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Welcome back to Physics 211 Today s agenda: • Energy • Elastic collisions • Work Physics 211– Fall 2012 Lecture 08-1 1 Current assignments • HW#8 due this Friday at 5 pm. – Solutions will be posted by Saturday to help you study • Midterm 2 in ONE week – on Tuesday, Oct 23: – Chapters 5-9 – Forces, circular motion, momentum – Self-study problems from textbook posted online – Videos working through sample problems and clicker questions from lectures available online – Extra office hours tomorrow (Friday) 3:30-5pm Physics 211– Fall 2012 Lecture 08-1 2 Total Energy E = Ug + K Gravitational Potential energy: for an object of mass m: Ug = mgh • h is height above arbitrary reference line Kinetic energy: For an object of mass m moving with speed v: K = (1/2)mv2 Total Energy is conserved: Ei = Ef Physics 211– Fall 2012 Lecture 08-1 3 8-2.1 If the velocity at B is v, then what is the velocity at C? 1. 2. 3. 4. 5. 2v v sqrt(2) v v/sqrt(2) None of the above Physics 211– Fall 2012 A B h h/2 C Lecture 08-1 4 8-2.2: Starting from a height h, a ball rolls down a frictionless shallow ramp of length l1 = h/sin(30) with an angle 30 degrees, and then up a steep ramp of height h with angle 60 degrees and length l2 = h/sin(60). How far up the steep ramp does the ball go before turning around? 1. ½ l2 2. l1 3. l1 sin(60)/sin(30) 4. l2 5. None of the above Physics 211– Fall 2012 l1 l2 h 60 h 30 Lecture 08-1 5 Demo: the “nose” pendulum Physics 211– Fall 2012 Lecture 08-1 6 Collisions If two objects collide and the net force exerted on the system (consisting of the two objects) is zero, the sum of their momenta is constant. pA,initial + pB,initial = pA,final + pB,final The sum of their kinetic energies may or may not be constant. Physics 211– Fall 2012 Lecture 08-1 7 Elastic and inelastic collisions • If K is conserved – collision is said to be elastic e.g., cue balls on a pool table KA,i + KB,i KA,f + KB,f • Otherwise termed inelastic e.g., lump of putty thrown against wall KA,i + KB,i KA,f + KB,f • Extreme case = completely inelastic -- objects stick together after collision Physics 211– Fall 2012 Lecture 08-1 8 8-2.3 Cart A moving to the right at speed v collides with an identical stationary cart (cart B) on a low-friction track. The collision is elastic (i.e., there is no loss of kinetic energy of the system). What is each cart’s velocity after colliding (considering velocities to the right as positive)? Cart A Cart B 1 -v 2v 2 - 1/3 v 0 3 4 Physics 211– Fall 2012 1 /3 v 4 /3 v v 2 /3 v Lecture 08-1 9 Check conservation of momentum and energy 1 2 3 4 Cart A (m) -v - 1/3 v 0 1 /3 v Physics 211– Fall 2012 Cart B Final Final (m) momentum kin. energy 2v 4 /3 v v 2 /3 v Lecture 08-1 10 Demo • Experiment –> zero sensors Physics 211– Fall 2012 Lecture 08-1 11 Elastic collision of two masses v1i m1 v2i = 0 v1f v2f m2 m1 m2 Momentum à m1v1i + 0 = m1v1f + m2v2f Energy à (1/2)m1v1i2 + 0 = (1/2)m1v1f2 + (1/2)m2v2f2 Physics 211– Fall 2012 Lecture 08-1 12 Special cases: (i) m1 = m2 v1i m1 Physics 211– Fall 2012 v2i = 0 m2 v1f m1 v2f m2 Lecture 08-1 13 Special cases: (ii) m1 << m2 v1i v2i = 0 v1f v2f m1 Physics 211– Fall 2012 Lecture 08-1 14 Midterm sample problem 1: In Sydney Australia there is a ride at an amusement park called the rotor. The Rotor is a large, upright barrel, rotated at 33 revolutions per minute. Once the barrel has attained full speed, the floor is retracted, leaving the riders stuck to the wall of the drum. What is the static coefficient of friction needed to keep a 60 kg man from slipping off the ride? Physics 211– Fall 2012 Lecture 08-1 15 Midterm sample problem 2: A 20 g ball of clay traveling east at 2.0 m/s collides with a 30 g ball of clay travelling 30 degrees south of west at 1.0 m/s. What are the speed and direction of the resulting 50g blob of clay? Physics 211– Fall 2012 Lecture 08-1 16 Springs -- Elastic potential energy frictionless table Force F = -kx (Hooke s law) F Area of triangle lying under straight line graph of F vs. x = (1/2)(+/-x)(-/+kx) Us = (1/2)kx2 Physics 211– Fall 2012 x F = -kx Lecture 08-1 17 (Horizontal) Spring frictionles s table • x = displacement from relaxed state of spring • Elastic potential energy stored in spring: Us = (1/2)kx2 (1/2)kx2 + (1/2)mv2 = constant Physics 211– Fall 2012 Lecture 08-1 18 8.2-4 A 0.5 kg mass is attached to a spring on a horizontal frictionless table. The mass is pulled to stretch the spring 5.0 cm and is released from rest. When the mass crosses the point at which the spring is not stretched, x = 0, its speed is 20 cm/s. If the experiment is repeated with a 10.0 cm initial stretch, what speed will the mass have when it crosses x = 0 ? 1. 40 cm/s 2. 0 cm/s 3. 20 cm/s 4. 10 cm/s Physics 211– Fall 2012 Lecture 08-1 19 Work, Energy • Newton s Laws are vector equations • Sometimes easier to relate speed of a particle to how far it moves under a force – a single equation can be used – need to introduce concept of work Physics 211– Fall 2012 Lecture 08-1 20 What is work? • Assume constant force in 1D • Consider: vF2 = vI2 + 2a s • Multiply by m/2 à (1/2)mvF2 - (1/2)mvI2 = m a s • But: F = ma à (1/2)mvF2 - (1/2)mvI2 = F s Physics 211– Fall 2012 Lecture 08-1 21 Work-Kinetic Energy theorem (1) (1/2)mvF2 - (1/2)mvI2 = F s Points: • W = Fs à defines work done on particle = force times displacement • K = (1/2)mv2 à defines kinetic energy =1/2 mass times square of v Physics 211– Fall 2012 Lecture 08-1 22 Improved definition of work • For forces, write F à FAB • Thus W = F s à WAB = FAB ΔsA is work done on A by B as A undergoes displacement ΔsA • Work-kinetic energy theorem: Wnet,A = ΣBWAB = ΔK Physics 211– Fall 2012 Lecture 08-1 23 The Work - Kinetic Energy Theorem Wnet = ΔK = Kf - Ki The net work done on an object is equal to the change in kinetic energy of the object. Physics 211– Fall 2012 Lecture 08-1 24 8.2-5 Suppose a tennis ball and a bowling ball are rolling toward you. The tennis ball is moving much faster, but both have the same momentum (mv), and you exert the same force to stop each. Which of the following statements is correct? 1. It takes equal distances to stop each ball. 2. It takes equal time intervals to stop each ball. 3. Both of the above. 4. Neither of the above. Physics 211– Fall 2012 Lecture 08-1 25 8.2-6 Suppose a tennis ball and a bowling ball are rolling toward you. Both have the same momentum (mv), and you exert the same force to stop each. It takes equal time intervals to stop each ball. The distance taken for the bowling ball to stop is 1. less than. 2. equal to 3. greater than the distance taken for the tennis ball to stop. Physics 211– Fall 2012 Lecture 08-1 26 8.2-7 Two carts of different mass are accelerated from rest on a low-friction track by the same force for the same time interval. Cart B has greater mass than cart A (mB > mA). The final speed of cart A is greater than that of cart B (vA > vB). After the force has stopped acting on the carts, the kinetic energy of cart B is 1. 2. 3. 4. less than the kinetic energy of cart A (KB < KA). equal to the kinetic energy of cart A (KB = KA). greater than the kinetic energy of cart A (KB > KA). Can t tell. Physics 211– Fall 2012 Lecture 08-1 27