* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Version PREVIEW – Semester 1 Review – Slade – (22222) 1 This
Survey
Document related concepts
Classical mechanics wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Coriolis force wikipedia , lookup
Equations of motion wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Center of mass wikipedia , lookup
Centrifugal force wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Jerk (physics) wikipedia , lookup
Fictitious force wikipedia , lookup
Seismometer wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Transcript
Version PREVIEW – Semester 1 Review – Slade – (22222) This print-out should have 48 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Holt SF 02Rev 10A 001 (part 1 of 2) 10.0 points Consider the position-time graph for a squirrel running along a clothesline. b position (m) 4 vavg = 1 ∆x 3.0 m = = 2 m/s . ∆t 1.5 s Holt SF 02A 02 003 10.0 points If Joe rides south on his bicycle in a straight line for 25 min with an average speed of 12.6 km/h, how far has he ridden? Correct answer: 5.25 km. 3 Explanation: b b 2 1 Let : 0 b 1 −1 2 3 4b ∆t = 25 min and vavg = 12.6 km/h . 5 b −2 time (s) What is the squirrel’s displacement at 1.5 s? Your answer must be within ± 5.0% Correct answer: 3 m. ∆x = vavg ∆t = (12.6 km/h)(25 min) · 1h 60 min = 5.25 km toward the south. Explanation: Let : xi = 0 m . The final position is 2 m + (4 m) x1 + x2 = 2 2 = 3.0 m xf = and the displacement is ∆x = xf − xi = 3.0 m − (0 m) = 3.0 m . 002 (part 2 of 2) 10.0 points What is the squirrel’s average velocity during the time interval between 0.0 s and 1.5 s? Your answer must be within ± 5.0% Correct answer: 2 m/s. Explanation: Let : Velocity Relationships 01 004 10.0 points Consider three position curves between time points tA and tB . s sA A 3 2 1 sB B 0 tA tB t Choose the correct relationship among vA + vB quantities v 1 , v 2 , and v3 . v = when 2 a is a constant. 1. v 1 = v 2 = v 3 correct 2. v 1 > v 2 > v 3 3. v 1 < v 2 < v 3 ∆t = 1.5 s . Explanation: Version PREVIEW – Semester 1 Review – Slade – (22222) 2 The average velocity of an object is 007 (part 3 of 4) 10.0 points Find the instantaneous velocity at 4.5 s. sB − sA displacement = . time tB − tA v= All three curves have exactly the same change in position ∆s = sB − sA in exactly the same time interval ∆t = tB − tA , so all three average velocities are equal: Correct answer: 0 m/s. Explanation: v= v1 = v2 = v3 . 008 (part 4 of 4) 10.0 points Find the instantaneous velocity at 8 s. Serway CP 02 15 005 (part 1 of 4) 10.0 points The position versus time for a certain object moving along the x-axis is shown. The object’s initial position is −4 m. 10 Correct answer: 2 m/s. Explanation: b position (m) 8 v= 6 b 4 b 2 b 0 −2 b −4 −6 0 b 1 2 3 4 5 time (s) 6 7 8 9 Find the instantaneous velocity at 0.5 s. Correct answer: 13 m/s. Explanation: The instantaneous velocity is the slope of the tangent line at that point. v= 9 m − (−4 m) = 13 m/s . 1s−0s Explanation: v= 4 m − (9 m) = −2.5 m/s . 3s−1 s 0 m − (−4 m) = 2 m/s . 9s−7s Holt SF 02Rev 54 009 10.0 points A tennis ball with a velocity of 11.3 m/s to the right is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a velocity of −8.50 m/s to the left. If the ball is in contact with the wall for 0.011 s, what is the average acceleration of the ball while it is in contact with the wall? Correct answer: −1800 m/s2 . Explanation: Let : vi = +11.3 m/s , vf = −8.50 m/s , ∆t = 0.011 s . 006 (part 2 of 4) 10.0 points Find the instantaneous velocity at 2 s. Correct answer: −2.5 m/s. 4 m − (4 m) = 0 m/s . 6s−3s ā = and vf − vi −8.5 m/s − 11.3 m/s = ∆t 0.011 s = −1800 m/s2 . Holt SF 02B 03 Version PREVIEW – Semester 1 Review – Slade – (22222) 010 10.0 points With an average acceleration of −1.5 m/s2 , how long will it take a cyclist to bring a bicycle with an initial speed of 14.5 m/s to a complete stop? Let : v0 = 10.1 m/s and g = 9.8 m/s2 . Basic Concept: For constant acceleration, we have Correct answer: 9.66667 s. Explanation: Let : 3 v = v0 + a t . aavg = −1.5 m/s2 , vi = 14.5 m/s , and vf = 0 m/s . (1) Solution: The velocity at the top is zero. Since we know velocities and acceleration, Eq. 1 containing v, a, and t . Choose the positive direction to be up; then a = −g and 0 = v0 + (−g) tup Ball N 03 011 10.0 points A ball is thrown upward. Its initial vertical speed is 10.1 m/s , acceleration of gravity is 9.8 m/s2 , and maximum height hmax are shown in the figure below. Neglect: Air resistance. The acceleration of gravity is 9.8 m/s2 . b b b b b b b b b 9.8 m/s2 b hmax 10.1 m/s bbb bb b b b b b What is its time interval, tup , between the release of the ball and the time it reaches its maximum height? or v0 g (10.1 m/s) = (9.8 m/s2 ) = 1.03061 s . tup = Ball Thrown Up 12 012 10.0 points A ball is thrown straight up and passes point B (at a height of 48.8 m above its starting point O) in 5.2 s. The acceleration of gravity is 9.8 m/s2 . tA is the time b b b hA b b b b b to reach its A b b y b b maximum b b height hA b b B 48.8 m vf − vi ∆v vi = =− ∆t ∆t ∆t −14.5 m/s −vi = = 9.66667 s . ∆t = a −1.5 m/s2 a= b b b v0 b b b b b O 5.2 s tA Figure is not drawn to scale. What was its initial speed k~v0 k ? Correct answer: 34.8646 m/s. Explanation: Basic Concept: Motion under uniform acceleration Correct answer: 1.03061 s. Explanation: t y = v0 t + 1 2 at . 2 Version PREVIEW – Semester 1 Review – Slade – (22222) Solution: The initial velocity is directed upward and gravity acts downward, so y = v0 t − 1 2 gt , 2 Basic Concepts: Horizontally, ∆x = vx ∆t and 1 2 g t , and 2 y + 21 g t2 v0 = t 48.8 m + 21 (9.8 m/s2 ) (5.2 s)2 = 5.2 s = 34.8646 m/s . v0 t = y + Drop a Rock 013 10.0 points If you drop a rock from a height of 12 m , it accelerates at g and strikes the ground 1.56492 s later. The acceleration of gravity is 9.8 m/s2 . If you drop the same rock from half that height, its acceleration will be since ax = 0 m/s2 . Vertically, 1 ∆y = − g (∆t)2 2 since vi,y = 0 m/s. Given: ∆y = −0.809 m ∆x = 18.3 m g = 9.81 m/s2 Solution: From the horizontal motion, ∆t = 1. more. 2. 0. 3. unable to determine. 4. about half. 5. the same. correct Explanation: The acceleration of an object due to gravity is virtually constant at the earth’s surface and is not affected by height. 4 ∆x vx so that 2 1 ∆x ∆y = − g 2 vx s −g(∆x)2 vx = 2∆y s − (9.81 m/s2 ) (18.3 m)2 = 2(−0.809 m) = 45.0605 m/s . Holt SF 03Rev 34 014 10.0 points The fastest recorded pitch in Major League Baseball was thrown by Nolan Ryan in 1974. If this pitch were thrown horizontally, the ball would fall 0.809 m (2.65 ft) by the time it reached home plate, 18.3 m (60 ft) away. The acceleration of gravity is 9.81 m/s2 . How fast was Ryan’s pitch? Accelerating Elevator 015 10.0 points An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.9 s. A passenger in the elevator is holding a 4.8 kg bundle at the end of a vertical cord. The acceleration of gravity is 9.8 m/s2 . What is the tension in the cord as the elevator accelerates? Correct answer: 45.0605 m/s. Correct answer: 49.6993 N. Explanation: Explanation: Version PREVIEW – Semester 1 Review – Slade – (22222) v 2 − vo2 −vo2 = . 2 ∆x 2 ∆x a= T g aelevator 5 X ~ = m~a to express the force exApply F erted on the bullet by the wood: mg Let h be the distance traveled and a the acceleration of the elevator. With the initial velocity being zero, we simplify the following expression and solve for acceleration of the elevator: 1 1 h = v0 t + a t2 = a t2 2 2 2h =⇒ a = 2 . t The equation describing the forces acting on the bundle is Fnet = m a = T − m g T = m (g + a) 2h =m g+ 2 t 2 (1 m) 2 = (4.8 kg) 9.8 m/s + (1.9 s)2 = 49.6993 N . Tipler PSE5 04 29 016 10.0 points A bullet of mass 0.0024 kg moving at 550 m/s impacts a large fixed block of wood and travels 6.8 cm before coming to rest. Assuming that the acceleration of the bullet is constant, find the force exerted by the wood on the bullet. Correct answer: −5.33824 kN. Fwood = m a m vo2 =− 2 ∆x 1 kN (0.0024 kg) (550 m/s)2 × =− 2 (0.068 m) 1000 N = −5.33824 kN . Jumping From the Fire 017 10.0 points A person of mass 73.4 kg escapes from a burning building by jumping from a window situated 39.4 m above a catching net. The acceleration of gravity is 9.8 m/s2 . If air resistance exerts a force of 106.2 N on him as he falls, determine his speed just before he hits the net. Correct answer: 25.6559 m/s. Explanation: The forces acting on him are the gravitational force m g acting downward, and the air resistance, Fa , acting upward. The acceleration is downward and the net force is Fnet = m a = m g − Fa a= The person is in free fall, so his final speed is defined by Explanation: m = 0.0024 kg , vo = 550 m/s , and ∆x = 6.8 cm = 0.068 m . The deceleration of the bullet is constant, so v 2 = vo2 + 2 a ∆x = 0 √ 2ah r 2 (m g − Fa ) h = m s vf = Let : m g − Fa m = 2 [(73.4 kg)(9.8 m/s2 ) − 106.2 N] (39.4 m) 73.4 kg = 25.6559 m/s. Version PREVIEW – Semester 1 Review – Slade – (22222) 6 3. No; it is balanced. Hewitt CP9 04 E11 018 10.0 points In the orbiting space shuttle you are handed four identical boxes. The first one is filled with sand. The second one is filled with iron. The third one is filled with water. The last one is filled with feathers. Shake the boxes. Which one offers the greatest resistance and which one offers the smallest resistance? 1. iron, water 2. sand, water 3. iron, feathers correct 4. Yes; upward. Explanation: The horizontal components of the two 30 N forces cancel, leaving an upward force that is less than 60 N. Thus, the net force on the box is down, causing it to accelerate downward. Conceptual 05 Q8 020 10.0 points The Earth exerts an 760 N gravitational force on a man. What force does the man exert on the Earth? Correct answer: 760 N. 4. feathers, iron Explanation: By Newton’s third law, the man exerts an equal but opposite force on the Earth. 5. All are wrong. Explanation: Among these three materials, iron has the largest density and feathers have the smallest density. So the box filled with iron has largest mass and offers greatest resistance while the box filled with feathers has the smallest mass and offers the smallest resistance. Conceptual 04 Q17 019 10.0 points Two 30 N forces and a 60 N force act on a hanging box as shown. 30 N 30 N Hewitt CP9 05 E33 021 10.0 points Consider a stone at rest on the ground. There are two interactions that involve the stone. One is between the stone and the Earth; Earth pulls down on the stone and the stone pulls up on the Earth. What is the other interaction? 1. between the ground and air 2. between the stone and the ground correct 3. between the ground and the Earth 4. All are wrong. 5. between the Earth and air 60 N Will the box experience acceleration? 1. Unable to determine without the angle. 2. Yes; downward. correct Explanation: If the action is the stone pushing down on the ground surface, the reaction is the ground pushing up on the stone. This upward force on the stone is called the normal force. Force and Motion 15 022 (part 1 of 2) 10.0 points Version PREVIEW – Semester 1 Review – Slade – (22222) The following 2 questions refer to the collisions between a car and a truck whose weight is much heavier than the car (M ≫ m). For each description of a collision below, choose the one answer from the possibilities that best describes the size (or magnitude) of the forces between the car and the truck. Assume: Friction is so small that it can be ignored. M v m v They are both moving at the same speed when they collide. 1. Neither exerts a force on the other; the car gets smashed simply because it is in the way of the truck. 2. The truck exerts a greater amount of force on the car than the car exerts on the truck. 3. Not enough information is given to pick one of these answers. 4. The truck exerts the same amount of force on the car as the car exerts on the truck. correct 5. None of these answers describes the situation correctly. 6. The car exerts a greater amount of force on the truck than the truck exerts on the car. Explanation: By Newton’s third law, action and reaction are of the same magnitude and in the opposite direction. 7 on the car than the car exerts on the truck. 2. The truck exerts the same amount of force on the car as the car exerts on the truck. correct 3. None of these answers describes the situation correctly. 4. The car exerts a greater amount of force on the truck than the truck exerts on the car. 5. Neither exerts a force on the other; the car gets smashed simply because it is in the way of the truck. 6. Not enough information is given to pick one of these answers. Explanation: The same reason as Part 1. Holt SF 04Rev 63 024 (part 1 of 5) 10.0 points Three blocks are in contact with each other on a frictionless horizontal surface. A 335 N horizontal force is applied to the block with mass of 2.9 kg as shown in the figure below. The acceleration of gravity is 9.8 m/s2 . F 2.9 kg 5.7 kg 7.5 kg a) What is the net force on the block with mass 2.9 kg? Correct answer: 60.3416 N. Explanation: 023 (part 2 of 2) 10.0 points The car is moving much faster than the heavier truck when they collide. 1. The truck exerts a greater amount of force F m1 m2 m3 Version PREVIEW – Semester 1 Review – Slade – (22222) Given : F m1 m2 m3 = 335 N , = 2.9 kg , = 5.7 kg , = 7.5 kg , 8 026 (part 3 of 5) 10.0 points c) What is the resultant force on the block with mass 7.5 kg? and Correct answer: 156.056 N. 2 g = 9.8 m/s . Explanation: Solution: Basic Concepts: F3,net = m3 a Fnet = m a = (7.5 kg) 20.8075 m/s2 = 156.056 N mtotal = m1 + m2 + m3 Solution: N F m1 + m2 + m3 g (m1 + m2 + m3 ) g 027 (part 4 of 5) 10.0 points d) What is the magnitude of the force between the block with mass 5.7 kg and 7.5 kg? Correct answer: 156.056 N. Explanation: a F = (m1 + m2 + m3 ) a . F2,3 The acceleration of the system is m3 F a= m1 + m2 + m3 335 N = 2.9 kg + 5.7 kg + 7.5 kg = 20.8075 m/s2 m3 g Basic Concept: F3,net = m3 a = F2,3 Solution: to the right. F2,3 = m3 a F1,net = m1 a = (2.9 kg) 20.8075 m/s2 = 60.3416 N 025 (part 2 of 5) 10.0 points b) What is the resultant force on the block with mass 5.7 kg? = (7.5 kg) 20.8075 m/s2 = 156.056 N 028 (part 5 of 5) 10.0 points e) What is the magnitude of the force between the block with mass 2.9 kg and 5.7 kg? Correct answer: 274.658 N. Correct answer: 118.602 N. Explanation: a Explanation: Solution: F1,2 = (5.7 kg) 20.8075 m/s2 = 118.602 N F2,3 m2 F2,net = m2 a m2 g Version PREVIEW – Semester 1 Review – Slade – (22222) Basic Concept: F2,net = m2 a = F1,2 − F2,3 Solution: Since F2,3 = 156.056 N is provided in the previous part, we have F1,2 = m2 a + F2,3 = (5.7 kg) 20.8075 m/s2 + 156.056 N = 274.658 N Down a Smooth Incline 029 (part 1 of 3) 10.0 points A 3.98 kg block slides down a smooth, frictionless plane having an inclination of 24◦ . The acceleration of gravity is 9.8 m/s2 . 9 Motion has a constant acceleration. Recall the kinematics of motion with constant acceleration. Solution: Because the block slides down along the plane of the ramp, it seems logical to choose the x-axis in this direction. Then the y-axis must emerge perpendicular to the ramp, as shown. Let us now examine the forces in the xdirection. Only the weight has a component along that axis. So, by Newton’s second law, X Fx = m a = m g sin θ Thus a = g sin θ With our particular value of θ, a = (9.8 m/s2 ) sin 24◦ = 3.98602 m/s2 2. 8 8m 8 3. 9 kg 24◦ Find the acceleration of the block. Correct answer: 3.98602 m/s2 . m = 3.98 kg θ = 24◦ . and Consider the free body diagram for the block N θ sin g m N = Correct answer: 4.79161 m/s. Explanation: Since v0 = 0, a = 3.98602 m/s2 and L = 2.88 m, Explanation: Given : 030 (part 2 of 3) 10.0 points What is the block’s speed when, starting from rest, it has traveled a distance of 2.88 m along the incline. θ cos g m W = mg Basic Concepts: Fx,net = F cos θ − W|| = 0 W|| = m g sin θ = m a vf2 = v02 + 2 a (x − x0 ) = 2 (3.98602 m/s2 ) (2.88 m) vf = 4.79161 m/s . It is very important to note at this point that neither of these values depended on the mass of the block. This may seem odd at first, but recall what Galileo discovered 300 years ago – objects of differing mass fall at the same rate. 031 (part 3 of 3) 10.0 points What is the magnitude of the perpendicular force that the block exerts on the surface of the plane at a distance of 2.88 m down the incline? Correct answer: 35.6319 N. Explanation: | FN | = m g cos θ = 3.98 kg (9.8 m/s2 ) cos 24◦ = 35.6319 N. Atwood Machine 15 032 10.0 points A light, inextensible cord passes over a light, frictionless pulley with a radius of 9.5 cm. It has a(n) 16 kg mass on the left and a(n) 7.6 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 1.2 m apart. The acceleration of gravity is 9.8 m/s2 . 9.5 cm ω 7.6 kg a T 10 m1 g 16 kg m2 g =⇒ a By examining the free body diagram again, we see that the force in the y direction is given by X F = −m g cos θ = FN . T Version PREVIEW – Semester 1 Review – Slade – (22222) Since the larger mass will move down and the smaller mass up, we can take motion downward as positive for m2 and motion upward as positive for m1 . Apply Newton’s second law to m1 and m2 respectively and then combine the results: For mass 1: X F1 : T − m1 g = m1 a (1) For mass 2: X F2 : m2 g − T = m2 a (2) We can add Eqs. (1) and (2) above and obtain: m2 g − m1 g = m1 a + m2 a m2 − m1 g m1 + m2 16 kg − 7.6 kg = (9.8 m/s2 ) 16 kg + 7.6 kg = 3.48814 m/s2 . a= 16 kg 1.2 m 7.6 kg At what rate are the two masses accelerating when they pass each other? Correct answer: 3.48814 m/s2 . T = m1 (g + a) = (7.6 kg) (9.8 m/s2 + 3.48814 m/s2 ) = 100.99 N . Explanation: Let : R = 9.5 cm , m1 = 7.6 kg , m2 = 16 kg , h = 1.2 m , and v = ωR. Consider the free body diagrams Centripetal Acceleration 033 10.0 points A car rounds a curve while maintaining a constant speed. Is there a net force on the car as it rounds the curve? 1. It depends on the sharpness of the curve and speed of the car. Version PREVIEW – Semester 1 Review – Slade – (22222) 2. No – its speed is constant. 3. Yes. correct Explanation: Acceleration is a change in the speed and/or direction of an object. Thus, because its direction has changed, the car has accelerated and a force must have been exerted on it. 11 Suppose an automobile has a kinetic energy of 2500 J. When it moves with five times the speed, what will be its kinetic energy? Correct answer: 62500 J. Explanation: Let Eold = 2500 J n = 5. and Horizontal Circle 04 034 10.0 points A 3.08 kg mass attached to a light string rotates on a horizontal, frictionless table. The radius of the circle is 0.691 m, and the string can support a mass of 30 kg before breaking. The acceleration of gravity is 9.8 m/s2 . What maximum speed can the mass have before the string breaks? The kinetic energy K ∝ v 2 , so for vnew = n · vold , Correct answer: 8.12152 m/s. and Explanation: The string will break when the tension exceeds the weight corresponding to 25 kg, so Tmax = M g = (30 kg)(9.8 m/s2 ) . When the smaller mass rotates in a horizontal circle, the string tension provides the centripetal force, so m v2 r r rT . v= m Thus the maximum velocity before the string breaks is r r Tmax vmax = m s (0.691 m)(30 kg)(9.8 m/s2 ) = 3.08 kg T = Knew (n · vold )2 v2 = = n2 = new 2 2 Kold vold vold Knew = n2 Kold = (5)2 (2500 J) = 62500 J . Dragging a Block 02 036 (part 1 of 5) 10.0 points An 18.1 kg block is dragged over a rough, horizontal surface by a constant force of 178 N acting at an angle of angle 28.1 ◦ above the horizontal. The block is displaced 96.2 m, and the coefficient of kinetic friction is 0.211. 178 N 18.1 kg µ = 0.211 Find the work done by the 178 N force. The acceleration of gravity is 9.8 m/s2 . = 8.12152 m/s . Correct answer: 15105.2 J. Kinetic Energy Comparison 035 10.0 points ◦ 28. 1 Explanation: Consider the force diagram Version PREVIEW – Semester 1 Review – Slade – (22222) 12 What is the sign of the work done by the frictional force? n F θ fk 1. zero 2. positive 3. negative correct mg Explanation: ~ =F ~ · ~s, where ~s is the distance Work is W traveled. In this problem ~s = 5 x̂ is only in the x̂ direction. WF = Fx sx = F cos θ sx = (178 N) cos 28.1◦ (96.2 m) = 15105.2 J . 037 (part 2 of 5) 10.0 points Find the magnitude of the work done by the force of friction. Correct answer: 1898.69 J. Explanation: To find the frictional force, Ff riction = µ N , we need to find N from vertical force balance. N is in the same direction as the y component of F and opposite the force of gravity, so F sin θ + N = m g and N = m g − F sin θ . 039 (part 4 of 5) 10.0 points Find the work done by the normal force. Correct answer: 0 J. Explanation: Since the normal force points in the ŷ direction and ~s is in the x̂ direction ~ · ~s = N ŷ · x̂ = 0 . WN = N 040 (part 5 of 5) 10.0 points What is the net work done on the block? Correct answer: 13206.5 J. Explanation: The net work done on the body is the algebraic sum of the work done by the external force F and the work done by the frictional force Wnet = WF + Wµ = 15105.2 J + −1898.69 J = 13206.5 J . Thus the friction force is ~ f riction = −µ N x̂ = −µ (m g − F sin θ x̂ F and the work done by friction is ~ f riction ~s = −|Ff | |s| Wµ = F = −µ (m g − F sin θ) sx = −(0.211)[(18.1 kg)(9.8 m/s2 ) −(178 N) sin 28.1◦ ](96.2 m) = −1898.69 J |Wµ | = 1898.69 J . 038 (part 3 of 5) 10.0 points Holt SF 06Rev 55 041 (part 1 of 2) 10.0 points A constant force of 3.0 N to the right acts on a 1.4 kg mass for 0.48 s. a) Find the final velocity of the mass if it is initially at rest. Correct answer: 1.02857 m/s. Explanation: Let to the right be positive. Let : F = 3.0 N , m = 1.4 kg , t = 0.48 s . and Version PREVIEW – Semester 1 Review – Slade – (22222) Since vi = 0 m/s, ~ ∆t = m ∆~v = m~vf − m~vi = m~vf F F ∆t m (3 N) (0.48 s) = 1.4 kg vf = = 1.02857 m/s to the right. 042 (part 2 of 2) 10.0 points b) Find the final velocity of the mass if it is initially moving along the x-axis with a velocity of 3.9 m/s to the left. Correct answer: −2.87143 m/s. Explanation: Let : vi = −3.9 m/s ~ ∆t = m ∆~v = m~vf − m~vi F F ∆t + m vi vf = m F ∆t + vi = m (3 N) (0.48 s) = + (−3.9 m/s) 1.4 kg = −2.87143 m/s , Applying impulse, ~ ∆t = m~vf − m~vi = m~vf F since vi = 0 m/s. m vf ∆t (0.57 kg) (9 m/s) = 0.22 s = 23.3182 N . F = Spring Between Blocks 02 044 10.0 points Two blocks of masses M and 3 M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. Correct answer: 23.3182 N. 3M M Before (a) v 3M M After (b) which is 2.87143 m/s to the left. Serway CP 04 02 043 10.0 points A football punter accelerates a 0.57 kg football from rest to a speed of 9 m/s in 0.22 s. What constant force does the punter exert on the ball? 13 A cord holding them together is burned, after which the block of mass 3 M moves to the right with a speed of 49 m/s. What is the speed of the block of mass M? Correct answer: 147 m/s. Explanation: Explanation: Given : m = 0.57 kg , vf = 9 m/s, and ∆t = 0.22 s . Let : m1 = M , m2 = 3 M , v2 = 49 m/s . and Version PREVIEW – Semester 1 Review – Slade – (22222) 14 From conservation of momentum ∆p = 0, so 0 = m1 v1 + m2 v2 m2 v2 v1 = − m1 (3 M ) (49 m/s) = M = 147 m/s . Pitching Machine Recoil 045 10.0 points A baseball player uses a pitching machine to help him improve his batting average. He places the 66.1 kg machine on a frozen pond. The machine fires a 0.108 kg baseball horizontally at a speed of 31.9 m/s. What is the magnitude of the recoil velocity of the machine? Correct answer: 0.052121 m/s. Explanation: Blocks Compress a Spring 02 046 10.0 points Two blocks of masses 4 kg and 15 kg are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord holding them together is burned, after which the block of mass 15 kg moves to the right with a speed of 27 m/s. What is the velocity of the other mass in m/s? Correct answer: −101.25 m/s. Explanation: By momentum conservation m1 v1 + m2 v2 = 0 so the velocity of m1 is Let : m1 = 0.108 kg , m2 = 66.1 kg , and v1f = 31.9 m/s . We take the system to consist of the baseball and the pitching machine. Because of the force of gravity and the normal force, the system is not really isolated. However, both of these forces are directed perpendicularly to the motion of the system. Therefore, momentum is constant in the x direction because there are no external forces in this direction (assuming the surface is frictionless). The total momentum of the system before firing is zero. Therefore, the total momentum after firing must be zero, that is, v1 = − m2 v2 m1 Serway CP 04 15 047 (part 1 of 2) 10.0 points Consider a 620 N cat burglar supported by a cable as in the figure. 36.1◦ m1 v1f + m2 v2f = 0 , or, in components, m1 v1f − m2 v2f = 0 . v2f = m1 0.108 kg v1f = (31.9 m/s ) m2 66.1 kg = 0.052121 m/s . Find the tension in the inclined cable. Correct answer: 1052.28 N. Explanation: Version PREVIEW – Semester 1 Review – Slade – (22222) Let : W = 620 N and θ = 36.1◦ . y T2 θ T1 x W Since the burglar is held in equilibrium, X Fy = 0 T2 sin θ − W = 0 620 N W = sin θ sin 36.1◦ = 1052.28 N . T2 = 048 (part 2 of 2) 10.0 points Find the tension in the horizontal cable. Correct answer: 850.232 N. Explanation: Since the burglar is held in equilibrium, X Fx = 0 T1 − T2 cos θ = 0 T1 = T2 cos θ = (1052.28 N) cos 36.1◦ = 850.232 N . 15