1 / 8

10/14 Energy Practice

10/14 Energy Practice. Text: Chapter 6 Energy Lab: “Ballistic Pendulum” Energy example (Like “Ramp Launch”) No HW assigned Exam 2 Thursday, 10/17 5-7 Wit 116 6-8 Wit 114 (only if needed) Please send email if other time needed Also let me know if you are a “6-8” person. v. h. v.

mura
Download Presentation

10/14 Energy Practice

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 10/14 Energy Practice • Text: Chapter 6 Energy • Lab: “Ballistic Pendulum” • Energy example (Like “Ramp Launch”) • No HW assigned • Exam 2 Thursday, 10/17 5-7 Wit 116 6-8 Wit 114 (only if needed)Please send email if other time neededAlso let me know if you are a “6-8” person

  2. v h v Your Lab “Ballistic Pendulum” Use conservation of energy for the swing You will find the initial velocity of the ball by measuring the height. (and some math) v = 0 You will also find the initial velocity of the ball by measuring its velocity with projectile motion ideas. Use conservation of momentum for the collision

  3. Your Lab “Ballistic Pendulum” Here is a quick primer on momentum. When two blocks collide, they exert equal and opposite forces on each other. (Third Law) Since the contact time is the same for both we can say that: FA,Bt = -FB,AtWhere the minus sign means direction. We call FA,Bt the “change in momentum” and the changes in momentum for colliding objects are equal and opposite. If we consider the system of both objects, then these third law pairs are “internal forces” and the momentum of the system as a whole does not change. It is conserved.

  4. The symbol for momentum is p and it is a vector. Changes in p are handled just like changes in v. p = mv and p = mv Your Lab “Ballistic Pendulum” Note that FA,Bt can be seen hiding in the 2nd law, Fnet = ma. Fnet = ma = mv/t. So Fnett = mv and for the system, mv is 0 and mv is constant. (conserved) Momentum of the system is the vector sum of the individual momenta.

  5. pA+B,i = pA+B,f mBvB,i +mAvA,i = (mB + mA)vA+B,f p = m(v) = Ft Bullet sticks in block A vB,i = 100 m/s mA = 2 kg mB = 0.05 kg A vA+B,f = ? A For system, Fnet = 0 5 + 0 = (2.05)vA+B,f (All directions to the right) vA+B,f = 2.4 m/s Is energy conserved? No. Mechanical energy was lost in friction between the bullet and block.

  6. Block up a ramp with friction: A block is held against a compressed spring then released. Find:The velocity at BThe height at CThe velocity at D NR,B fR,B WY WX WE,B KE KE KE KE PEg PEg PEg PEg PEs PEs PEs PEs ? 0.5m Lost energy = -fR,Bx = kNR,B x m = 0.5kg k = 320N/m k = 0.25 g = 10N/kg C B 3.0m ? D A PEs,A - lost energy = KEB + PEg,B 5.0m 40 5 20J 15J PE and h = 0 4.0m PEs,A = 1/2kx2 = 40J PEg,B = mghB = 15J NR,B = WY = 4/5WE.B = 4/5mg = 4N = 0.25(4)(5) = 5J KEB = 20J = 1/2mv2 vB = 8.9m/s

  7. Block up a ramp with friction: A block is held against a compressed spring then released. Find:The velocity at BThe height at CThe velocity at D vC = ? KE KE KE KE PEg PEg PEg PEg PEs PEs PEs PEs m = 0.5kg k = 320N/m k = 0.25 g = 10N/kg Energy conserved 20J 15J C vB = 8.9m/s B Projectile motion! vC = vX,B 3.0m 40J KEB + PEg,B = KEC + PEg,C D A 20J 15J 12.8J ? 5.0m PE and h = 0 4.0m vC = vX,B = 4/5vB = 7.16m/s KEC = 1/2mv2 = 12.8J 0.5m PEC = mghC = 22.2J hC = 4.44m

  8. Block up a ramp with friction: A block is held against a compressed spring then released. Find:The velocity at BThe height at CThe velocity at D KE KE KE KE PEg PEg PEg PEg PEs PEs PEs PEs 7.16 The angle  = Arccos 11.8 m = 0.5kg k = 320N/m k = 0.25 g = 10N/kg 20J 15J 12.8J 22.2J C vB = 8.9m/s B 35J 3.0m 40J KEB + PEg,B = KED D A 20J 15J 35J 5.0m PE and h = 0 4.0m  KED = 1/2mv2 = 35J vD = 11.8m/s 0.5m Also note: vX,D = 7.16m/s = 53

More Related