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Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. Chapter 9 problems. 001 (part 1 of 1) 0 points An object of mass m is moving with speed v0 to the right on a horizontal frictionless surface, as shown, when it explodes into two 2 pieces. Subsequently, one piece of mass m 5 v0 to the left. moves with a speed v2/5 = 2 v0 m The speed k~v3/5 k of of the other piece the object is v0 . 2 v0 2. k~v3/5 k = . 3 7 v0 . 3. k~v3/5 k = 5 3 v0 4. k~v3/5 k = . 2 1. k~v3/5 k = 5. k~v3/5 k = 2 v0 . correct Explanation: The horizontal component of the momentum is conserved, so 2 3 m v2/5 + m v3/5 5 5 2 ³ v0 ´ 3 0 + m v0 = m − + m v3/5 5 2 5 1 3 mv0 = − m v0 + m v3/5 5 5 3 6 v = v0 5 3/5 5 k~v3/5 k = 2 v0 . 1 engine pulling the car. It begins to rain. The rain falls straight down and begins to fill the train car. The speed of the car 1. stays the same. 2. decreases. correct 3. increases. Explanation: Using Newton’s second law, we have dP = 0, dt since no external forces act on the train in the horizontal direction. With no rain, the train will move at a constant velocity; however, when it starts to rain, and the rain starts to fill the car, the mass of the train changes. Thus, dm dv m = −v . dt dt dm is positive; i.e., the mass of the Since dt train is increasing with accumulating rain, dv should be negative; i.e., the speed of the dt train should decrease. 003 (part 1 of 2) 5 points Consider the set up of a ballistic pendulum where m1 = 11.4 g, m2 = 3.97 kg, h = 0.243 m. The acceleration of gravity is 9.8 m/s2 . 0 + m v0 = 002 (part 1 of 1) 0 points An open train car moves with speed 19 m/s on a flat frictionless railroad track, with no m1+ m 2 v1 m1 m2 vf h Find the final velocity of the system (m1 + m2 ) immediately after the collision and before the pendulum starts to swing upwards. Correct answer: 2.18238 m/s. Explanation: From the setup, the final velocity of the collision is the initial velocity of the subsequent Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004 2 swing. During the swinging process the total energy is conserved after he catches the ball back. That is, when the man throws the ball away, Ei = E f M v1 = m v 1 (m1 + m2 ) vf2 = (m1 + m2 ) g h . 2 Therefore p 2gh q = 2 (9.8 m/s2 ) (0.243 m) vf = and when the man catches the ball bounced back from the wall, (M + m) V1 = M v1 + m v = 2 m v . Solving for V1 , we get, = 2.18238 m/s . V1 = 004 (part 2 of 2) 5 points Find v1 , the initial speed of m1 . Correct answer: 762.188 m/s. Explanation: The linear momentum is conserved in a collision pi = p f m1 v1 = (m1 + m2 ) vf Therefore m1 + m 2 vf m1 (0.0114 kg) + (3.97 kg) (2.18238 m/s) = (0.0114 kg) = 762.188 m/s . v1 = 005 (part 1 of 2) 0 points A 83 kg man holding a 0.336 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 15.7 m/s (relative to the ground) and then catches the ball after it rebounds from the wall. Hint: Ignore the projectile motion of the ball, and assume that it loses no energy in its collision with the wall. How fast is he moving after he catches the ball? Correct answer: 0.126601 m/s. Explanation: Let M and m be the mass of the man and the ball respectively. Let v be the speed of the ball. Using momentum conservation twice, we can get the speed the man moves 2mv M +m 006 (part 2 of 2) 0 points Note: Remember that the ball is thrown with a speed of 15.7 m/s relative to the ground each time it is thrown. How many times does the man have to go through this process before his speed reaches at least 2.7 m/s relative to the ground? Correct answer: 21 . Explanation: If he repeats this n times, he’ll move at a speed 2mnv Vn = . M +m If he is to reach V , he should repeat this µ ¶ M +m n=V 2mv times. 007 (part 1 of 4) 3 points A massless spring with force constant k =441 N/m is fastened at its left end to a vertical wall, as shown below. The acceleration of gravity is 9.8 m/s2 . k MC MD C D Initially, block C (mass mC =3.57 kg) and block D (mass mD =2.86 kg) rest on a horizontal surface with block C in contact with the spring (but not compressing it) and with block D in contact with block C. Block C is then moved to the left, compressing the spring Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004 a distance of 0.4 m, and held in place while block D remains at rest as shown in below. d C µ D Determine the elastic energy U stored in the compressed spring. Correct answer: 35.28 J. Explanation: Should use the expression for the energy stored in a spring: 1 U = k x2 2 1 = (441 N/m) (0.4 m)2 2 = 35.28 J . 008 (part 2 of 4) 3 points Block C is then released and accelerates to the right, toward block D. The surface is rough and the coefficient of friction between each block and the surface is µ =0.421. The two blocks collide instantaneously, stick together, and move to the right. Remember that the spring is not attached to block C. Find the speed vC of block C just before it collides with block D. Correct answer: 4.05759 m/s. Explanation: Apply conservation of energy or workenergy theorem to this problem. Kinetic en1 ergy of block C is K = mC vC2 ; work done (or 2 energy dissipated by) friction is Wf = µ Fn d. Therefore we have 1 mC vC2 = U − µ mC g d . 2 Solving for vC s 2 (U − µ mC g d) vC = mC = 4.05759 m/s . Correct alternate Z solution Computing Fnet dx, where Fnet = k x − µ Fn , to find the kinetic energy, and then computing the speed. 3 009 (part 3 of 4) 2 points Find the speed vf of blocks C and D just after they collide. Correct answer: 2.25282 m/s. Explanation: In this configuration, we have the conservation of momentum: mC vC = (mC + mD ) vf . Solving for vf mC vC mC + m D = 2.25282 m/s . vf = 010 (part 4 of 4) 2 points Find the horizontal distance the blocks move before coming to rest. Correct answer: 0.615054 m. Explanation: The blocks come to rest when all their kinetic energy has been dissipated; i.e., ∆KE =Work done by frictional force 1 (mC + mD ) vf2 = µ (mC + mD ) g ` 2 Solving for ` `= vf2 2µg (2.25282 m/s)2 = 2 (0.421) (9.8 m/s2 ) = 0.615054 m . Alternate Solution: X F = ma P F a= m µ (mC + mD ) g = (mC + mD ) = µg = 4.1258 m/s2 v 2 = v02 + 2 a ` v 2 − v02 `= 2a = 0.615054 m . Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004 m v 2 k x2 m vi2 + − 2 2 2 (0.0055 kg) (177.182 m/s)2 = 2 (1400 N/m) (0.012 m)2 + 2 (0.0055 kg) (315 m/s)2 − 2 = (86.3323 J) + (0.1008 J) − (272.869 J) = −186.436 J |∆E| = 186.436 J . = 011 (part 1 of 2) 0 points A bullet of mass 5.5 g moving with an initial speed 315 m/s is fired into and passes through a block of mass 2.85 kg, as shown in the figure. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 1400 N/m. v If the block moves a distance 1.2 cm to the right after the bullet passed through it, find the speed v at which the bullet emerges from the block. Correct answer: 177.182 m/s. Explanation: From conservation of energy 1 1 M V02 = k x2 . 2 2 4 013 (part 1 of 2) 0 points Consider the general case of collisions of two masses m1 = m with m2 = 2m along a frictionless horizontal surface. Denote the initial and the final center of mass momenta to be picm = p1 + p2 , and pfcm = p01 + p02 . And the initial and final kinetic energies to be Hence, V0 = r k x2 = 0.265964 m/s . M Then from momentum conservation m v0 = M V0 + m v , or M V0 v = v0 − m (2.85 kg) (0.265964 m/s) = (315 m/s) − (5.5 g) (0.001 kg/g) = 177.182 m/s . Ki = K 1 + K 2 , and Kf = K10 + K20 . v1 m1 v2 m2 For an elastic collision which pair of statements is correct? 1. picm < pfcm , Ki < Kf 2. picm > pfcm , Ki = Kf 012 (part 2 of 2) 0 points Find the magnitude of the energy lost in the collision. Correct answer: 186.436 J. Explanation: The energy lost is ∆E = Kf + Uf − Ki 3. picm > pfcm , Ki > Kf 4. picm = pfcm , Ki > Kf 5. picm > pfcm , Ki < Kf 6. picm = pfcm , Ki = Kf correct Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004 7. picm < pfcm , with the magnitude of the momentum conserved. Note: In the CM frame the total momentum is still zero if each velocity changes sign. Hence, the final velocity of m2 in the CM frame is Ki = Kf 8. picm < pfcm , Ki > Kf 9. picm = pfcm , Ki < Kf Explanation: Momentum and energy are always conserved in elastic collision. 014 (part 2 of 2) 0 points Now consider an elastic head-on collision again with initial velocity of m1 to be v1 and of m2 to be v2 = 0. Find the final speed v2f of m2 . 1. v2f = 1 v1 3 (CM ) v2,f (CM ) lab = v2,f + VCM = 2 VCM v2,f 2m1 v1 2 = = v1 . m1 + m 2 3 v2,f = 2vcm − v2,i . 3 v1 4 2 4. v2f = v1 7 3 5. v2f = v1 5 1 6. v2f = v1 2 v1 For the present case,v2,i = 0, vcm = , so 3 2 v2,f = v1 3 7. v2f = 2 v1 1 v1 5 2 9. v2f = v1 correct 3 1 10. v2f = v1 4 Explanation: The CM velocity is given by P mi vi m1 v1 = VCM = Pi m1 + m 2 i mi 8. v2f = as v2 = 0. Thus, in the CM frame, m2 has a velocity = v2 − VCM = −VCM = m1 v1 m1 + m 2 To sum up, we have the following general expression, 3. v2f = (CM ) = VCM = lab The velocity v2,f (v2f ) in the lab frame is 2. v2f = v1 v2 5 −m1 v1 . m1 + m 2 After a one dimensional elastic collision, each body changes the direction of its momentum 015 (part 1 of 3) 0 points Two balls with masses m1 and m2 are on the x-axis. Ball m1 has an initial velocity v1 > 0 along the positive x-axis and ball m2 is initially at rest. The balls collide elastically and remain on the x-axis after the collision. If m1 = m2 , what is the final velocity v10 of the ball m1 ? 1. v10 = +v1 2. +2 v1 < v10 < ∞ 3. v10 = −v1 4. v10 = 0 correct 5. v10 = +2 v1 Explanation: Basic Concepts: Conservation of Energy, where v2 = 0: 1 1 1 2 2 m1 v1 2 = m1 v10 + m2 v20 2 2 2 Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004 Conservation of Momentum, where v2 = 0: m1 v1 = Solving for v10 m1 v10 + m2 v20 3. +2 v1 < v10 < ∞ 4. v10 = −v1 correct v20 and yields: µ ¶ m1 − m 2 0 v1 = v1 m1 + m 2 ¶ µ 2 m1 0 v1 v2 = m1 + m 2 5. v10 = +2 v1 (1) (2) All solutions can be determined using the above equations (1) and (2). Think of playing pool. You hit the cue ball with velocity v1 dead center on the 8 ball v2 = 0. The cue ball stops and the 8 ball proceeds with the same velocity as the cue ball originally had. Part 1: Using eqn (1), where m2 = m1 = m, ¶ µ m−m 0 v1 = 0 v1 = m+m Part 2: Using eqn (2), where m2 = m1 = m, µ ¶ 2m 0 v2 = v1 = +v1 m+m 016 (part 2 of 3) 0 points 0 If m1 = m2 , what is the final velocity v2 of the ball m2 ? 1. v20 = −v1 2. +2 v1 < v20 < ∞ 3. v20 = +2 v1 4. v20 = 0 5. v20 = +v1 correct Explanation: Explanation: Think of throwing a golf ball with velocity v1 against a concrete wall v2 = 0. The golf ball hits the wall, bounces back (reverses direction) and has the same speed as it originally had (elastic collision). Part 3: Using m1 → 0, eqn (1), where 1 À m2 ! Ãm 1 − 1 2 v1 = −v1 v10 = m m1 m2 + 1 Part 4: Using eqn (2), where 1 À v20 = 2. v10 = +v1 Ã m1 2m 2 m1 m2 +1 ! m1 → 0, m2 v1 = 0 018 (part 1 of 1) 0 points A 22.5 g bullet is fired horizontally into a 1 kg wooden block resting on a horizontal surface ( µ = 0.257 ). The bullet goes through the block and comes out with a speed of 186 m/s. The acceleration of gravity is 9.8 m/s2 . If the block travels 5.64 m before coming to rest, what was the initial speed of the bullet? Correct answer: 422.893 m/s. Explanation: Let m be the mass of the bullet and M the mass of the block. From momentum conservation, we have, mv1 = mv + M v2 From energy conservation, we have, 017 (part 3 of 3) 0 points In the limit, when m1 ¿ m2 , what is the final velocity v10 of the ball m1 ? 1. v10 = 0 6 Thus we have, 1 M v22 = µM gd 2 p v2 = 2µgd v1 = v + Mp M v2 = v + 2µgd m m Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004 11 m/s 60 ◦ 87 kg 524 kg 9.8 m/s2 What is the speed of the two blocks immediately after the collision? Correct answer: 9.43372 m/s. Explanation: 9.8 m/s2 611 kg h 019 (part 1 of 2) 5 points A(n) 524 kg mass is sliding on a horizontal frictionless surface with a speed of 11 m/s when it collides with a 87 kg mass initially at rest, as shown in the figure. The masses stick together and slide up a frictionless track at 60◦ from horizontal. The acceleration of gravity is 9.8 m/s2 . To what maximum height h above the horizontal surface will the masses slide? Correct answer: 4.54056 m. Explanation: The kinetic energy K of the combined system is K= m21 1 2 (m1 + m2 ) v12 = v2 2 2 (m1 + m2 ) 1 At the maximum height, all the kinetic energy is converted to potential energy, and so (m1 + m2 ) g h = K = Let : v12 m1 m2 v1 g = speed of blocks after collision = 524 kg , = 87 kg , = 11 m/s , and = 9.8 m/s2 . Therefore, the speed of the combined masses is m1 v1 m1 + m 2 (524 kg) = (11 m/s) (524 kg) + (87 kg) = 9.43372 m/s . v12 = m21 v2 2 (m1 + m2 ) 1 Solving for h, we get · ¸2 v12 m1 h= 2 g m1 + m 2 (11 m/s)2 = 2 (9.8 m/s2 ) · ¸2 (524 kg) × (524 kg) + (87 kg) = 4.54056 m . The initial momentum of block 1 is m1 v1 . When m1 sticks with m2 , the resulting momentum p12 is equal to that initial momentum; thus, p12 ≡ (m1 + m2 ) v12 = m1 v1 . 021 (part 1 of 2) 0 points A 30-06 caliber hunting rifle fires a bullet of mass 0.0158 kg with a velocity of 284 m/s to the right. The rifle has a mass of 6.24 kg. What is the recoil speed of the rifle as the bullet leaves the rifle? Correct answer: 0.719103 m/s. Explanation: From conservation of momentum ∆p = 0 0 = ∆pb + ∆pr or 020 (part 2 of 2) 5 points 7 0 = m b vb + m r vr . Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004 Hence mb vb mr (0.0158 kg) (284 m/s) =− (6.24 kg) = 0.719103 m/s . V =− 022 (part 2 of 2) 0 points If the rifle is stopped by the hunter’s shoulder in a distance of 3.88 cm, what is the magnitude of the average force exerted on the shoulder by the rifle? Correct answer: 41.5819 N. Explanation: The average force is Fav = ∆pr . ∆t The time of contact is ∆t = 2∆s = 0.107912 s , vr since the average velocity from the guns initial velocity (v = vr ) to when it stops (v = 0) is vr v = assuming constant de-acceleration. 2 Then the average force is Fav mr vr = ∆t (6.24 kg) (0.719103 m/s) = (0.107912 s) = 41.5819 N . 023 (part 1 of 2) 5 points Given: G = 6.67259 × 10−11 N m2 /kg2 Two hypothetical planets of masses 3.8 × 23 10 kg and 6.5 × 1023 kg and radii 3.6 × 106 m and 5.9 × 106 m, respectively, are at rest when they are an infinite distance apart. Because of their gravitational attraction, they head toward each other on a collision course. When their center-to-center separation is 6.5 × 108 m, find their relative velocity. Correct answer: 459.459 m/s. Explanation: 8 At infinite separation the potential energy U is zero, and at rest the kinetic energy K is zero. Since energy is conserved we have 0= 1 1 G m 1 m2 m1 v12 + m2 v22 − . 2 2 d The initial momentum is zero and momentum is conserved, so 0 = m 1 v1 − m 2 v2 . Combine these two equations to find s 2G v1 = m 2 d (m1 + m2 ) = 6.5 × 1023 kg s 2 (6.67259 × 10−11 N m2 /kg2 ) × (6.5 × 108 m) (1.03 × 1024 kg) = 289.95 m/s s 2G v2 = m 1 d (m1 + m2 ) = 3.8 × 1023 kg s 2 (6.67259 × 10−11 N m2 /kg2 ) × (6.5 × 108 m) (1.03 × 1024 kg) = 169.509 m/s . where m1 + m2 = (3.8 × 1023 (6.5 × 1023 kg) = 1.03 × 1024 kg. The relative velocity is kg) + vr = v1 − (−v2 ) = v1 + v2 = 289.95 m/s + 169.509 m/s = 459.459 m/s . 024 (part 2 of 2) 5 points Find the total kinetic energy of the planets just before they collide. Hint: Both energy and momentum are conserved. Correct answer: 1.73487 × 1030 J. Explanation: From conservation of energy the total kinetic energy of the system just before the Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004 planets collide is the opposite of the gravitational potential energy calculated for a distance d = r1 + r2 : G m 1 m2 r1 + r 2 = (6.67259 × 10−11 N m2 /kg2 ) (3.8 × 1023 kg) (6.5 × 1023 kg) × (3.6 × 106 m) + (5.9 × 106 m) = 1.73487 × 1030 J . Ktot = 025 (part 1 of 2) 0 points For this problem, which consists of 2 parts, we assume that we are on Planet-I. The radius of this planet is R =4290 km, the gravitational acceleration at the surface is gI =2.48 m/s2 , and the gravitational constant G = 6.67259 × 10−11 N m2 /kg2 . The mass of Planet-I is not given. Not all the quantities given here will be used. Suppose a rocket of mass m =5900 kg is projected vertically upward from the surface of this planet. It stops at a point h =20935.2 km from the surface of the planet. Caution: Here the gravitational acceleration decreases as the rocket travels away from Planet-I. Determine the kinetic energy (in Joules) of the rocket immediately after it is fired off. Correct answer: 5.20959 × 1010 J. Explanation: For r > R, the potential energy of the GM m GM rocket is U = − . Since gI = , r R2 m gI R 2 U = − . Apply conservation of enr ergy: E = Ki + Ui = Kf + Uf ⇒ Ki = Uf − Ui = −m gI R2 /r − (−m gI R) = mgI R(1 − 1/n). 026 (part 2 of 2) 0 points Find the minimum initial velocity of the rocket such that it will escape the gravitational field of Planet-I. Note: You can work on this part even if you did not get Part 1. Correct answer: 4612.85 m/s. Explanation: 9 For the minimum velocity case, the final kinetic energy of the rocket is zero. Since at the end the rocket will be infinitely far away from Planet-I, the final potential energy of the rocket is also zero. Conservation of energy implies Ki + Ui = Kf + Uf = 0. Thus Mm Ki = 0 − U i = G = m gI R R p ⇒ vi = 2 g R