( m/s) 2 4(4.9 m/s 2 )( 52.7 m)

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1 Version 072 idterm 2 OConnor (05141) 1 This print-out should have 18 questions ultiple-choice questions may continue on the next column or page find all choices before answering V1:1, V2:1, V3:3, V4:5, V5:3 BE SURE TO CIRCLE THE ANSWERS ON THE FRONT PAGES AND ATTACH YOUR WORK TO THE BACK YOU UST SHOW YOUR WORK! DO THE EASY PROBLES FIRST AND SAVE THE HARD ONES FOR LAST 001 (part 1 of 2) 10 points A stone is thrown from the top of a building upward at an angle of 233 to the horizontal and with an initial speed of 10 m/s, as in the figure The height of the building is 527 m The acceleration of gravity is 98 m/s 2 y θ0 v 0 x s The initial x and y components of the velocity are v x0 = v 0 cos θ 0 = m/s v y0 = v 0 sin θ 0 = m/s To find t, we can use the relation y = v y0 t 1 2 g t2 with y = 527 m and v y0 = m/s (we have chosen the top of the building as the origin): ( m/s) t 1 2 (98 m/s2 ) t 2 = 527 m (49 m 2 /s 2 )t 2 ( m/s) t 527 m = 0 Applying the quadratic formula, since h How long is the stone in flight? s s s s correct s s s s s ( m/s) 2 4(49 m/s 2 )( 527 m) then = m 2 /s 2, t = m/s ± m 2 /s 2 98 m/s 2 = s 002 (part 2 of 2) 10 points What is the speed of the stone just before it strikes the ground? m/s m/s m/s m/s m/s correct m/s m/s

2 Version 072 idterm 2 OConnor (05141) m/s m/s m/s The y component of the velocity just before the stone strikes the ground can be obtained using the equation with t = s : v y = v y0 g t v y = ( m/s) (98 m/s 2 ) ( s) = m/s Since v x = v x0 = m/s, the required speed is v = vx 2 + vy 2 = ( m/s) 2 + ( m/s) 2 = m/s 003 (part 1 of 3) 10 points A block of mass kg lies on a frictionless table, pulled by another mass kg under the influence of Earth s gravity The acceleration of gravity is 98 m/s N N N correct N N N N Given : m 1 = kg, m 2 = kg, µ = 0 N a T m 1 m 2 m 1 g T and m 2 g The net force on the system is simply the weight of m 2 F net = m 2 g = ( kg) (98 m/s 2 ) = N a kg µ = kg 004 (part 2 of 3) 10 points What is the magnitude of the acceleration a of the two masses? m/s m/s 2 What is the magnitude of the net external force F acting on the two masses? N N N m/s m/s m/s m/s 2

3 Version 072 idterm 2 OConnor (05141) m/s 2 correct m/s m/s m/s 2 From Newton s second law, F net = m 2 g = (m 1 + m 2 ) a Solving for a, m 2 a = g m 1 + m kg = kg kg (98 m/s2 ) = m/s (part 3 of 3) 10 points What is the magnitude of the tension T of the rope between the two masses? N N N N N N N correct N 006 (part 1 of 3) 10 points A block is at rest on the incline shown in the figure The coefficients of static and kinetic friction are µ s = 075 and µ k = 064, respectively The acceleration of gravity is 98 m/s 2 42 kg µ 33 What is the frictional force acting on the 42 kg mass? N N N N N N N N N correct N N N Analyzing the horizontal forces on block m 1, we have Fx : T = m 1 a = ( kg) ( m/s 2 ) = N N m g F f 33

4 Version 072 idterm 2 OConnor (05141) 4 The forces acting on the block are shown in the figure Since the block is at rest, the magnitude of the friction force should be equal to the component of the weight on the plane of the incline F f = g sin θ = (42 kg) (98 m/s 2 ) sin 33 = N 007 (part 2 of 3) 10 points What is the largest angle which the incline can have so that the mass does not slide down the incline? correct The largest possible value the static friction force can have is F f,max = µ s N, where the normal force is N = g cos θ Thus, since F f = g sin θ, g sin θ m = µ s g cos θ m tan θ m = µ s θ m = tan 1 (µ s ) = tan 1 (075) = (part 3 of 3) 10 points What is the acceleration of the block down the incline if the angle of the incline is 44? m/s m/s m/s m/s m/s m/s m/s m/s m/s m/s 2 correct When θ exceeds the value found in part 2, the block starts moving and the friction force is the kinetic friction F k = µ k N = µ k g cos θ Newton s equation for the block then becomes and a = g sin θ F f = g sin θ µ k g cos θ a = g [sin θ µ k cos θ] = (98 m/s 2 ) [sin 44 (064) cos 44 ] = m/s (part 1 of 1) 10 points As viewed by a bystander, a rider in a barrel of fun at a carnival finds herself stuck with her back to the wall ω

5 Version 072 idterm 2 OConnor (05141) 5 Which diagram correctly shows the forces acting on her? 1 correct 2 None of the other choices The acceleration of gravity is 980 cm/s 2 What is the coefficient of static friction between the coin and the turntable? correct The normal force on the coin is N = m g The force provided by friction immediately before slippage is given by f = µ N = µ m g 6 The normal force of the wall on the rider provides the centripetal acceleration necessary to keep her going around in a circle The downward force of gravity is equal and opposite to the upward frictional force on her Note: Since this problem states that it is viewed by a bystander, we assume that the free-body diagrams are in an inertial frame 010 (part 1 of 1) 10 points A coin is placed 31 cm from the center of a horizontal turntable, initially at rest The turntable then begins to rotate When the speed of the coin is 120 cm/s (rotating at a constant rate), the coin just begins to slip At this moment the centripetal acceleration (there is no tangential acceleration) provides a force F = m v2 r Summation of forces yields F = f = µ m g = m v2 r, therefore µ = v2 r g (120 cm/s) 2 = (31 cm) (980 cm/s 2 ) = (part 1 of 2) 10 points Consider the loop-a-loop setup, where a mass m is sliding along a frictionless track and the radius of the loop is R

6 Version 072 idterm 2 OConnor (05141) 6 m C B 4 h = 2 3 R h D Given h = 5 R, determine the speed of the mass at A 1 v A = 4 2 g R 2 v A = 3 2 g R R 3 v A = 2 2 g R correct 4 v A = 3 g R 5 v A = 4 g R 6 v A = 2 g R Applying the work energy theorem from C to A we get A m g h m g R = 1 2 m v2 A 5 h = 1 2 R The critical condition at B implies that m v 2 R = m g Conservation of energy between C and B implies m g (h 2 R) = 1 2 m v2 = 1 2 m g R m g = m v2 R h = 2 R + R 2 = 5 2 R 013 (part 1 of 1) 10 points A(n) 632 g ball is dropped from a height of 658 cm above a spring of negligible mass The ball compresses the spring to a maximum displacement of cm The acceleration of gravity is 98 m/s 2 h mg(5 R R) = 4 R m g = 1 2 m v2 A v 2 A = 8 g R v A = 2 2 g R 012 (part 2 of 2) 10 points Find the critical initial height of the mass, such that it would just barely pass the point B 1 h = 1 3 R 2 h = 5 2 R correct 3 h = 3 2 R Calculate the spring force constant k N/m N/m correct N/m N/m N/m N/m N/m x

7 Version 072 idterm 2 OConnor (05141) N/m N/m N/m Let : m = 632 g, h = 658 cm, and x = cm Using conservation of energy, we have m/s m/s Let : m = 163 kg, = 2000 kg, v = 140 m/s and The cannon s velocity immediately after it was fired is found by using conservation of momentum along the horizontal direction: k = from which 2 m g (h + x) x k x2 = m g (h + x) = 2 (00632 kg)(98 m/s2 )(0658 m m) ( m) 2 = N/m 014 (part 1 of 2) 10 points A revolutionary war cannon, with a mass of 2000 kg, fires a 163 kg ball horizontally The cannonball has a speed of 140 m/s after it has left the barrel The cannon carriage is on a flat platform and is free to roll horizontally What is the speed of the cannon immediately after it was fired? m/s m/s m/s V + m v = 0 V = m v where is the mass of the cannon, V is the velocity of the cannon, m is the mass of the cannon ball and v is the velocity of the cannon ball Thus, the cannon s speed is V = m v = 163 kg (140 m/s) 2000 kg = 1141 m/s 015 (part 2 of 2) 10 points The same explosive charge is used, so the total energy of the cannon plus cannonball system remains the same Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all other parameters remained the same? m/s m/s correct m/s m/s m/s m/s m/s m/s m/s correct m/s m/s

8 Version 072 idterm 2 OConnor (05141) m/s m/s m/s m/s By knowing the speeds of the cannon and the cannon ball, we can find out the total kinetic energy available to the system K net = 1 2 m v V 2 This is the same amount of energy available as when the cannon is fixed Let v be the speed of the cannon ball when the cannon is held fixed Then, 1 2 m v 2 = 1 2 (m v2 + V 2 ) v = = v v 2 + m V m = (140 m/s) = m/s 1 + Thus, the velocity difference is 163 kg 2000 kg v v = m/s 140 m/s = m/s 016 (part 1 of 1) 10 points Bill (mass m) plants both feet solidly on the ground and then jumps straight up with velocity v The earth (mass ) then has velocity ( ) 1 V Earth = + v m ( m ) v 2 V Earth = correct m 3 V Earth = + v 4 V Earth = v ( ) 5 V Earth = v m 6 V Earth = + v m 7 V Earth = v ( m ) v 8 V Earth = + The momentum is conserved We have So m v + V Earth = 0 ( m ) v V Earth = 017 (part 1 of 1) 10 points A child bounces a 49 g superball on the sidewalk The velocity change of the superball is from 21 m/s downward to 17 m/s upward If the contact time with the sidewalk is 1/800 s, what is the magnitude of the force exerted on the superball by the sidewalk? N N N correct N N N N N N N

9 Version 072 idterm 2 OConnor (05141) 9 Let : m = 49 g = 0049 kg, v u = 17 m/s, v d = 21 m/s, and t = s Choose the upward direction as positive Then the impulse is I = F t = P = m v u m ( v d ) = m (v u + v d ) F = m (v u + v d ) t (49 g) (17 m/s + 21 m/s) = s = N 018 (part 1 of 1) 10 points As shown in the top view above, a disc of mass m is moving horizontally to the right with speed v on a table with negligible friction when it collides with a second disc of mass 6 m The second disc is moving horizontally to the right with speed v at the moment of 6 impact The two discs stick together upon impact v m after v 6 6 m before v f 7 m 2 v f = 1 3 v 3 v f = v 4 v f = 1 2 v 5 v f = 4 9 v 6 None of these are correct 7 v f = 2 v correct 7 8 v f = v 9 v f = 3 5 v 10 v f = 3 7 v The total momentum of the system is conserved because there is no exterior force So what we have is (m + 6 m) v f = m v + 6 m v ( 6 7 m v f = m v ) 6 ( 6 7 v f = ) v 6 7 v f = 12 6 v v f = (12) (6) (7) v = v, so = 2 7 v The speed of the composite body immediately after the collision is 1 v f = 3 10 v