Physics 122 Exam #2 Spring 2015

Transcription

1 Physics 122 Exam #2 Spring 2015 You are subject to the Bryn Mawr College honor code while taking this exam. This is a closed book exam. You may use a simple calculator and a pencil during this exam, as well as both sides of a single page of notes. Do all questions on the exam. Do all work directly on the examination, in the space provided. If you run out of room, you may use the back of the page. Be sure to answer all parts of each question. Some problems will take longer than others, but all will be weighted equally when graded. When providing a sketch, please be as neat, accurate, and quantitative as possible. A sloppy sketch indicates misunderstanding and will receive a reduced grade. The final pages of this exam provide several potentially useful equations. You may remove these page for easy reference. This exam will be available between Friday, 3 April 2015 and Wednesday, 8 April 2015, at the reserve desk in the Collier Science Library during normal library operating hours. I will collect all of the exams Thursday morning when the library opens. Take the exam during a single 90 minute sitting in the library. Before you begin, write your name in the space below and at the top of EVERY page of this exam, note the date in the space provided on this page, and note the time. When you finish, note the time and return the exam to the circulation desk. The difference between your start and finish times should be no more that 90 minutes. Name: Date: Start time: Finish time:

2 h m 1 1. A block of mass m 1 slides down the frictionless ramp shown below. The block starts from rest at a height h above the bottom of the ramp. At the bottom it collides elastically with a block of mass m 2 = 2m 1. Derive an expression for the speed of each block after the collision. After the collision, m 1 will travel back up the incline to a height h. What is the ratio of the two heights, h /h? h' m 2 = 2m 1 The speed v 1 of m 1 at the bottom of the ramp, but before the collision can be found using energy conservation, m 1 gh = m 1 v1 2 /2 (1) v 1 = 2gh. (2) The momentum before and after the collision is conserved, m 1 v 1 = m 1 v 1 +m 2v 2. (3) Also, for a one dimensional elastic collision the relative velocity before and after the collision has the same magnitude, but opposite direction, v 1 = v 2 v 1. (4) To find v 2 I will solve (4) for v 1 and substitute into (3), m 1 v 1 = m 1 (v 2 v 1 )+m 2 v 2 (5) 2m 1 v 1 = (m 1 +m 2 )v 2 (6) v 2 = 2m 1 v 1 /(m 1 +m 2 ). (7) Finally, I will substitute in v 1 from (2) and m 2 = 2m 1, v 2 = 2m 1 v 1 /3m 1 (8) v 2 = 2 3 2gh. (9) Substituting (9) into (4) we get, v 1 = 1 3 2gh. (10) Since m 1 travels to the left after the collision, it will go some distance back up the ramp. To find out how high I again use energy conservation, m 1 gh = 1 2 m 1(v 1) 2 (11) gh = 1 ( 1 ) 2 2gh (12) 2 3 gh = 1 gh (13) 9 h h = 1 9 (14)

3 2. A pendulum of length l is held with its string horizontal, and then released. The string runs into a peg a distance d below the pivot, as shown below. What is the smallest value of d for which the string remains taut at all times? d Theradius ofthe circleafterhittingthepeg isr = l d. Theheightchangefrom where itisreleasedtothetopofthemotionafterhittingthepegisl 2r = l 2(l d) = 2d l. Now we can use conservation of energy to find the velocity of the mass at the top of the motion after it hits the peg mg(2d l) = mv 2 /2. In general, both gravity and tension pull down on the object at this point and keep it moving in a circle. The minimum speed is when the tension at the top of the motion is zero. In this case, just gravity provides the centripetal acceleration needed to keep the object moving in a circle. mg = mv 2 /r. Putting in v and r from earlier we have mg = m[2g(2d l)]/(l d). Solving for d we find d = 3l/5.

4 3. Suppose a light weight golf ball is hurled at a heavy bowling ball, which is initially at rest. Assume that the collision is head-on and elastic. After the collision, which ball has the greater momentum? Which has the greater kinetic energy? Why? Before After golf golf bowling After the collision the momentum of the bowling ball is greater. Since the golf ball has a much smaller mass than the bowling ball, it will reverse direction after the collision. To conserve momentum the bowling ball must have as much momentum as the golf ball originally had plus enough extra to counter the new momentum of the golf ball that is now in the opposite direction. After the collision the kinetic energy of the golf ball is bigger. Since the golf ball is much less massive, its velocity will reverse and be nearly equal in magnitude to the original velocity. So, the kinetic energy of the golf ball will not change by much. That means that not much energy will be transferred to the bowling ball. Alternatively, we can consider the kinetic energy of the two balls after the collision using K = p 2 /2m. While the momentum (which appears in the numerator) of the bowling ball after the collision is larger than that of the golf ball, this difference is at most a factor of two. However, the mass (which appears in the denominator) of the bowling ball is much larger than that of the golf ball. This large mass ultimately makes the kinetic energy of the bowling ball smaller.

5 before 4. A massless spring with spring constant k is placed between a block of mass m and a block of mass 3m. Initially the blocks are at rest on a frictionless surface and they are held together so that the spring between them is compressed by an amount D from its equilibrium length. The blocks are then released and the spring pushes them off in opposite directions. The spring is not attached to either block so it falls to the ground after the blocks move apart. Find the final speeds of the two blocks. after v 1 v 3 m 3m m 3m Using conservation of momentum we can write, Using conservation of energy, p before = p after 0 = 3mv 3 mv 1 v 1 = 3v 3. K before +U before = K after +U after kd2 = 1 2 mv mv kd 2 m = v2 1 +3v 2 3. Substituting v 1 from above and solving for v 3, we find, So, kd 2 m = (3v 3) 2 +3v3 2 kd 2 m = 12v2 3 kd 2 v 3 = 12m. kd 2 v 1 = 3 12m.

6 Potentially useful equations v x = v 0x +a x t x = x 0 +v 0x t+ 1 2 a xt 2 v 2 x = v2 0x +2a x(x x 0 ) r = rˆr v = ṙˆr+r θˆθ a = ( r r θ 2 )ˆr+(r θ+2ṙ θ)ˆθ a c = v2 r = ω2 r v = rω ω = 2πf = 2π T F = ma F friction = µn W = mg F s = kx F grav = GM am b r 2 for q +ω 2 q = 0, q(t) = Asin(ωt+φ) A B = A B cosθ A B = A x B x +A y B y +A z B z A B = A B sinθ A B = î(a y B z A z B y ) ĵ(a x B z A z B x )+ ˆk(A x B y A y B x ) g = 10 m/s 2 G = m 3 kg 1 s 2 for ax 2 +bx+c = 0, x = b± b 2 4ac 2a

7 More potentially useful equations p = mv F = dp dt p = R = 1 M F dt m i r i i W ba = b a F dr K = 1 2 mv2 W ba = K b K a b U b U a = F dr a U spring = 1 2 kx2 U grav = mgh U G = GMm r p 1 +p 2 = p 1 +p 2 K 1 +K 2 = K 1 +K 2 v 1 v 2 = v 2 v 1 K +U = K +U W nc = K + U