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Download Ch 8 HW Day 5 (Collisions and Ballistic Pendulum): p 254 – 265, #`s
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Transcript
Ch 8 HW Day 6: p 254 – 265, #’s 48, 109, 110, 114, 115, 116, 117, 129, 132 – 134 *48 •• Picture the Problem Assume that the object whose mass is m1 is moving downward and take that direction to be the positive direction. We’ll use Newton’s 2nd law for a system of particles to relate the acceleration of the center of mass to the acceleration of the individual particles. (a) Relate the acceleration of the center of mass to m1, m2, mc and their accelerations: Because m1 and m2 have a common acceleration a and ac = 0: From Problem 4-81 we have: Ma cm m1a1 m2 a 2 mc a c acm a ag m1 m2 m1 m2 mc m1 m2 m1 m2 Substitute to obtain: m m2 m1 m2 acm 1 g m m m m m 2 2 c 1 1 (b) Use Newton’s 2nd law for a system of particles to obtain: Solve for F and substitute for acm from part (a): m1 m2 2 g m1 m2 m1 m2 mc F Mg Macm where M = m1 + m2 + mc and F is positive upwards. F Mg Ma cm 2 m1 m2 Mg g m1 m2 4m1 m2 mc g m1 m2 (c) From Problem 4-81: Substitute in our result from part (b) to obtain: T 2m1m2 g m1 m2 2m1m2 F 2 mc g m1 m2 T 2 mc g 2T mc g g *114 •• Picture the Problem Take the origin to be at the initial position of the right-hand end of raft and let the positive x direction be to the left. Let w denote the woman and r the raft, d be the distance of the end of the raft from the pier after the woman has walked to its front. The raft moves to the left as the woman moves to the right; with the center of mass of the woman-raft system remaining fixed (because Fext,net = 0). The diagram shows the initial (xw,i) and final (xw,f) positions of the woman as well as the initial (xr_cm,i) and final (xr_cm,f) positions of the center of mass of the raft both before and after the woman has walked to the front of the raft. CM x xr_cm,i x w i 6 m , xC M 0 0.5 m CM x xr_cm,f xr_cm,i 0 xw f , (a) Express the distance of the raft from the pier after the woman has walked to the front of the raft: Express xcm before the woman has walked to the front of the raft: P I E R d d 0.5 m xf, w (1) xcm mw xw,i mr xr_cm, i mw mr Express xcm after the woman has walked to the front of the raft: xcm mw xw,f mr xr_cm, f mw mr Because Fext,net = 0, the center of mass remains fixed and we can equate these two expressions for xcm to obtain: mw xw ,i mr xr_cm, i mw xw,f mr xr_cm, f Solve for xw,f: xw,f xw,i From the figure it can be seen that xr_cm,f – xr_cm,i = xw,f. Substitute xw,f for xr_cm,f – xr_cm,i and to obtain: xw,f mr xr_cm, f xr_cm, i mw mw xw,i mw mr 60 kg 6 m Substitute numerical values and evaluate xw,f: xw,f Substitute in equation (1) to obtain: d 2.00 m 0.5 m 2.50 m (b) Express the total kinetic energy of the system: K tot 12 mw vw2 12 mr vr2 60 kg 120 kg 2.00 m Noting that the elapsed time is 2 s, find vw and vr: vw x w,f x w,i t 2m 6m 2 m/s 2s relative to the dock, and vr xr,f xr,i t 2.50 m 0.5 m 1 m/s 2s also relative to the dock. Substitute numerical values and evaluate Ktot: K tot 60 kg 2 m/s 2 2 12 120 kg 1m/s 1 2 180 J Evaluate K with the raft tied to the pier: K tot 12 mw vw2 1 2 60 kg 3 m/s 2 270 J All the kinetic energy derives from the chemical energy of the woman and, (c) assuming she stops via static friction, the kinetic energy is transform ed into her internal energy. After the shot leaves the woman' s hand, the raft - woman system constitute s an inertial reference frame. In that frame the shot has the same initial (d) velocity as did the shot that had a range of 6 m in the reference frame of the land. Thus, in the raft - woman frame, the shot also has a range of 6 m and lands at the front of the raft. 115 •• Picture the Problem Let the zero of gravitational potential energy be at the elevation of the 1-kg block. We can use conservation of energy to find the speed of the bob just before its perfectly elastic collision with the block and conservation of momentum to find the speed of the block immediately after the collision. We’ll apply Newton’s 2nd law to find the acceleration of the sliding block and use a constant-acceleration equation to find how far it slides before coming to rest. (a) Use K U 0 conservation of energy to find or the speed of the bob just before Kf Ki U f U i 0 its collision with the block: Because Ki = Uf = 2 1 m v ball ball mball gh 0 2 0: and vball 2 gh Substitute numerical values and evaluate vball: vball 2 9.81 m/s 2 2 m 6.26 m/s Because the collision is perfectly elastic and the ball and block have the same mass: vblock vball 6.26 m/s (b) Using a constantacceleration equation, relate the displacement of the block to its acceleration and initial speed and solve for its displacement: Apply F ma to the sliding block: vf2 vi2 2ablockx Since vf 0, 2 vi2 vblock x 2ablock 2ablock F x f k mablock and F y Fn mblock g 0 Using the definition of fk (kFn) eliminate fk and Fn between the two equations and solve for ablock: ablock k g Substitute for ablock to obtain: 2 2 vblock vblock x 2 k g 2 k g Substitute numerical values and evaluate x: 2 6.26 m/s x 2 20.1 9.81 m/s 20.0 m 129 •• Picture the Problem Choose the zero of gravitational potential energy at the location of the spring’s maximum compression. Let the system include the spring, the blocks, and the earth. Then the net external force is zero as is work done against friction. We can use conservation of energy to relate the energy transformations taking place during the evolution of this system. Apply conservation of energy: K U g U s 0 Because K = 0: U g U s 0 Express the change in the gravitational potential energy: U g mgh Mgx sin Express the change in the potential energy of the spring: U s 12 kx2 Substitute to obtain: mgh Mgx sin 12 kx2 0 Solve for M: kx2 mgh kx 2mh M gx sin 30 g x 1 2 Relate h to the initial and rebound positions of the block whose mass is m: h 4 m 2.56 msin 30 0.720 m Substitute numerical values and evaluate M: 1110 M N/m 0.04 m 21kg 0.72 m 8.85 kg 2 9.81m/s 0.04 m 3