Faculty of Engineering and Department of Physics Engineering Physics 131 Midterm Examination February 27, 2006; 7:00 pm 8:30 pm

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Faculty f Engineering and Department f Physics Engineering Physics 131 Midterm Examinatin February 27, 2006; 7:00 pm 8:30 pm N ntes r textbks allwed. Frmula sheet is n the last page (may be remved). Calculatrs apprved by the Faculty f Engineering permitted. The exam has 6 questins. Attempt ALL questins. The value f each questin is indicated n the table belw. Budget yur time accrdingly. Shw all wrk in a neat and lgical manner. Indicate clearly if yu use the backs f pages fr material t be marked. DO NOT separate the pages f the exam. NAME: ID#: Please circle the name f yur instructr: B1: Hedayatipr B2: Sigurdsn B3: Mertiny B4: Behzadipur B5: Rpchan B6: Le Questin Value Mark 1 8 (2 each fr a,b,c;d) 2 16 (4 each fr a,b,c,d) 3 12 (4 each fr a,b; 2 each fr c,d) 4 20 5 20 6 24 Ttal 100

1. The Figure belw gives the velcity cmpnent v as a functin f time fr a particle mving in rectilinear mtin alng an axis. It is part f a sine curve. Pint 1 has the maximum value n the curve; pint 4 has the minimum value; and pints 2 and 6 have the same value. v 1 2 6 3 4 5 t a) What is the directin f travel at time t = 0? b) What is the directin f travel at pint 4? c) At which f the numbered pints des the particle reverse its directin f travel? d) Rank the six pints accrding t the magnitude f the acceleratin, greatest first.

2. The small blck is sliding n the inclined wedge as shwn. Fr each f the fllwing scenaris, draw the free-bdy and kinetic diagrams (FBD/KD) f i) the blck; ii) the wedge. Use fr kinetic frictin and fr static frictin wherever necessary and clearly indicate the directins f frces and acceleratins whenever pssible. The frce f gravity is acting straight dwn in the Figure. a) The incline is smth, the hrizntal surface is rugh enugh t prevent the wedge frm mving. FBD Blck KD Blck FBD Wedge KD Wedge b) The incline is rugh, the hrizntal surface is smth. FBD Blck KD Blck

FBD Wedge KD Wedge c) All surfaces are rugh, with the frictin in the hrizntal surface sufficient t prevent the wedge frm mving. FBD Blck KD Blck FBD Wedge KD Wedge d) All surfaces are smth FBD Blck KD Blck FBD Wedge KD Wedge

3. Persn A is mving frward with velcity vectr v as shwn in the figure lking dwn frm verhead. The ttal acceleratin vectr f the persn may be either f fur vectrs a 1, a 2, a 3, a 4. a) Check the bx that crrectly explains what is happening t the directin f mtin at this mment: Acceleratin vectr Directin remains unchanged Turning t the persn s left Turning t the persn s right a 1 a 2 a 3 a 4 b) Check the bx that crrectly explains the change f the speed f the persn at this mment: cnstant Increasing Decreasing a 1 a 2 a 3 a 4 c) Rank the radius f curvature in increasing rder (mentin any ties and justify yur answer) d) Rank the rate f change f speed in increasing rder (mentin any ties and justify yur answer)

4. A jet plane travels dwn a runway with the initial speed f 50.0 ft/s at s = 0 and is then subjected t the acceleratin shwn. a(ft/s ) 75 a = 75 0.15s a) Determine the speed v f the plane when it has traveled 200 ft. 500 s(ft) b) Hw much time t is required fr it t travel 200 ft?

5. During actin mvies, cars ften leave the grund fr a variety f dramatic reasns. In the scene under examinatin, the stunt driver has been asked t get the car up t a sufficient speed s that it can g up a 20 incline, launch int the air, and land n a platfrm lcated ahead f the car. The relevant dimensins are shwn in the Figure. Treat the car as a mass particle. a) Slve fr the minimum speed v that will allw the stunt t be successfully undertaken. b) Determine the magnitude and directin f the velcity immediately befre landing. The directin shuld be reprted as an angle with respect t the nearest hrizntal.

6. In the figure shwn belw, all blcks are mving. Blck B is mving t the left with a speed f 1 m/s and its speed is decreasing at a rate f 0.5 m/s 2. Blck C is mving t the right with a speed f 2 m/s and the speed f C is decreasing at a rate f 0.2 m/s 2. Determine the fllwing: a) The cnstraint equatin relating the velcities and acceleratins f the three blcks. b) The velcity and acceleratin f blck A, expressed as a magnitude and directin. c) The relative velcity and acceleratin f blck B with respect t blck C, expressed as a magnitude and directin.

Extra Page

1. The Figure belw gives the velcity cmpnent v as a functin f time fr a particle mving in rectilinear mtin alng an axis. It is part f a sine curve. Pint 1 has the maximum value n the curve; pint 4 has the minimum value; and pints 2 and 6 have the same value. a) What is the directin f travel at time t = 0? Frward (r psitive) b) What is the directin f travel at pint 4? Backward (r negative) c) At which f the numbered pints des the particle reverse its directin f travel? 3,5 d) Rank the six pints accrding t the magnitude f the acceleratin, greatest first. 2,6 > 3,5 > 1,4

3. Persn A is mving frward with velcity vectr v as shwn in the figure (verhead picture). The ttal acceleratin vectr f the persn may be either f fur vectrs a 1, a 2, a 3,a 4. a) Check the bx that crrectly explains what is happening t the directin f mtin at this mment: Directin remains unchanged Turning t the persn s left Turning t the persn s right a 1 a 2 a 3 a 4 b) Check the bx that crrectly explains the change f the speed f the persn at this mment: cnstant Increasing Decreasing a 1 a 2 a 3 a 4 c) Rank the radius f curvature in increasing rder (mentin any ties and justify yur answer) 2, 1=4, 3 Since the velcity is the same fr all fur cases, the nrmal cmpnent f acceleratin 2 2 determines the radius f curvature: v v an = ρ = ρ a therefre, larger nrmal cmpnent results in smaller radius f curvature. n d) Rank the rate f change f speed in increasing rder (mentin any ties and justify yur answer) 3=4, 2, 1 Speed change is determined by the tangential cmpnent f acceleratin. In 3 and 4, tangential acceleratin is 2 units in the negative directin which make the speed decrease. In 2, there is n tangential acceleratin therefre, the speed remains unchanged and in 1, the speed is increasing because f the psitive tangential acceleratin.

2. The small blck is sliding n the inclined wedge as shwn. Fr each f the fllwing scenaris, draw the free-bdy and kinetic diagrams (FBD/KD) f i) the blck; ii) the wedge. Use F f fr kinetic frictin and k f s F fr static frictin wherever necessary and clearly indicate the directins f frces and acceleratins whenever pssible. The frce f gravity is acting straight dwn in the Figure. a) The incline is smth, the hrizntal surface is rugh enugh t prevent the wedge frm mving. FBD Blck KD Blck FBD Wedge KD Wedge b) The incline is rugh, the hrizntal surface is smth. FBD Blck KD Blck

FBD Wedge KD Wedge c) All surfaces are rugh, with the frictin in the hrizntal surface sufficient t prevent the wedge frm mving. FBD Blck KD Blck FBD Wedge KD Wedge d) All surfaces are smth FBD Blck KD Blck FBD Wedge KD Wedge

5. During actin mvies, cars ften leave the grund fr a variety f dramatic reasns. In the scene under examinatin, the stunt driver has been asked t get the car up t a sufficient speed s that it can g up a 20 incline, launch int the air, and land n a platfrm lcated ahead f the car. The relevant dimensins are shwn in the sketch. Treat the car as a mass particle, a. slve fr the minimum speed v that will allw the stunt t be successfully undertaken and b. determine the magnitude and directin f the velcity immediately befre landing. The directin shuld be reprted as an angle with respect t the hrizntal Slutin: Assume the car will launch at the edge f the platfrm, then V 0 is the minimum initial speed, and V is the velcity befre landing. a) t In hrizntal directin: V0 cs 20 t =79 ft (1) 1 2 In vertical directin: V0 sin 20 t gt = 21 10 = 11 ft (2) 2 79 ft Substitute t = V cs 20 int (2): V 0 79 1 79 2 0 sin 20 g( ) = 11 ft V0cs 20 2 V0cs 20 2 79 g V0 = = 80.1 ft/s 2 2(cs 20 ) (79 tan 20 11) 2

b) V = V = V cs 20 = 80.1*cs 20 = 75.3 ft/s tx 0 0x 0 79 ft 79 ft t = = = 1.05 s V cs 20 75.3 ft/s V = V gt = V sin 20 gt ty 0y 0 = 80.1*sin 20 32.2*1.05 = 6.41 ft/s = 6.41 ft/s ( ) V = V + V = 75.6 ft/s 2 2 t tx ty 1 6.41 ft/s θ = tan ( ) = 4.87 (clckwise frm hrizntal directin) 75.3 ft/s