V Mechanics Lecture 11, Slide 54
Breaking it down into steps before during after m v M V H splat Which quantities are conserved after the collision A) momentum B) mechanical energy C) both momentum and mechanical energy V 2 = 2gH Mechanics Lecture 11, Slide 55
Ballistic Pendulum m v M H mv = ( m + M ) V V 2 = 2gH V = m ( m + M ) v m + M v = 2gH v = 2gH 1 + m m M Mechanics Lecture 11, Slide 56
Ballistic Pendulum http://hyperphysics.phy-astr.gsu.edu/hbase/balpen.html Mechanics Lecture 11, Slide 57