Physics 2210 Homework 12 Spring 2015

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Physics 220 Homework 2 Spring 205 Charles Jui March 23, 205 IE Spring Loaded collision Figure : IE Spring Loaded collision A cart with mass m = 3.2 kg and initial velocity of v,i = 2.

1 Physics 220 Homework 2 Spring 205 Charles Jui March 23, 205 IE Spring Loaded collision Figure : IE Spring Loaded collision A cart with mass m = 3.2 kg and initial velocity of v,i = 2. m/s collides with another cart of mass M 2 = 4.3 kg which is initially at rest in the lab frame. The collision is completely elastic, and the wheels on the carts can be treated as massless and frictionless. What is the velocity of m in the center of mass frame after the collision? See Figure. Center of mass is moving with v cm = m v + M 2 v 2 m + M 2 The velocity of m before the collision in the center of mass frame is thus v,i = v,i v cm Since the collision is elastic, the velocity will reverse in the center of mass frame, i.e. v f = v i v f =.204 m/s

2 Bumper Cars Figure 2: Bumper Cars A bumper car with mass m = 03 kg is moving to the right with a velocity of v = 4 m/s. A second bumper car with mass m 2 = 92 kg is moving to the left with a velocity of v 2 = 3.4 m/s. The two cars have an elastic collision. Assume the surface is frictionless. See Figure 2.. What is the velocity of the center of mass of the system? 2. What is the initial velocity of car in the center-of-mass reference frame? 3. What is the final velocity of car in the center-of-mass reference frame? 4. What is the final velocity of car in the ground (original) reference frame? 5. What is the final velocity of car 2 in the ground (original) reference frame? 6. In a new (inelastic) collision, the same two bumper cars with the same initial velocities now latch together as they collide. What is the final speed of the two bumper cars after the collision? 7. Compare the loss in energy in the two collisions. ) Center of mass velocity is v cm = m v + m 2 v 2 m + m 2 v cm = m/s 2) Initial velocity of the car in the center-of-mass frame v (cm),i = v v cm v (cm),i = 3.49 m/s 2

3 3) Since the collision is elastic, the car will bounce off with opposite velocity, i.e. v (cm),f = 3.49 m/s 4) In the ground frame, we have v (G),f = v cm + v (cm),i v (G),f = m/s 5) The momemtum is conserved, i.e. Thus m v + m 2 v 2 = m v + m 2 v 2 v 2 = m m 2 (v v ) + v 2 where the values are taken in the ground frame. v (G) 2,f = 4.47 m/s 6) Momemtum is still conserved m v + m 2 v 2 = (m + m 2 )v 2 So we get v 2 = fracm v + m 2 v 2 m + m2 v 2,f = m/s 7) In the first case, there is no energy loss - the collision is elastic! In the second case, we have E = KE f KE i = 2 (m + m 2 )v m v 2 2 m 2v

4 Tipler6 8.P.06. A proton of mass m is moving with initial speed v 0 directly toward the center of a nucleus of mass 9m, which is initially at rest. Because both carry positive electrical charge, they repel each other. Find the speed of the nucleus for the following conditions.. when the distance between the two is at its smallest value 2. when the distance between the two is large (a) Conservation of momentum gives p p + p n = p p + p n When the distance is the smallest, proton and nucleus are moving with the same velocity v := v p = v n mv = mv + 9mv which yields v = 0. v 0 (b) Here again, we have conservation of momentum which yields mv 0 = p p + p n = p p + p n = mv p + 9mv n v p = v 0 9v n But this time, also conservation of energy applies and gives 2 mv2 0 = KE p + KE n = KE p + KE n = 2 m(v p ) (9m)(v n) 2 Combining the two results that gives non-zero solution v 2 0 = (v p ) 2 + 9(v n) 2 0 = (5v n v 0 )v n v n = 0.2 v 0 Zero solution is basically the starting velocity. 4

5 Tipler6 8.P.062. An electron collides elastically with a hydrogen atom that is initially at rest. Assume all the motion occurs along a straight line. What fraction of the electron s initial kinetic energy is transferred to the atom? (Take the mass of the hydrogen atom to be 840 times the mass of an electron.) Conservation of momentum mv e = mv e + 840mv H v H = v e v e 840 Conservation of energy 2 mv2 e = 2 mv 2 e + 2 (840m)v2 H If we divide by the initial kinetic energy of the electron, we get ( ) v 2 = e + η where η is the fraction. Also v 2 e = v 2 e v e + 840v 2 H Combining previous equations, we obtain ( ) v 2 ( ) e v e 839 = 0 2 will be put into Tipler6 8.P.03. v e v e ( v η = e v e η = 0.27 % A steel ball of mass m = 0.9 kg and a cord of length of L = 2 m of negligible mass make up a simple pendulum that can pivot without friction about the point O, as in the figure below. This pendulum is released from rest in a horizontal position, and when the ball is at its lowest point it strikes a block of mass m 2 = 0.9 kg sitting at rest on a shelf. Assume that the collision is perfectly elastic and that the coefficient of kinetic friction between the block and shelf is 0.0. See Figure 3.. What is the velocity of the block just after impact? 2. How far does the block slide before coming to rest (assuming that the shelf is long enough)? ) 2 5

6 Figure 3: Tipler6 8.P.03. (a) Velocity right before the impact is determined by conservation of the energy m gl = 2 m v 2 v = 2gL During the collision, the momemtum is conserved, i.e. m v = m v + m 2 v 2 as is the energy Eliminating v gives 2 m v 2 = 2 m v m 2v 2 2 v 2 = 2m m + m 2 2gL v 2 = m/s (b) Since this is a motion with a constant acceleration, we can use vf 2 vi 2 = 2aD where v f = 0, v i = v 2. Acceleration is given by friction m 2 a = µ k N = µ k m 2 g which yields D = v 2 2 2µ k g D = 20 m 6

7 Billiard Balls Figure 4: Billiard Balls A white billiard ball with mass m w =.32 kg is moving directly to the right with a speed of v = 2.9 m/s and collides elastically with a black billiard ball with the same mass m b =.32 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of θ w = 69 and the black ball ends up moving at an angle below the horizontal of θ b = 2. See Figure 4.. What is the final speed of the white ball? 2. What is the final speed of the black ball? 3. What is the magnitude of the final total momentum of the system? 4. What is the final total energy of the system? The momemtum is conserved p + p 2 = p + p 2 So in components And so is the energy Solving for speeds gives m w v = m w v w cos θ w + m b v b cos θ b 0 = m w v w sin θ w m b v b sin θ b 2 m wvw 2 = 2 m wv w m bv b 2 ) v w =.039 m/s 7

8 2) v b = m/s 3) Total momentum of the system is(conserved) P = mv w P = kg m/s 4) Total energy is(conserved) E = 2 m wv 2 w E = J 8

Exam 2 is at 7 pm tomorrow Conflict is at 5:15 pm in 151 Loomis

* By request, but I m not vouching for these since I didn t write them Exam 2 is at 7 pm tomorrow Conflict is at 5:15 pm in 151 Loomis There are extra office hours today & tomorrow Lots of practice exams