Oslund (hno56) – Homework 6 – sai – (58015) This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Two blocks, A and B, are connected by a compressed spring. The mass of block A is twice that of block B. Immediately after release, compare the net force on block A to the net force on block B. The table is frictionless and the mass of the spring is zero. 1. FA,net and FA,net are both directed toward the left with FA,net larger than FA,net , but not twice as large. Block A is pulling block B. 1 Note: This is correct only when the mass of the spring can be neglected as mentioned in the problem. 002 (part 2 of 2) 10.0 points Which of the following best describes the situation after the blocks are released? 1. The total momentum of the system of blocks is conserved because there is no net external force. correct 2. The total momentum of the system of blocks is not conserved because there is an external velocity acting on the system. 2. FA,net is equal in magnitude and opposite in direction to FB,net , and not zero. correct 3. The total momentum of the system is not conserved because there are external forces; e.g., gravity. 3. FA,net will be in the opposite direction and half as large as FB,net . 4. Momentum for any body is always conserved. 4. There is no net force on block A or block B. 5. The momentum of block A is conserved and the momentum of block B is conserved. However, the total momentum of the system of blocks is not conserved. 5. FA,net is equal in magnitude and direction to FB,net , and not zero. 6. FB,net will be in the opposite direction and half as large as FA,net . Explanation: FA,net is the force on block A from the spring, it’s reaction force is the force acting on the spring by block A, which we denote as FS,A . They have the same magnitude but they are in opposite directions. Similarly, FB,net has a reaction force FS,B . The total force on the spring ! S,net = F ! S,A + F ! S,B F ! A,net − F ! B,net = −F " ! ! A,net + F ! B,net =− F = mspring · a =0 · a = 0. ! ! B,net must have the So that FA,net and F same magnitude but in opposite directions. 6. The momentum of block A is conserved and the momentum of block B is conserved. It follows that the total momentum of the system of blocks is also conserved. Explanation: Neither block A nor B conserves its momentum since there is a force on each of them. However, once we consider the two blocks and the spring as a whole system, its momentum is conserved since there is no external force. The contribution of momentum from the spring is zero since its mass is zero. 003 10.0 points A 62.8 kg astronaut is on a space walk when the tether line to the shuttle breaks. The astronaut is able to throw a 11.0 kg oxygen tank in a direction away from the shuttle with a speed of 13.7 m/s, propelling the astronaut back to the shuttle. Oslund (hno56) – Homework 6 – sai – (58015) Assuming that the astronaut starts from rest, find the final speed of the astronaut after throwing the tank. Correct answer: 2.39968 m/s. Explanation: Let toward the spacecraft be positive. Let : m1 = 62.8 kg , m2 = 11.0 kg , and v2,f = −13.7 m/s . The wrench has the same initial speed as the astronaut, and vi = 0 m/s, so (m1 + m2 ) !vi = m1 !vf,1 + m2 !vf,2 0 = m1 !vf,1 + m2 !vf,2 m2 vf,2 m1 (11 kg) (−13.7 m/s) =− 62.8 kg vf,1 = − = 2.39968 m/s . 004 10.0 points A spring is compressed between two cars on a frictionless airtrack. Car A has four times the mass of car B, MA = 4 MB , while the spring’s mass is negligible. Both cars are initially at rest. When the spring is released, it pushes them away from each other. Which of the following statements correctly describes the velocities, the momenta, and the kinetic energies of the two cars after the spring is released? Note: Velocities and momenta are given below as vectors. 1 1. !vA = − !vB 3 2 !pA = − !pB 3 4 KA = KB 3 2. !vA = −4 !vB !pA = −16!pB KA = 64 KB 1 3. !vA = − !vB 4 2 !pA = −!pB 1 KA = KB correct 4 4. !vA = −!vB !pA = −!pB KA = KB 1 5. !vA = + !vB 5 4 !pA = + !pB 5 4 KB KA = 25 1 6. !vA = + !vB 4 !pA = +!pB KA = 4 KB 1 7. !vA = − !vB 4 !pA = −!pB KA = 4 KB 8. !vA = −!vB !pA = −4!pB KA = 16 KB 9. !vA = −2 !vB !pA = −8!pB KA = 16 KB 1 10. !vA = − !vB 2 !pA = −2!pB KA = KB Explanation: Let : MA = 4 MB . There are no external forces acting on the cars, so their net momentum !pA + !pB is conserved. The initial net momentum is obviously zero, hence after the spring is released, !pA + !pB = 0; i.e., !pA = −!pB . The velocities and the kinetic energies follow from the momenta: !v = !p . M Oslund (hno56) – Homework 6 – sai – (58015) So given MA = 4 MB , it follows that 1 !vA = − !vB . 4 Likewise, K= !p2 1 M !v 2 = . 2 2M Since !pA = −!pB , KA = 3 Explanation: Solution: Denote the smaller mass initially moving to the right as m and the larger mass initially moving to the left as 4m. By assumption of the problem, before the collision !vm = +v ı̂ and !v4m = −v ı̂ where ı̂ is the unit vector in the positive x direction. Hence, the particle’s initial momenta are respectively 1 KB . 4 !pm = +m v ı̂ 005 (part 1 of 2) 10.0 points Two particles of masses m and 4 m are moving toward each other along the x-axis with the same speed v. They undergo a headon elastic collision and rebound along the xaxis. v v and !p4m = −4 m v ı̂, and the total momentum of the two particles is !ptotal = !pm + !p4m = −3 m v ı̂. In magnitude, "!ptotal " = 3 m v . m 4m Determine the magnitude of the total momentum of the system "!ptotal " at the instant when the two particles are touching each other; i.e., at the moment of collision. 006 (part 2 of 2) 10.0 points Determine the final speed v4m of the heavier object. 1. v4m = 2 v 3 1. "!ptotal " = 4 m v 2. v4m = 2 v 2. "!ptotal " = 5 m v 3. v4m = v 3. "!ptotal " = ∞ 4. v4m = 4. "!ptotal " = 0 5. v4m = 4 v 5. "!ptotal " = 2 m v 6. v4m = 6. "!ptotal " = 3 m v correct 7. v4m = 3 v 7. "!ptotal " = 8 m v 8. v4m = ∞ 8. "!ptotal " = m v 9. "!ptotal " = 6 m v 10. "!ptotal " = 7 m v 9. v4m = 1 v 3 3 v 2 1 v correct 5 10. v4m = 0 Explanation: Oslund (hno56) – Homework 6 – sai – (58015) From the relative velocities formula, ! ! vm − v4m = v4m − vm . (1) From conservation of momentum, ! ! m vm + 4 m v4m = m vm + 4 m v4m . (2) Combining equations (1) and (2) gives m vm + 4 m v4m ! ! = m (v4m + v4m − vm ) + 4 m v4m , or m vm + 4 m v4m − m (v4m − vm ) ! ! = m v4m + 4 m v4m , so ! v4m m vm + 4 m v4m − m (v4m − vm ) = m +4m m v + 4 m (−v) − m (−v − v)) = m+ 4m 1 = − v . 5 Alternative Solution: The compact form of the head-on elastic collision formula, can be derived from coordinate transformations between the lab frame and the center-of-mass frame, since What is the kinetic energy of the second ball after the collision? Correct answer: 0.196967 J. Explanation: Basic Concepts For an elastic, head-on collision, both the linear momentum and the kinetic energy are conserved. The balls have the same mass m and v2 = 0, so by conservation of momentum, m1 v1 + m2 v2 = m1 v1! + m2 v2! m v1 = m v1! + m v2! v1 = v1! + v2! v1! = v1 − v2! . By conservation of energy, 1 1 1 1 m1 v12 + m2 v22 = m1 v1!2 + m2 v2!2 2 2 2 2 1 1 1 m v12 = m v1!2 + m v2!2 2 2 2 v12 = v1!2 + v2!2 v12 = (v1 − v2! )2 + v2!2 ! v4m = 2 vcm − v4m v12 = v12 − 2 v1 v2! + v2!2 + v2!2 0 = −2v1 v !2 + 2 v2!2 Part one implies that vcm ptotal = m+ 4m 3mv =− 5m 3 = − v. 5 This gives 4 0 = −v1 + v2! v2! = v1 and v1! = v1 − v2! = 0 . Thus 1 m v2!2 2 1 = m v12 2 $ # 1 = (0.842 kg) (0.684 m/s)2 2 = 0.196967 J . K2! = ! v4m = 2 vcm − v4m $ # 3 = 2 − v − (−v) 5 1 = − v . 5 007 (part 1 of 2) 10.0 points A billiard ball of mass 0.842 kg and speed 0.684 m/s makes a head-on elastic collision with an identical billiard ball at rest. 008 (part 2 of 2) 10.0 points What is the magnitude of its linear momentum? Correct answer: 0.575928 kg m/s. Oslund (hno56) – Homework 6 – sai – (58015) Explanation: Since 5 Solution: p!2 = m v2! = m v1 = (0.842 kg) (0.684 m/s) = 0.575928 kg m/s . 009 (part 1 of 2) 10.0 points A 42.1 kg student runs down the sidewalk and jumps with a horizontal speed of 4.03 m/s onto a stationary skateboard. The student and skateboard move down the sidewalk with a speed of 3.81 m/s. a) Find the mass of the skateboard. Correct answer: 2.43097 kg. Explanation: Basic Concept: mb!vi,b + ms!vi,s = (mb + ms )!vf mb!vi,b = (mb + ms )!vf since vi,s = 0 m/s. Given: Let to the right be positive: (mb + ms )vf mb (42.1 kg + 2.43097 kg)(6.52 m/s) = 42.1 kg = 6.89648 m/s vi,b = 011 10.0 points Use 5.98 × 1024 kg as the mass of the Earth. A 1080 kg meteorite has a speed of 147 m/s just before colliding head-on with the Earth. Determine the recoil speed of the Earth. Correct answer: 2.65485 × 10−20 m/s. Explanation: From conservation of momentum ∆p = 0 m v + (m + Me ) V = 0 . Therefore mv [M + m] (1080 kg) (147 m/s) = [(5.98 × 1024 kg) + (1080 kg)] = 2.65485 × 10−20 m/s . !"= "V mb = 42.1 kg vi,b = 4.03 m/s vf = 3.81 m/s Solution: mb vi,b = mb vf + ms vf mb (vi,b − vf ) ms = vf (42.1 kg)(4.03 m/s − 3.81 m/s) = 3.81 m/s = 2.43097 kg 012 10.0 points A(n) 5 g bullet is fired into a 255 g block that is initially at rest at the edge of a frictionless table of height 1.4 m. The bullet remains in the block, and after impact the block lands 2.5 m from the bottom of the table. The acceleration of gravity is 9.8 m/s2 . 1.4 m 010 (part 2 of 2) 10.0 points b) How fast would the student have to jump to have a final speed of 6.52 m/s? Correct answer: 6.89648 m/s. 2.5 m Find the initial speed of the bullet. Correct answer: 243.208 m/s. Explanation: Given: vf = 6.52 m/s Explanation: Oslund (hno56) – Homework 6 – sai – (58015) Given : m1 = 5 g , m2 = 255 g , ∆y = −1.4 m , ∆x = 2.5 m . Explanation: Given : m1 v1 m2 v2 m3 v3 and The bullet-block combination is a projectile, with the vertical motion defining the time: 1 ∆y = − g (∆t)2 %2 −2 ∆y ∆t = g m1 v1 + m2 v2 + m3 v3 = (m1 + m2 + m3 ) v ! Since m! = m1 + m2 + m3 = 6 kg + 12 kg + 7 kg = 25 kg , ! = ∆x g −2 ∆y Applying conservation of momentum to the collision, m1 v1 = (m1 + m2 ) v ! (m1 + m2 ) v ! v1 = m1 & (m1 + m2 ) ∆x g = m1 −2∆y % 9.8 m/s2 (5 g + 255 g) (2.5 m) = 5g −2 (−1.4 m) = 243.208 m/s . 013 10.0 points Three carts of masses 6 kg, 12 kg, and 7 kg move on a frictionless horizontal track with speeds of 8 m/s, 2 m/s, and 2 m/s, as shown in figure. The carts stick together after colliding. 8 m/s 6 kg = 6 kg , = 8 m/s , = 12 kg , = 2 m/s , = 7 kg , and = −2 m/s . From conservation of momentum Horizontally, the constant velocity is v ! from just after the collision, so ∆x = v ∆t ∆x v! = ∆t & 6 2 m/s 2 m/s 12 kg 7 kg we find that m1 v1 + m2 v2 + m3 v3 v! = m1 + m2 + m3 (6 kg) (8 m/s) (12 kg) (2 m/s) = + 25 kg 25 kg (7 kg) (−2 m/s) + 25 kg = 2.32 m/s . toward the right. 014 (part 1 of 3) 10.0 points Two small spheres of mass 417 g and 601 g are suspended from the ceiling at the same point by massless strings of equal length 10.7 m . The lighter sphere is pulled aside through an angle of 48◦ from the vertical and let go. The acceleration of gravity is 9.8 m/s2 . 417 g . 10 7 m 48◦ Find the final velocity of the three carts. Correct answer: 2.32 m/s. 9.8 m/s2 Before 601 g Oslund (hno56) – Homework 6 – sai – (58015) 7 After lighter sphere is let go and collides with the heavier sphere at the bottom of its swing, the two spheres immediately bind together. What is the speed of the combined system just after the collision? Correct answer: 3.41222 m/s. θf After 9.8 m/s2 At what speed will the lighter mass m1 hit the heavier mass m2 ? Correct answer: 8.33006 m/s. Explanation: Let : m1 = 417 g , m2 = 601 g , θi = 48◦ , # = 10.7 m , vi = before collision , Vf = after collision , and g = 9.8 m/s2 . The velocity just before the collision vi can be determined by energy conservation. When particle 1 is at its initial condition, it is at rest and displaced by an angle θi from the vertical. The total energy is all potential and is given by Ui = m1 g # (1 − cos θi ) where # (1 − cos θi ) is the distance above the lowest point. Just before the collision, the en1 2 ergy of sphere 1 is all kinetic energy, m1 v1i . 2 Equating the two energies gives 1 2 m1 v1i = m1 g # (1 − cos θi ) . 2 Solving for V1i gives v1i = {2 g # [1 − cos θi ]}1/2 ' = 2 (9.8 m/s2 ) (10.7 m) × [1 − cos(48◦ )]}1/2 = 8.33006 m/s . 015 (part 2 of 3) 10.0 points Explanation: This is a perfectly inelastic collision. The speed of the two spheres after collision is determined by momentum conservation m1 v1i = (m1 + m2 ) Vf (1) where m1 is the mass and v1i is the initial velocity of sphere 1 just before the collision, m2 is the mass of sphere 2, and Vf is the velocity of the combined spheres just after the collision. Note: v2i = 0. Now, calculating Vf of the composite system from Eq. (1), m1 Vf = v1i (2) m1 + m2 (417 g) (8.33006 m/s) = (417 g) + (601 g) = 3.41222 m/s . 016 (part 3 of 3) 10.0 points What is the maximum angle of deflection of the two bound objects? Correct answer: 19.1816◦ . Explanation: The composite system has an initial energy 1 (m1 + m2 ) Vf2 2 ( )2 m1 1 2 v1i = (m1 + m2 ) 2 m1 + m2 m21 [2 g # (1 − cos θi )] . = 2 (m1 + m2 ) The composite system comes to rest at an angular displacement θf , where the potential energy equals the initial kinetic energy of the composite system, (m1 +m2 ) g # (1 − cos θf ) 1 = (m1 + m2 ) Vf2 2 m21 [2 g # (1 − cos θi )] . = 2(m1 + m2 ) Oslund (hno56) – Homework 6 – sai – (58015) Therefore cos θf = 1 − # m1 m1 + m2 $2 Solving for θf we have * # θf = cos−1 1 − m1 m1 + m2 [1 − cos θi ] $2 [(− cos θi ) + ( 1 − (0.409627)2 ) ◦ × (1 − cos 48 ) = cos −1 = 19.1816◦ . Alternative Solution: From conservation of energy after the collision, we have 8 1 1 3. v1! = v1 , v2! = v1 2 2 1 3 4. v1! = − v1 , v2! = v1 2 2 1 1 5. v1! = − v1 , v2! = v1 2 2 6. v1! = 0, v2! = v1 correct 7. v1! = v1 , v2! = −v1 8. 0 < v2! < v1! , v1! < 2v2! 9. v1! = −2v1 , v2! = 0 10. v1! = 2 v1 , v2! = −v1 1 (m1 + m2 ) g h = (m1 + m2 ) Vf2 , 2 Explanation: Basic Concepts: Conservation of Energy, where v2 = 0: where Vf has been determined in Eq. (2). Therefore 1 1 1 2 2 m1 v1 2 = m1 v1! + m2 v2! 2 2 2 2 1 Vf h= 2 g 1 (3.41222 m/s)2 = 2 (9.8 m/s2 ) = 0.594042 m . #−h Since cos θ = , we have # $ # −1 # − h θ = cos # # $ −1 (10.7 m) − (0.594042 m) = cos (10.7 m) ◦ = 19.1816 . 017 10.0 points You are playing pool. You hit the cue ball with velocity v1 dead center on another ball. All balls including the cue ball have the same mass. What is the final velocity of the cue ball v1! and the final velocity of the other ball v2! ? 1. v1! = v1 , v2! = 0 2. v1! = 0, v2! = 0 Conservation of Momentum, where v2 = 0: m1 v1 = m1 v1! + m2 v2! Solving for v1! and v2! yields: # $ m1 − m2 ! v1 = v1 m1 + m2 # $ 2 m1 ! v2 = v1 m1 + m2 (1) (2) All solutions can be determined using the above equations (1) and (2). Part 1: Using eqn (1), where m2 = m1 = m, # $ m−m ! v1 = v1 = 0 m+m Part 2: Using eqn (2), where m2 = m1 = m, $ # 2m ! v1 = +v1 v2 = m+m 018 (part 1 of 2) 10.0 points A 1790 kg car skidding due north on a level frictionless icy road at 212.52 km/h collides with a 3168.3 kg car skidding due east at Oslund (hno56) – Homework 6 – sai – (58015) # 3 $# $ 138 km/h in such a way that the two cars 10 m 1h × stick together. km 3600 s = 1.21451 × 105 kg m/s , and θ pf y = m1 v1 vf 138 km/h = (1790 kg) (212.52 km/h) # 3 $# $ N 10 m 1h 3168.3 kg × km 3600 s 5 212.52 km/h 1790 kg = 1.0567 × 10 kg m/s . At what angle (−180◦ ≤ θ ≤ +180◦ ) East of North do the two coupled cars skid off at? Correct answer: 48.9749◦. Explanation: Let : m1 v1 m2 v2 = 1790 kg , = 212.52 km/h , = 3168.3 kg , and = 138 km/h . θ pf = mf vf p2 N m2 v2 p1 m1 v1 Basic Concepts: Momentum Conserva1 tion, K = mv 2 , !p = m!v . 2 During the collision, the total momentum of the two car system will be conserved !pf = !p1 + !p2 = m1 !v1 + m2 !v2 pf x ı̂ + pf y ̂ = m1 v1 ̂ + m2 v2 ı̂ . Looking at the x and y components of momentum, pf x = m2 v2 = (3168.3 kg) (138 km/h) 9 Since we are asked to find the angle from the y-axis instead of the x, $ pf x tan θ = pf y $ # 1.21451 × 105 kg m/s = 1.0567 × 105 kg m/s = 1.14935# $ 1.21451 × 105 kg m/s θ = arctan 1.0567 × 105 kg m/s = 48.9749◦ . # 019 (part 2 of 2) 10.0 points How much kinetic energy is lost in the collision? Correct answer: 2.83339 × 106 J. Explanation: To find the energy lost in the collision, we need to know the total kinetic energy of the system before and after the collision. (There is no change in potential energy since the cars remain on level ground.) The initial and final energies are given by Ei = Ki = 1 1 m1 v12 + m2 v22 2 2 1 (1790 kg) (59.0333 m/s)2 2 1 + (3168.3 kg) (38.3333 m/s)2 2 = 5.44684 × 106 J , and Ef = Kf 1 = (m1 + m2 ) vf2 . 2 = Oslund (hno56) – Homework 6 – sai – (58015) To find vf , we need to go back to the momentum equation; !pf , m1 + m2 or in component form, pf x vf x = , m1 + m2 pf y vf y = . m1 + m2 Explanation: Let : M = 7 kg , v = 9.63 m/s , θ = 33◦ . !vf = and and The y component of the momentum is unchanged. The x component of the momentum is changed by Now that we know the components of !vf , we can find its magnitude ∆Px = −2 M v cos θ . Therefore, using impulse formula, vf2 = vf2x + vf2y . We can substitute this into the equation for the final kinetic energy to get 2 10 2 1 pf x + pf y 2 m1 + m2 = 2.61345 × 106 J . Ef = ∆P ∆t −2 M v cos θ = ∆t −2 (7 kg) (9.63 m/s) cos 33◦ = 0.318 s ! " = 355.565 N . "F F = The energy lost is then Note: The direction of the force is in negative x direction, as indicated by the minus sign. Elost = −∆E = Ei − Ef = (5.44684 × 106 J) − (2.61345 × 106 J) = 2.83339 × 106 J . 020 10.0 points A 7 kg steel ball strikes a wall with a speed of 9.63 m/s at an angle of 33◦ with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. 9. 6 3m y /s ◦ 7 kg 33 x 33◦ 7 kg 9. 6 /s 3m If the ball is in contact with the wall for 0.318 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 355.565 N. 021 10.0 points If you throw a raw egg against a wall, you’ll break it, but if you throw it with the same speed into a sagging sheet it won’t break. Why? 1. The sheet is much slicker than the wall. 2. The breaking egg causes a larger impact time, decreasing the force. 3. The impact time when the egg strikes a sagging sheet is long, so the impact force is small. correct 4. The velocity of the egg decreases faster in the sheet than on the wall. Explanation: Although the impulses may be the same for the two cases, the times of impact are not. When the egg strikes the wall, impact time is short and impact force is correspondingly large. But when the egg strikes the sagging Oslund (hno56) – Homework 6 – sai – (58015) sheet, impact time is long and the force correspondingly small. 022 10.0 points A(n) 4.3 lb hammer head, traveling at 5 ft/s strikes a nail and is brought to a stop in 0.00079 s. The acceleration of gravity is 32 ft/s2 . What force did the nail receive? Correct answer: 850.475 lb. Applying impulse, ! ∆t = m!vf − m!vi = m!vf F since vi = 0 m/s. m vf ∆t (0.33 kg) (12 m/s) = 0.21 s = 18.8571 N . F = Explanation: Let : W = 4.3 lb , vo = 5 ft/s , and t = 0.00079 s . Impulse is the change in momentum: F ∆t = m ∆v = m (vf − vo ) . 11 024 (part 1 of 3) 10.0 points The force shown in the force-time diagram acts on a 622 g object. Fx (N) The hammer’s final velocity is 0, so Ft= W (0 − vo ) g Thus the force the hammer felt was −W vo F = gt (it had to be a negative force to stop it) and by Newton’s Third Law of Motion, the force the nail received was F = W vo (4.3 lb) (5 ft/s) = gt (32 ft/s2 ) (0.00079 s) = 850.475 lb . 023 10.0 points A football punter accelerates a 0.33 kg football from rest to a speed of 12 m/s in 0.21 s. What constant force does the punter exert on the ball? Correct answer: 18.8571 N. Explanation: Given : m = 0.33 kg , vf = 12 m/s, and ∆t = 0.21 s . 2 1 t (s) 0 0 1 2 3 4 5 Find the impulse of the force. Correct answer: 2 N s. Explanation: The impulse is the area under the F versus t graph; i.e, the sum of the area of the rectangle plus the triangle, so Impulse = (1 N) (1 s) + 1 (1 N) (2 s) 2 = 2 Ns. 025 (part 2 of 3) 10.0 points Find the final velocity of the object if it is initially at rest. Correct answer: 3.21543 m/s. Explanation: Oslund (hno56) – Homework 6 – sai – (58015) Given : m = 622 g = 0.622 kg , vi = 0 m/s . and Impulse = ∆p F ∆t = m (vf − vi ) m vf = F ∆t + m vi F ∆t vf = + vi m (2 N s) + (0 m/s) = (622 g) = 3.21543 m/s . 026 (part 3 of 3) 10.0 points Find the final velocity of the object if it is initially moving along the x axis with average velocity of −15 m/s. Correct answer: −11.7846 m/s. Explanation: Given : vf = vi + vi = −15 m/s . F ∆t m (2 N s) (0.622 kg) = −11.7846 m/s . = (−15 m/s) + 12