Physics Homework Answer Key

1 This print-out should have 40 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points The Joule and the kilowatt-hour are both units of energy. 19.2 kW · h is equivalent to how many Joules? Correct answer: 455.786 W. Explanation: Let : F = 709 N , d = 9 m , and t = 14 s . The power expended is Correct answer: 6.912 × 107 J. F d W = t t (709 N) (9 m) = 14 s P = Explanation: 1 W = 1 J/ s 1000 W 3600 s · kW 1 h = 6.912 × 107 W · s 19.2 kW · h = (19.2 kW · h) · = 455.786 W . = 6.912 × 107 J . 004 10.0points You leave your 75 W portable color TV on for 7 hours each day and you do not pay attention 002 10.0points to the cost of electricity. A cheerleader lifts his 59.9 kg partner straight If the dorm (or your parents) charged you up off the ground a distance of 0.557 m before for your electricity use and the cost was releasing her. $0.1 /kW · h, what would be your monthly The acceleration of gravity is 9.8 m/s2 . (30 day) bill? If he does this 27 times, how much work has he done? Correct answer: 1.575 dollars. Correct answer: 8828.19 J. Explanation: The work done in lifting the cheerleader once is W1 = m g h = (59.9 kg)(9.8 m/s2 )(0.557 m) = 326.97 J . Explanation: Let : P = 75 W and t = 7 h/day , The energy consumed in each day is W =P t The work required to lift her n = 27 times is = (75 W) (7 h/day) · W = n W1 = (27)(326.97 J) = 8828.19 J. = 0.525 kW · h/day. kW 1000 W In 30 days, you would use 003 10.0points (30 day) (0.525 kW · h/day) = 15.75 kW · h, A student weighing 709 N climbs at constant speed to the top of an 9 m vertical rope in which would cost you 14 s. What is the average power expended by the (15.75 kW · h) ($0.1/kW · h) = $1.575 . student to overcome gravity? 2 Correct answer: 120.004 s. Explanation: The work against gravity which is necessary to raise the professor to the rim of the canyon is given by W = m g ∆h = (105 kg) (9.8 m/s2 ) (87 m) = 89523 J . If the horse’s power output is one horsepower, then the the time for the horse to provide the necessary work is given by W Poutput 89523 J = 746 W = 120.004 s . 5.heat Explanation: 007 10.0points A Joule is a measure of 1.density. 2.distance. 3.volume. 4.momentum. 5.energy. correct Explanation: 008 10.0points Shawn and his bike have a total mass of 58.7 kg. Shawn rides his bike 0.77 km in 13.9 min at a constant velocity. The acceleration of gravity is 9.8 m/s2 . What is Shawn’s kinetic energy? Correct answer: 25.0184 J. Explanation: Shawn’s velocity is v= tup = 1.chemical Thus his kinetic energy is K= = 1 mv2 2 1  006 10.0points Potential energy and kinetic energy are forms of what kind of energy? d . t (58.7 kg) = 25.0184 J . 2 13.9 0.77 min km · 1000 m 1 min · 1 km 60 s  005 10.0points A 105 kg physics professor has fallen into the Grand Canyon. Luckily, he managed to grab a branch and is now hanging 87 m below the rim. A student (majoring in linguistics and physics) decides to perform a rescue/experiment using a nearby horse. After lowering a rope to her fallen hero and attaching the other end to the horse, the student measures how long it takes for the horse to pull the fallen physicist to the rim of the Grand Canyon. The acceleration of gravity is 9.8 m/s2 . If the horse’s output power is truly 1 horsepower (746 W), and no energy is lost to friction, how long should the process take? 2 3.mechanical correct 009 10.0points Tim, with mass 71.6 kg, climbs a gymnasium rope a distance of 2 m. The acceleration of gravity is 9.8 m/s2 . How much potential energy does Tim gain? 4.electromagnetic Correct answer: 1403.36 J. 2.nuclear 3 Explanation: Let : m = 71.6 kg , h = 2 m , and g = 9.8 m/s2 . Potential energy is U = m gh = (71.6 kg) (9.8 m/s2 ) (2 m) = 1403.36 J . 010(part1of3)10.0points A 40.0 kg child is in a swing that is attached to ropes 2.30 m long. The acceleration of gravity is 9.81 m/s2 . Find the gravitational potential energy associated with the child relative to the child’s lowest position under the following conditions: a) when the ropes are horizontal. Correct answer: 902.52 J. Explanation: Given: θ = 36.0◦ Solution: The child is a distance ℓ cos θ from the attachment point, so h2 = ℓ − ℓ cos θ and Ug = mgh2 = mgℓ(1 − cos θ ) = (40 kg)(9.81 m/s2 )(2.3 m) · (1 − cos 36◦ ) = 172.366 J 012(part3of3)10.0points c) at the bottom of the circular arc. Correct answer: 0 J. Explanation: Solution: The child is at the lowest point, so h3 = 0, and Ug = mgh3 = (40 kg)(9.81 m/s2 )(0 m) = 0 J Explanation: BasicConcept: 013 Ug = mgh Given: ℓ = 2.30 m m = 40.0 kg g = 9.81 m/s2 Solution: The child is at a height h1 = ℓ from the lowest point, so Ug = mgh1 = (40 kg)(9.81 m/s2 )(2.3 m) = 902.52 J 011(part2of3)10.0points b) when the ropes make a 36.0◦ angle with the vertical. Correct answer: 172.366 J. 10.0points A block of mass m slides on a horizontal frictionless table with an initial speed v0 . It then compresses a spring of force constant k and is brought to rest. v k m m µ= 0 How much is the spring compressed x from its natural length? 1.x = v0 m k g correct r m k 2.x = vv02 0 3.x = 2g k 4.x = v0 g m 4 r k m m g k m 7.x = v0 k g Let : R = 3 m and h = 11.6 m . 5.x = v0 6.x = v0 0 + m gh = r m gk 9.x = vv0 2 0 10.x = 2m Explanation: Total energy is conserved (no friction). The spring is compressed by a distance x from its natural length, so 1 1 m v02 = Ei = Ef = k x2 , or 2 2 m 2 2 v , therefore x = k 0 r . mk x=v Anyone who checks 0to see if the units are correct should get this problem correct. 014 10.0points A bead slides without friction around a loopthe-loop. The bead is released from a height of 11.6 m from the bottom of the loop-theloop which has a radius 3 m. The acceleration of gravity is 9.8 m/s2 . A 3 m 11.6 m Therefore p v = 2 g (h − 2 R) r h i = 2 (9.8 m/s2 ) 11.6 m − 2 (3 m) = 10.4766 m/s . 015(part1of3)10.0points A block starts at rest and slides down a frictionless track. It leaves the track horizontally, flies through the air, and subsequently strikes the ground. 570 g v 2.4 m 8.x = v0 K i + Ui = K f + Uf m v2 + m g (2 R) 2 2 v = 2 g (h − 2 R) . km g 4.9 m s From conservation of energy, we have x What is the speed of the ball when it leaves the track? The acceleration of gravity is 9.81 m/s2 . Correct answer: 7.00357 m/s. Explanation: What is its speed at point A ? Correct answer: 10.4766 m/s. Explanation: Let : g = −9.81 m/s2 , m = 570 g , and h1 = 2.5 m . 5 h1 m 017(part3of3)10.0points What is the speed of the block when it hits the ground? h v h2 Correct answer: 9.805 m/s. Explanation: x Choose the point where the block leaves the track as the origin of the coordinate system. While on the ramp, K b = Ut 1 m vx2 = −m g h1 2 vx2 = −2 g h1 Let : Now choose ground level as the origin. Energy conservation gives us K f = Ui 1 m vf2 = −m g h 2 p vf = −2 g h q 2 −2 (−9.81 = = 9.805 m/s .m/s ) (4.9 m) p vx = −2 g h1 q 2 −2 (−9.81 = = 7.00357 m/sm/s . ) (2.5 m) 016(part2of3)10.0points What horizontal distance does the block travel in the air? Correct answer: 4.89898 m. AlternateSolution: p vy = −2 g h2 q = 6.86207 m/sm/s , 2 )so(2.4 m) −2 (−9.81 Explanation: q Let : h2 = −2.4 m . With the point of launch as the origin, h2 = t= s x = vx t = vx vf = x + v 2y v2 q = 9.805 m/s m/ . s)2 + (6.86207 m/s)2 (7.00357 1 g t2 2 s Thus h = 4.9 m . 2 h2g . 2 h2 g s = (7.00357 m/s) = 4.89898 m . 2 2−9.81 (−2.4m/s m) 018(part1of3)10.0points A block is pushed against the spring with spring constant 9.1 kN/m (located on the lefthand side of the track) and compresses the spring a distance 4.5 cm from its equilibrium position (as shown in the figure below). The block starts at rest, is accelerated by the compressed spring, and slides across a frictionless track (as shown in the figure below). It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.81 m/s2 . jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) 6 4.5 cm 9.81 m/s2 522 g 2.4 m Solution: Using Eqs. 1, 2, and 3, we have 9.1 kN/m v 1 1 m vx2 = k d2 2 2 vx2 = x What is the speed v of the block when it leaves the track? vx = Correct answer: 5.94153 m/s. Let : g = 9.81 m/s2 , (5) r (6) 2 k dm s 019(part2of3)10.0points What is the horizontal distance x the block travels in the air? m = 0.522 kg , k = 9100 N/m , vx = v , and Correct answer: 4.15608 m. d = 0.045 m . Explanation: At ∆y = h = −2.4 m (below the jump off height), m g k k d2 m 0.522 kg m)2 (9100 N/m) (0.045 = = 5.94153 m/s . Explanation: 5.94 m/s (4) h=− h Conservation of Me- Ui = Uf + K f , (1) x = vx t since vi = 0 m/s. r (2) 1 k d2 . (3) 2 Choosing the point where the block leaves the track as the origin of the coordinate system, Us = ∆ x = v x ∆t ∆y = − 1 2 g ∆t2 , since axi = 0 m/s2 and vyi = 0 m/s. = s k d2 m = −  1 m v2 K= 2 (7) s . (9) −2 hg t= Since x = vx t , therefore using Eq. 6 and 8, we have 4.16 m Basic Concepts: chanical Energy 1 g t2 2 = s −2 hg (10) 2 k dm2 hg 0.522 N/m) kg 2 (9100 − (0.045 m)2 (−2.4 m) × = 4.15608 9.81 m . m/s2 020(part3of3)10.0points  d 1/2 7 What is the total speed of the block when it hits the ground? Correct answer: 20.0938 m/s. Correct answer: 9.07688 m/s. Explanation: Explanation: BasicConcept: r K f = Ui , v= since vi = 0 m/s and hf = 0 m. 1 m vy2 = m g h , we have Solution: Using 2 p vy = −2 g h q = 6.86207 −2 (9.81m/s m/s, 2 ) so (−2.4 m) x q vf = v 2 q = 9.07688 (5.94153m/s m/.s)2 + (6.86207 m/s)2 (12) 021(part1of2)10.0points While skiing in Jackson, Wyoming, your friend Ben (of mass 72.9 kg) started his descent down the bunny run, 20.6 m above the bottom of the run. If he started at rest and converted all of his gravitational potential energy into kinetic energy, what is Ben’s kinetic energy at the bottom of the bunny run? Correct answer: 14717.1 J. and h = 20.6 m . Kbottom = Utop = m gh = (72.9 kg) (9.8 m/s2 ) (20.6 m) 022(part2of2)10.0points What is his final velocity? 023(part1of4)10.0points A 162 g particle is released from rest at point A along the diameter on the inside of a frictionless, hemispherical bowl of radius 47.5 cm. The acceleration of gravity is 9.8 m/s2 . Calculate its gravitational potential energy at point A relative to point B . Correct answer: 0.75411 J. Explanation: The potential energy at A is Explanation: = 14717.1 J . s 72.9 kgJ) 2 (14717.1 = = 20.0938 m/s . (11) + v 2y Let : m = 72.9 kg , g = 9.8 m/s2 , 2 Km U A = m g hA = (162 g)(0.001 kg/g)(9.8 m/s2 ) × (47.5 cm)(0.01 m/cm) = 0.75411 J . 024(part2of4)10.0points Calculate its kinetic energy at point B . Correct answer: 0.75411 J. Explanation: 8 The kinetic energy at B is equal to the potential energy at A, since all the energy went to kinetic energy. K B = UA = 0.75411 J . Basic Concepts: Conservation of Mechanical Energy Ui = K f since vi = 0 m/s and hf = 0 m. 1 mv2 2 Ug = mgh K= 025(part3of4)10.0points Calculate its kinetic energy at point C at 2R . height 3 hi = ℓ − ℓ cos θ Given: ℓ = 31.0 m θ = 36.0◦ g = 9.81 m/s2 Correct answer: 0.25137 J. Explanation: The kinetic energy at C is Solution: KC = m g (hA − hC ) 1 mv2f 2 vf2 = 2ghi = 2gℓ(1 − cos θ ) = 2(9.81 m/s2 )(31 m)(1 − cos 36◦ ) = 116.16 m2 /s2 mghi =   2R =m UAg R − 3 = 3 0.75411 J = 3 = 0.25137 J . 026(part4of4)10.0points Calculate its potential energy at point C . so that q 2 /s 116.16 vf = 10 .7777 mm/ s2 Correct answer: 0.50274 J. 028(part2of2)10.0points b) pushes off with a speed of 6.00 m/s? Explanation: The potential energy at C is Correct answer: 12.3353 m/s. UC = UA − K C = 0.75411 J − 0.25137 J = 0.50274 J . 027(part1of2)10.0points Tarzan swings on a 31.0 m long vine initially inclined at an angle of 36.0 ◦ with the vertical. The acceleration of gravity if 9.81 m/s2 . What is his speed at the bottom of the swing if he a) starts from rest? Correct answer: 10.7777 m/s. Explanation: Explanation: BasicConcept: Ui + K i = K f since hf = 0 m. Given: vi = 6.00 m/s Solution: 1 1 2 mv f = mghi + mv2 2 2 i 2 vf = 2gℓ(1 − cos θ) + vi 2 = 2(9.81 m/s2 )(31 m) · (1 − cos 36◦ ) + (6 m/s)2 = 152.16 m2 /s2 9 so that A block of mass m slids up a ramp making an angle θ with the horizontal. The block has 1 an initial KE of m v02 as it starts up the 2 ramp and travels a distance L along the ramp before coming to momentary rest. q 152.16 mm2//ss 2 vf = 12.3353 029 10.0points A 200 kg block is released at a 4.2 m height as shown. The track is frictionless. The block travels down the track, hits a spring of force constant k = 1623 N/m . The acceleration of gravity is 9.8 m/s2 . 200 kg 4.2 m L v0 How much work did kinetic friction do on the block between its starting point and the point it came to momentary rest? 1623 N/m 0   v2 1.m g L sin θ − correct 2. Zero, since friction is perpendicular to 2 mg x Determine the compression of the spring x from its equilibrium position before coming to rest momentarily. Correct answer: 3.18499 m. Explanation: 4.m g (sin θ − cos θ) 5.m g L Let : m = 200 kg , k = 1623 N/m , h = 4.2 m , and g = 9.8 m/s2 . From conservation of energy ∆E = Wf or 3.m g L tan θ 6.m g L sin θ 1 m v02, since the block comes to rest: 7. 2 W = ∆KE 8.m g L cos θ Explanation: k x2 −m gh = 0 . 2 L θ Hence, r x= If we take the gravitational potential energy to be zero at the point where the block has its initial speed, then from E = K + Ug 2 m gkh s N/m) 2 ) (4.2 m) 2 (200 kg)(1623 (9.8 m/s = = 3.18499 m . 030 L sin θ 10.0points Ei = 1 m v02 + 0 and 2 Ef = 0 + m g L sin θ . 10 The work done by friction, conservative force, is thus a non- = F sx cos θ = (81.6 N) (95.4 m) cos 29◦ = 6808.6 J .  0  v2 Wf = Ef − Ei = m g L sin θ − . 2 031 10.0points A 16.9 kg block is dragged over a rough, horizontal surface by a constant force of 81.6 N acting at an angle of 29 ◦ above the horizontal. The block is displaced 95.4 m, and the coefficient of kinetic friction is 0.176. The acceleration of gravity is 9.8 m/s2 . 6 81 . N To find the frictional force, Ff riction = µ N , we need to find N from vertical force balance. Note that N is in the same direction as the y component of F and opposite the force of gravity. Thus F sin θ + N = m g so that N = m g − F sin θ . ◦ 29 Thus the friction force is 16.9 kg ~ Ffriction = −µ N ıˆ = −µ (m g − F sin θ)ˆı . µ = 0.176 The work done by friction is then If the block was originally at rest, determine its final speed. Correct answer: 23.5641 m/s. Wµ = ~Ffriction · ~s = −|fµ | |s| = −µ (m g − fµ sin θ) sx h = −(0.176) (16.9 kg) (9.8 m/s2 ) i = −2116.59−(81.6 J . N) sin 29◦ (95.4 m) Explanation: Let : m = F = sx = µ= g= 16.9 kg , 81.6 N , 95.4 m , 0.176 , and 9.8 m/s2 . The net work done on the block is equal to the change in kinetic energy, so 1 m v 2 − 0 = WF + Wµ 2 Consider the force diagram N F θ fk m g Work is W = F~ · ~s, where ~s is the distance traveled. In this problem ~s = 5ˆı is only in the x direction. ⇒ WF = F x s x r v= 2 [WF +mWµ ] v i u h u 2 (6808.6 (16.9 J) + (−2116.59 J) kg) t = = 23.5641 m/s . 032 10.0points The spring has a constant of 10 N/m and the frictional surface is 0.4 m long with a coefficient of friction µ = 1.56 . The 2 kg block depresses the spring by 29 cm, then is 11 released. The first drop is 1 m and the second is 2 m. The acceleration of gravity is 9.8 m/s2 . the spring, then it gains kinetic energy by dropping a distance h1 , then it loses kinetic energy by doing work against friction. Us + Ug − Wfr = Kf = 1 2 m v2 Since 2 (Us + Ug − Wfr ) m 2 (0.4205 J + 19.6 J − 12.2304 J) = 2 kg = 7.7901 m2 /s2 v2 = How far from the bottom of the cliff does it land? then q 2 /s2. m/s 7.7901 m v = 2.79108 Correct answer: 1.78315 m. Explanation: Basicconcepts The potential energy of a spring is Kinematics gives the time for the mass to drop a distance h2 1 g t2 2 2 h2 t2 = g 1 k x2 2 1 = (10 N/m) (29 cm)2 2 = 0.4205 J . h2 = Us = s Gravitational potential energy is t= Ug = m g h = (2 kg) (9.8 m/s2 ) (1 m) s .8 m) m/s2 ) 2(9(2 = 0.638877 s . = = 19.6 J . Work done against friction on a flat surface is Wfr = µ m g L 2 = (1.56) (2 kg) (9.8 m/s ) (0.4 m) = 12.2304 J . Thus the mass will land a distance s=v t = (2.79108 m/s) (0.638877 s) = 1.78315 m . from the bottom of the cliff. Kinetic energy is K= 2 h2g 1 m v2 . 2 The height of the cliff determines how long it takes for the mass to reach the bottom. Energy considerations give us the horizontal velocity as the mass leaves the cliff top. The mass gets its initial kinetic energy from 033 10.0points A solid metal ball and a hollow plastic ball of the same external radius are released from rest in a large vacuum chamber. When each has fallen 1 m, they both have the same 1.inertia. 12 What average force is exerted on the car? 2.momentum. Correct answer: 27247.7 N. 3.speed. correct Explanation: 4.kinetic energy. Let : m = 972 kg and v = 110 km/h = 30.5556 m/s . 5.change in potential energy. Explanation: All of the quantities mentioned above except terminal velocity depend upon mass. Note: If this experiment had not taken place in a vacuum chamber, none of these answers would have been correct. 034 10.0points A child bounces a 59 g superball on the sidewalk. The velocity change of the superball is from 20 m/s downward to 16 m/s upward. 1 If the contact time with the sidewalk is 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? Momentum is defined as ∆p = m · ∆ v = (972 kg) (30.5556 m/s − 0 m/s) = 29700 kg · m/s . Let : t = 1.09 s . Momentum and impulse are related by ∆p = F t , so Correct answer: 1699.2 N. Explanation: Let : m = 59 g = 0.059 kg , vu = 16 m/s , vd = 20 m/s , and ∆t = 0.00125 s Choose the upward direction as positive. The impulse is I = F ∆t = ∆P = m vu − m (−vd ) = m (vu + vd ) m (vu + vd ) F = ∆t (59 g) (16 m/s + 20 m/s) = 0.00125 s = 1699.2 N . 035 10.0points A(n) 972 kg drag race car accelerates from rest to 110 km/h in 1.09 s. F = 29700 kg · m/s ∆p = = 27247.7 N . t 1.09 s 036 10.0points A(n) 680 N man stands in the middle of a frozen pond of radius 12.1 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.3 kg physics textbook horizontally toward the north shore, at a speed of 8 m/s. The acceleration of gravity is 9.81 m/s2 . How long does it take him to reach the south shore? Correct answer: 80.6477 s. Explanation: Let : Wm = 680 N , r = 12.1 m , mb = 1.3 kg , and v ′b = 8 m/s . 13 The mass of the man is mm = 3. decrease because of conservation of momentum. correct Wm . g 4. increase because of conservation of momentum. From conservation of momentum, m m v m + m b v b = mm v ′ m + mb v ′ b 0 = mm vm′ + mb v ′ b 5. remain the same because the raindrops are falling perpendicular to the direction of cart’s motion. mb v′ mm b g mb ′ v =− Wm b vm′ = −   2 N 680 9.81 m/s (1.3 kg) − = −0.150035 m/s . (8 m/s) Explanation: This is an inelastic collision in the direction along which the cart is rolling. Only momentum ~p along that direction is conserved. Because the raindrops fall vertically, they do not carry momentum horizontally. Assume ∆m of rain water accumulates on the cart: pi = pf m v = (m + ∆m) v ′ . The time to travel the 12.1 m to shore is ∆x ′ | |vm 12.1 m = 0.150035 m/s t= Therefore m v m + ∆m v′ < v . v′ = = 80.6477 s . The speed of the cart will decrease because of conservation of momentum. 037 10.0points An open cart on a level surface rolls without frictional loss through a vertical downpour of rain, as shown below. As the cart rolls, an appreciable amount of rain water accumulates in the cart. rain 038 10.0points Two ice skaters approach each other at right angles. Skater A has a mass of 75 kg and travels in the +x direction at 2.58 m/s. Skater B has a mass of 38.2 kg and is moving in the +y direction at 1.68 m/s. They collide and cling together. Find the final speed of the couple. Correct answer: 1.80092 m/s. v cart The speed of the cart will 1. decrease because of conservation of mechanic energy. 2. increase because of conservation of mechanic energy. Explanation: From conservation of momentum ∆p = 0 mA vAıˆ + mB vBˆ = (mA + mB ) vf  rain water Therefore p vf = mBB vB )2 (mA vAm)2A++ (m p 75s)kg 38.2 kgkg m/s)2 2 ++(64.176 (193.5 kg m/ = 1.80092 m/s = 14 039 10.0points A 1.12 kg skateboard is coasting along the pavement at a speed of 6.33 m/s when a 0.660 kg cat drops from a tree vertically downward onto the skateboard. What is the speed of the skateboard-cat combination? Let : m1 = 7 kg , m2 = 16 kg , mf = m1 + m2 = 23 kg , v1 = 9.5 m/s , v2 = 10 m/s , p1 = m1 v1 = 66.5 kg m/s , p2 = m2 v2 = 160 kg m/s , p 1 p 2 = m1 v 1 m 2 v 2 = 21280 kg2 m2 /s2 , Correct answer: 3.98292 m/s. Explanation: BasicConcept: All of the motion related to the momentum is horizontal: px = p1 + p2 cos θ = 217.783 kg m/s , py = p2 sin θ = 52.0909 kg m/s , θ = 19◦ , and π − θ = 161◦ . vf mf v1 φ m1~v1,i = (m1 + m2 )~vf since v2,i = 0 m/s. Given: m1 = 1.12 kg vi,1 = 6.33 m/s m2 = 0.660 kg m1 θ Solution: m1 vi,1 vf = m1 + m2 (1.12 kg)(6.33 m/s) = 1.12 kg + 0.66 kg = 3.98292 m/s v2 m2 The final momentum is p f = ( m1 + m2 ) v f . (1) Momentum is conserved 040 10.0points (2) ~p1 + ~p2 = ~pf . A(n) 7 kg object moving with a speed of Using the law of cosines, we have 9.5 m/s collides with a(n) 16 kg object moving with a velocity of 10 m/s in a direction 19 ◦ p21 + p22 − 2 p1 p2 cos(π − θ) = p2f . from the initial direction of motion of the 7 kg Solving for vf , we have object. 9.5 m/s 16 kg /s 10 m What is the speed of the two objects after the collision if they remain stuck together? Correct answer: 9.73592 m/s. Explanation: vf = 1 p2 + p22 − 2 p1 p2 cos(π − θ ) m1 + m2 1 (7 kg) + (16 kg) h 2 2 2 2m ×+ (4422.25 (25600 kgkg m2 /s2/s)2 )  19◦ q  7 kg = − (21280 kg2 m2 /s2 ) cos(161◦ ) = 9.73592 m/s . i1/2 AlternateSolution: Since p2f = p2x 15 + p2y , we have 1 (7 kg) + (16 kg) h × (47429.4 kg2 m2 /s2 )   vf = + p2y q x 2 p m1 + m2 = + (2713.46 kg2 m2 /s2 ) = 9.73592 m/s . about:blank i1/2 15/15