Physics 2514 Lecture 34 P. Gutierrez Department of Physics & Astronomy University of Oklahoma Physics 2514 p. 1/13
Information Information needed for the exam Exam will be in the same format as the practice with the same number of questions Bring a # 2 pencil & eraser Calculators will be allowed No cell phones, no laptops,... Only exam, pencil, eraser, calculator allowed on desk. Bring student id with you You will need to know Student id number Discussion section # Your name Physics 2514 p. 2/13
Material to be covered This exam will cover chapters 9, 10, & 11 Impulse-momentum theorem J = p with tf J = t i F dt and p = m v Third law momentum conservation p i = p f if no external forces Work-Kinetic energy theorem W net = K with W net = i W i, and W i = s f s Fi i d s Conservation of mechanical energy E = T + U Potential energy W i = U Physics 2514 p. 3/13
Review Momentum Momentum given by p = m v Newton s second law defines forces in terms of momentum change d p dt = F net p = J = tf t i Fnet dt ( J is the impulse) Newton s third law leads to momentum conservation F 12 = F 21 leads to m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f This implies that no external forces are acting on the two objects Physics 2514 p. 4/13
Review Momentum Object released from rest near the Earth s surface J = p mg t = mvyf 0 = MV yf + mv yf v yf = g t Physics 2514 p. 5/13
Mechanical Energy Review Kinetic energy K = 1 2 mv2 energy due to motion, is always positive Potential energy gravitational U = mgy; spring U = 1 2 k( s)2 Stored energy K = U Total mechanical energy E = K i + U i constant assuming no frictional forces, kinetic & potential energies are mechanical energy Energy is a scalar (not a vector) Zero of potential energy is arbitrary you decide where to set it Only potential energy differences matter Energy units are kg-m 2 /s 2 = Joules Physics 2514 p. 6/13
Review of Work Introduced concept of work: Energy added due to forces acting on an object, W net = i W i = K; Work for an individual force given by W i = s f s i if conservative W i = U Fi d s and If the work done is independent of the path, the force is conservative Force can be written as a potential (true for gravity and spring forces), The mechanical energy is conserved. Friction is not a conservative force The work done depends on the path; Force from potential energy F s = du ds Physics 2514 p. 7/13
Steps in Problem Solving Steps in problem solving 1.) Rewrite the problem eliminating all extraneous information. (What are you given, what are you looking, what are the constraints); 2.) Draw a diagram along with a coordinate system, label each object with the variables associated with it (include forces, initial, final momentum, initial and final energy, and work and impulse where appropriate); 3.) What are the known and unknown quantities, which unknowns are you solving for; 4.) Write down the equations associated with the problem, and solve the problem algebraically 5.) Finally, substitute numbers into the equation, and calculate the numerical solution Physics 2514 p. 8/13
Example The skiing duo of Brian (80 kg) and Ashley (50 kg) is always a crowd pleaser. In one routine, Brian starts at the top of a 200 m long 20 slope. Ashley waits for him halfway down. As he skis past, she leaps into his arms and carries her the rest of the way down. What is their speed at the bottom of the slope (assume there is no friction)? An object (80 kg) starting from rest slides 100 m down a 20 frictionless PSfrag replacements incline. It collides inelastically with a second object (50 kg), which is at rest, and can continues for an additional 100 m. What is its final speed? n 2 m = 80 kg v, a mg m = 130 kg s 1 θ m = 130 kg 0 Physics 2514 p. 9/13
Example eplacements An object (m 1 = 80 kg) starting from rest slides 100 m down a θ = 20 frictionless incline. It collides inelastically with a second object (m 2 = 50 kg), which is at rest, and can continues for an additional 100 m. What is its final speed? 2 m = 80 kg v, a mg n 1 m = 130 kg s θ 0 Speed at 100 m: m 1 gy 2 = 1 2 m 1v 2 1 + m 1gy 1 y n = s n sin θ v 2 1 = 2g sin θ s, v 1 = 25.9 m/s s = s 2 s 1 Inelastic collision v m 1 v 1 = (m 1 + m 2 )v 1, v 1 = 15.9 m/s Speed at 200 m (s = 0, y = 0) (m 1 + m 2 )gy 1 + 1 2 (m 1 + m 2 )v 2 1 = 1 2 (m 1 + m 2 )v 2 0 v 2 0 = v 2 1 + 2gs 1 sin θ, v = 30.4 m/s Physics 2514 p. 10/13
Example Energy Conservation A 10 kg box slides 4.0 m down the frictionless ramp shown in the figure. It then collides with a spring whose spring constant is 250 N/m. 1. What is the maximum compression of the spring? 2. At what compression of the spring does the box have the maximum velocity? Physics 2514 p. 11/13
Example Energy Conservation What is the maximum compression of the spring (m = 10 kg, k = 250 N/m)? 1. Select coordinate system that simplifies the problem. (Selected to divide the problem into gravity only, and gravity plus spring) Use energy conservation (K = 0 initial replacements y s and final positions) Initial energy: E i = mgs i sin(30) = 196 J x Final energy: E f = mgs f sin(30) + 1 2 ks2 f Energy conservation: E i = E f mgs i sin(30) = mgs f sin(30) + 1 2 ks2 f s f = 1.46 m Physics 2514 p. 12/13
Example Energy Conservation At what compression of the spring does the box have the maximum velocity? PSfrag replacements Mass in contact w/spring n ks mg Net Force vs Position 1.4 0.7 200 0 200 Net force at initial contact along s: F s = mg sin(30) Net force at maximum displacement: F s = ks mg sin(30) > 0 (s < 0) Max speed (kinetic energy) at F s = 0 replacements Fs 100 0 50 1.4 0.7 s 0 100 0 50 F s = 0 = ks mg sin(30) s = mg sin(30) = 0.196 k Physics 2514 p. 13/13