PH Exam 2 Solutions

Transcription

1 Mar06 LeClair PH 05-3 Exam 2 Solutions Instructions: There are 8 problems below. Choose any 6 problems to solve. All questions have equal point values. Problem : Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (see below). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. Find the original elastic potential energy in the spring if M = kg. What we want in the end is to relate the initial energy (figure part a) stored in the spring to the final energy which is purely kinetic. Thus at some point conservation of total energy is needed. What we first have to realize is that once the cord has been burned, the spring pushes away both masses, and the masses are moving apart in the final state. This means that each block has kinetic energy after the cord is cut, not just the one with mass 3M. Therefore, we need to find the velocity of the block of mass M before we proceed. Finding the velocities after the cord has been cut can be most easily done by conservation of momentum:

2 p before = p after p before = 0 p after = 3Mv 3M + Mv M v M = 3v 3M = 6m/s After we have both velocities, we can use conservation of energy: Problem 2: K i + U i = K f + U f 0 + U spring = K f + 0 K f = 2 Mv2 M + 2 (3M)v2 3M K f = (0.5)(0.25)( 6) 2 J + (0.5)(3)(0.25)(2) 2 J K f = 4.5J +.5J = 6J U spring = 6J A bead slides without friction around a loop-the-loop (below). The bead is released from a height h = 2.95R. What is its speed at point A? Answer in terms of R and g, the acceleration of gravity. Since there is no friction present, we are only dealing with conservative forces, i.e., forces independent of the path taken. Basically, this means that we can use conservation of energy and make this a very simple calculation. Since the object starts out at rest at a height h off of the ground, its initial energy is purely potential: E i = K i + U i = mgh When it reaches the top of the loop-the-loop, it has some velocity v and a height off the floor of 2R. The total energy is now: 2

3 E f = K f + U f = 2 mv2 + mg(2r) The most common mistake here was to forget that at point A on the loop, the height off of the ground is not zero and miss the 2R. Equating E i and E f, mgh = 2 mv2 + 2mgR mg(h 2R) = 2 mv2 g(h 2R) = g(2.95r 2R) = 0.95gR = 2 v2 v 2 = 2(0.95)gR v =.9gR Note that if we had measured potential energy relative to the top of the loop instead of the floor, we could just write down E i = mg(h 2R) and E f = 2 mv2 and the result is the same. It has to be, gravitational potential energy only depends on height differences. Problem 3: Consider the ballistic pendulum shown below, an apparatus used to measure the speed of fast moving projectiles. A bullet of mass m = kg ( 00 grains) is fired at the initially stationary block of wood (m 2 = 2kg) hanging from light wires. The bullet, traveling with a velocity of v A, embeds itself in the block, and the entire block + bullet system swings up to some height h. If h = 0.5m, find v A. What we know in this problem are the masses and initial velocities of the bullet and block. What we want to know is the final height the bullet + block swings up to. Moving the bullet + block to a new height h implies we have gained a potential energy of U = (m + m 2 )gh, which has to be the 3

4 same as the kinetic energy of the bullet + block system just after the collision K = 2 (m + m 2 )v 2 B. The velocity v B is the first thing we need to find then. This is basically just a perfectly inelastic collision, i.e., a collision in which kinetic is not conserved. We can t just use conservation of energy between the initial state of the bullet and the final state of the bullet + block, because the collision is inelastic some mechanical energy is converted to internal energy. We can use conservation of momentum to first relate v A and v B, however, and then use conservation of energy after the collision. Since the block + bullet represent a closed system with no external forces acting, momentum has to be conserved : p before = p after p before = m v A p after = (m + m 2 )v B Equating the momentum before and after the collision, we can solve for v B : v B = m m + m 2 v A Note that this didn t depend on what we called the collision, inelastic or otherwise: we just wrote down the momentum p i = m i v i for every object in the system before and after, momentum is always conserved. Now that we have v B, we can use conservation of energy to find the maximum height the pendulum swings to. The system is still isolated from external forces, so total energy (K + U) is conserved for the block + bullet system. The total energy just after collision (point B in the figure) is purely kinetic, while the total energy at the maximum height (point C) is purely potential. K B + U B = K C + U C U B = K C = 0 so K B = U C The kinetic energy just after the collision can be found from v B above: K B = 2 m totv 2 B = 2 (m + m 2 )vb 2 = ( 2 (m m + m 2 ) v A m + m 2 = m2 v2 A 2(m + m 2 ) ) 2 4

5 Finding U C is easier: Setting K B = U C and solving for v A : U C = m tot gh = (m + m 2 )gh m 2 v2 A 2(m + m 2 ) = (m + m 2 )gh m 2 va 2 = 2(m + m 2 ) 2 gh ( ) va 2 = 2 m + m 2 2 gh m v A = m + m 2 m 2gh Using the numbers given, we find v A 900m/s ( 3000 fps). Problem 4: In the coefficient of restitution lab, we dropped a golf ball from a height h and in one part measured the height after a certain number of bounces. We presumed that we could model the collision between the ball and floor by saying that the velocity after rebound was related to the velocity before rebound by the formula ε = v after v before, where ε was a constant. What is the height of the golf ball after one bounce (h ) in terms of the starting height h and ε? Neglecting air resistance, conservation of mechanical energy dictates that after being dropped from a height h, the ball will reach the ground with a kinetic energy of: 2 mv2 before = mgh The coefficient of restitution defined above gives us the velocity after the ball rebounds from the ground once as v after = εv before. The maximum height after rebound can then be calculated from conservation of energy: mgh = 2 mv2 after or ε = h h 0 = 2 mε2 v 2 before = ε 2 ( 2 mv2 before ) = ε 2 (mgh) h = ε 2 h if you like. One can repeat the same analysis, and show that the height after the second rebound is ε 4 h, and in general h n = ε 2n h. 5

6 Problem 5: A block of mass 2.3 kg is attached to a horizontal spring that has a force constant of N/m. The spring is compressed 3.5 cm and is then released from rest. Calculate the speed of the block as it passes through the equilibrium position x = 0 if the surface is frictionless. Think about the initial and final states. Initially, the energy of the system is purely potential, stored in the spring. In the final state, the energy is purely kinetic: at the equilibrium position, the spring is neither compressed nor expanded (this defines equilibrium), and no energy is stored. Proceed then with conservation of energy: E i = K i + U i,spring = U i,spring E f = K f + U f,spring = K f U i,spring = K f Writing down the surviving energy terms (viz.,, K f and U i,spring ): U i,spring = 2 k( x)2 K f = 2 mv2 K f = U i,spring 2 mv2 = 2 k( x)2 k v = m x v =.26m/s (You did convert the x = 3.5cm into meters first, right? If not, you probably got 26 m/s...) Problem 6: A 000 kg car is traveling at 0 m/s and hits a 2000 kg SUV head-on. The SUV was at rest before the collision, but left in neutral, such that the collision is elastic. What is the car s final velocity (hint: magnitude and direction)? As usual with a collision, conservation of momentum is the starting point. Let the car be and the SUV 2. Then v i = 0m/s, v 2i = 0, m = 000kg, and m 2 = 2000kg. Conservation of momentum before and after the collision gives: 6

7 p before = p after p before = m v i + 0 p after = m v f + m 2 v 2f m v f + m 2 v 2f = m v i Since the collision is elastic, we know that kinetic energy is conserved as well: KE i = KE f 2 m vi = 2 m vf m 2v2f 2 Now we have two equations with two unknowns (v f and v 2f ) Solving for v f and v 2f, m v f + m 2 v 2f = m v i 2 m v 2 i + 0 = 2 m v 2 f + 2 m 2v 2 2f v f = m m 2 m + m 2 v i v 2f = 2m m + m 2 v i This result has already been derived as equations 9.22 and 9.23 in the text. Plugging in the numbers we have, the car s final velocity is: v f = (0m/s) = ( 3 )(0m/s) = 0 3 m/s Where in this case the negative sign indicates that the final motion is the opposite direction of the initial motion. Problem 7: A block of mass m is released from the top of a frictionless incline plane of height h and angle θ, which is sitting at the edge of a table of height H (see below). At the end of the incline, the mass falls off of the table, and hits the floor at a distance R from the edge. What is the speed of the block just before it hits the floor in terms of the variables given above? Again, since there is no friction present, we are only dealing with conservative forces, i.e., forces independent of the path taken. The essence of the problem is then that the initial potential energy of the block at the top of the ramp is converted completely into kinetic energy just before it hits 7

8 the floor. K i + U i = K f + U f () U i = K f (2) = 2 mv2 (3) The initial potential energy is just the total height off of the floor times the weight of the object (mg) the difference in height from initial to final states times mg, or mg(h + H). Note that R and θ do not matter. So: Problem 8: mg(h + H) = 2 mv2 v 2 = 2g(h + H) v = 2g(h + H) A pendulum consists of a mass m hanging from a cord of length L, shown below. Ignore the mass of the cord. The pendulum is released from a position such that the cord makes an angle θ with respect to the vertical. What will the speed of the mass be when it is at a new angle θ < θ? Once again, we are starting from rest at a certain height, and we want to find the speed at some lower height. This generally means conservation of energy... Only the gravitational force on the mass m is doing any work in this problem, so we need only worry about converting gravitational potential energy to kinetic energy. By the way: this differs 8

9 from the practice exam, in that we are moving the pendulum from θ to θ, not from θ to 0. We call the equilibrium position of the pendulum, when it is hanging perfectly vertical and at rest, our reference position at which y = 0, and U = 0. If the pendulum is inclined at an angle θ, simple geometry dictates that the mass m has moved upward by an amount h = L L cos θ = L( cos θ). For an angle θ, then h = L L cos θ = L( cos θ ) Applying conservation of energy, K i + U i = K f + U f 0 + mgh = 2 mv2 + mgh 2 mv2 = mg(h h ) [ 2 v2 = g (L L cos θ) (L L cos θ )] [ ] v 2 = 2g L cos θ L cos θ v = 2gL(cos θ cos θ) In the end, it doesn t matter for this problem that the mass is part of a pendulum at all, except that it constrains the path of the mass and therefore the change in height. The conservation of energy equations are exactly the same as those for a falling mass. The trickier part of this problem was getting the height difference correct from initial (θ) and final (θ ) states, or equivalently, noting that the potential energy in the final (θ ) state is not zero. 9