康涅狄格州大学：《普通物理》课程PPT教学课件（英文版）Lecture 16 Impulse-momentum theorem

Physics 121, Sections 9, 10, 11, and 12 Lecture 15 Today's Topics Homework 6 Chap6:#6,12,20,24,29,38,52,57,78,and83 Midterm 1 Average: 65% Chapter 7 Collision 2-D collisions Chapter 8 orque C M Physics 121: Lecture 16, Pg 1

Physics 121: Lecture 16, Pg 1 Physics 121, Sections 9, 10, 11, and 12 Lecture 15 Today’s Topics: Homework 6: Chap. 6: # 6, 12, 20, 24, 29, 38, 52, 57, 78, and 83. Midterm 1: Average: 65% Chapter 7: Collision 2-D collisions Chapter 8: Torque C.M

Impulse-momentum theorem: The impulse of the force action on an object equals the change in momentum of the object FAt=4p=mv,-mv Impulse has units of Ns mpule≡FAt=Ap The impulse imparted by a force during the time interval At is equal to area under the force-time graph from beginning to the end of the time interval F △t Physics 121: Lecture 16, Pg 2

Physics 121: Lecture 16, Pg 2 Impulse-momentum theorem: The impulse of the force action on an object equals the change in momentum of the object: Ft = p = mvf – mvi Impulse  Ft = p The impulse imparted by a force during the time interval t is equal to area under the force-time graph from beginning to the end of the time interval. F t t i t f t Favt F t t i t f t Impulse has units of Ns

Average Force and Impulse soft spring F Fav F tiff spring △tbig, F small △ t small, av big Physics 121: Lecture 16, Pg 3

Physics 121: Lecture 16, Pg 3 Average Force and Impulse t F t F t t t big, Fav small t small, Fav big soft spring stiff spring Fav Fav

Momentum Conservation 0 EXT 0 IXI △t The concept of momentum conservation is one of the most fundamental principles in physics This is a component (vector) equation We can apply it to any direction in which there is no external force applied You will see that we often have momentum conservation even when energy is not conserved Physics 121: Lecture 16, Pg 4

Physics 121: Lecture 16, Pg 4 Momentum Conservation The concept of momentum conservation is one of the most fundamental principles in physics. This is a component (vector) equation. We can apply it to any direction in which there is no external force applied. You will see that we often have momentum conservation even when energy is not conserved. t p EXT   F = = 0   t p FEXT = 0

Comment on Energy Conservation We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved Energy is lost 》Heat(bomb) >)Bending of metal (crashing cars) Kinetic energy is not conserved since work is done during the collision Momentum along a certain direction is conserved when there are no external forces acting in this direction In general, easier to satisfy than energy conservation Physics 121: Lecture 16, Pg 5

Physics 121: Lecture 16, Pg 5 Comment on Energy Conservation We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved. Energy is lost: »Heat (bomb) »Bending of metal (crashing cars) Kinetic energy is not conserved since work is done during the collision ! Momentum along a certain direction is conserved when there are no external forces acting in this direction. In general, easier to satisfy than energy conservation

Ballistic Pendulum v=0 H M+m M 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? Physics 121: Lecture 16, Pg 6

Physics 121: Lecture 16, Pg 6 Ballistic Pendulum 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? H L L L L m M M + m v V V=0

Ballistic Pendulum Two stage process m collides with M. inelastically. Both M and m then move together with a velocity V(before having risen significantly) 2. M and m rise a height H, conserving energy E (no non-conservative forces acting after collision Physics 121: Lecture 16, Pg 7

Physics 121: Lecture 16, Pg 7 Ballistic Pendulum... Two stage process: 1. m collides with M, inelastically. Both M and m then move together with a velocity V (before having risen significantly). 2. M and m rise a height H, conserving energy E. (no non-conservative forces acting after collision)

Ballistic Pendulum Stage 1: Momentum is conserved m in x-direction: mv=(m+M)V m+M Stage 2: Energy is conserved (EI=EF) 2(m+M)V2=(m+M)gH ) V2=2gH Eliminating v gives m/V<9 Physics 121: Lecture 16, Pg 8

Physics 121: Lecture 16, Pg 8 Ballistic Pendulum... Stage 1: Momentum is conserved in x-direction: mv = (m + M )V V m m M = v +       Stage 2: Energy is conserved (E E ) I = F 1 2 2 (m + M )V = (m + M )gH V gH 2 = 2 Eliminating V gives: 2gH m M v 1       = +

Ballistic Pendulum L H M+mi d If we measure the forward displacement d, not H (L-H) L-H H=L-√2-d Hd Physics 121: Lecture 16, Pg 9

Physics 121: Lecture 16, Pg 9 Ballistic Pendulum If we measure the forward displacement d, not H: H L L L L m M M + m v d L H d L-H L d (L H) H L L d 2 2 2 2 2 = + − = − −

Ballistic Pendulum L-H H for M 2gH d m m for d<< l Physics 121: Lecture 16, Pg 10

Physics 121: Lecture 16, Pg 10 Ballistic Pendulum L H d L-H d L for 1 2gH m M v 1       = + L g d m M v 1        = + for d << L