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Download Ch 8 HW Day 4: p 254 – 265, #`s 5, 11 – 15, 18, 21, 67, 71 – 74
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Ch 8 HW Day 4 (8.5 – 8.6): p 254 – 265, #’s 5, 11 – 15, 18, 21, 67, 71 – 74 21. Determine the Concept. We can find the loss of kinetic energy in these two collisions by finding the initial and final kinetic energies. We’ll use conservation of momentum to find the final velocities of the two masses in each perfectly elastic collision. (a) Letting V represent the velocity of the masses after their perfectly inelastic collision, use conservation of momentum to determine V: Express the loss of kinetic energy for the case in which the two objects have oppositely directed velocities of magnitude v/2: Letting V represent the velocity of the masses after their perfectly inelastic collision, use conservation of momentum to determine V: pbefore pafter or mv mv 2mV V 0 v 2 K K f K i 0 2 12 m 2 mv2 4 pbefore pafter or mv 2mV V 12 v Express the loss of kinetic energy for the case in which the one object is initially at rest and the other has an initial velocity v: K K f K i 2 mv v 12 2m 12 mv2 4 2 2 The loss of kinetic energy is the same in both cases. (b) Express the percentage loss for the case in which the two objects have oppositely directed velocities of magnitude v/2: 2 mv K 100% 2 K before mv Express the percentage loss for the case in which the one object is initially at rest and the other has an initial velocity v: 2 1 mv K 14 2 50% K before 2 mv 1 4 1 4 The percentage loss is greatest for the case in which t he two objects have oppositely directed velocitie s of magnitude v/2. 73 •• Picture the Problem. We can find the velocity of the center of mass from the definition of the total momentum of the system. We’ll use conservation of energy to find the maximum compression of the spring and express the initial (i.e., before collision) and final (i.e., at separation) velocities. Finally, we’ll transform the velocities from the center of mass frame of reference to the table frame of reference. (a) Use the definition of the total momentum of a system to relate the initial momenta to the velocity of the center of mass: P mi vi Mv cm i or m1v1i m2 v2i m1 m2 vcm Note: The RHS is true because of the definition of the velocity of the com of a system, not because it is a PIC. Solve for vcm: vcm Substitute numerical values and evaluate vcm: vcm m1v1i m2 v2i m1 m2 2 kg 10 m/s 5 kg 3 m/s 2 kg 5 kg 5.00 m/s (b) Find the kinetic energy of the system at maximum compression (u1 = u2 = 0): Use conservation of energy to relate the kinetic energy of the system to the potential energy stored in the spring at maximum compression: 2 K K cm 12 Mvcm 7 kg 5 m/s 2 87.5 J K U s 0 or K f K i U sf U si 0 K cm K i k x 0 Because Kf = Kcm and Usi = 0: Solve for x: 1 2 1 2 x 2 2K i K cm k 2 12 m1v12i 12 m2 v22i K cm k m1v12i m2v22i 2 K cm k Substitute numerical values and evaluate x: x 2 kg 10 m/s 2 5 kg 3 m/s 2 287.5 J 1120 N/m 1120 N/m 0.250 m (c) Find u1i, u2i, and u1f for this elastic collision: Use conservation of mechanical energy to set the velocity of recession equal to the negative of the velocity of approach and solve for u2f: Transform u1f and u2f to the table frame of reference: u1i v1i vcm 10 m/s 5 m/s 5 m/s, u 2i v 2i vcm 3 m/s 5 m/s 2 m/s, and u1f v1f vcm 0 5 m/s 5 m/s u2f u1f u2i u1i and u2f u2i u1i u1f 2 m/s 5 m/s 5 m/s 2 m/s v1f u1f vcm 5 m/s 5 m/s 0 and v 2f u 2f vcm 2 m/s 5 m/s 7.00 m/s *74 •• Picture the Problem. Let the system include the earth, the bullet, and the sheet of plywood. Then Wext = 0. Choose the zero of gravitational potential energy to be where the bullet enters the plywood. We can apply both conservation of energy and conservation of momentum to obtain the various physical quantities called for in this problem. (a) Use conservation of mechanical energy after the bullet exits the sheet of plywood to relate its exit speed to the height to which it rises: Solve for vm: Proceed similarly to relate the initial velocity of the plywood to the height to which it rises: (b) Apply conservation of momentum to K U 0 or, because Kf = Ui = 0, 2 1 m 2 mv mgh 0 vm vM 2 gh 2 gH pi pf or the collision of the bullet and the sheet of plywood: Substitute for vm and vM and solve for vmi: mvmi 0 mvm MvM mvmi m 2 gh M 2 gH vmi M 2 gh 2 gH m (c) Express the initial mechanical energy of the system (i.e., just before the collision): Express the final mechanical energy of the system (i.e., when the bullet and block have reached their maximum heights): KEi 12 mvm2 i (d) Use the work-energy theorem with Wext = 0 to find the energy dissipated by KEf KEi Wfriction 0 2M mg h m 2 M hH H m KEf mgh MgH g mh MH and friction in the inelastic collision: Wfriction KEi KEf h M gMH 2 1 H m