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Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 1 Q1. A particle moves in one dimension such that its position x(t) as a function of time t is given by x(t) = 2.0 + 27 t – t3, where t is in seconds and x(t) is in meters. At what position does the particle change its direction of motion? A) B) C) D) E) 56 m 54 m 27 m 81 m 100 m Ans: dx = 0 = 27t − 3t 2 dt ⇒ t = 3.0 s x(t) = 2.0 + 27 t − t 3 ⇒ x(3) = (2 + 27 × 3 − 27)m x(3) = (2 + 81 − 27)m = 56.0 m Q2. What is your total displacement when you follow directions that tell you to walk 40.0 m north then 30.0 m east? A) B) C) D) E) 50.0 m, 53.1o North of East 50.0 m, 53.1o East of North 50.0 m, 45.0o East of North 45.0 m, 45.0o East of North 45.0 m, 30.1o North of East Ans: North 30 m 40 cm West ��⃗ = �30 �i + 40 �j � m D ��⃗� = 50.0 m �D Qx = cos−1 � 30 � = 53.1° 50 �⃗ D θx South East Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 2 Q3. A projectile is fired from the ground with an initial velocity = vo (30.00 iˆ + 19.62 ˆj ) m/s. Find the horizontal distance the projectile travels before hitting the ground. A) B) C) D) E) 120.0 m 60.00 m 78.48 m 156. 9 m 400.3 m Ans: v0x = 30.0 m/s ; v0y = 19.62 m/s Time to reach ground: 0 ∆y = 19.62 t − Q4. 1 2 2 × 19.6 gt ⇒ t = = 4.0 s 2 9.8 ∆x = v0x xt = (30.0 × 4.0) = 120.0 m There are two forces on the 3.00 kg box in the overhead view of Figure 1, but only one is shown. For F 1 = 20.0 N, a = 12.0 m/s2, and θ = 30.0o, find the second force in unit vector notation. A) (−38.0 N ) iˆ + (−31.2 N ) ˆj B) (−51.2 N ) iˆ − (+18.0 N ) ˆj C) (−38.0 N ) iˆ + (+31.2 N ) ˆj D) (+38.0 N ) iˆ + (+31.2 N ) ˆj E) (−31.2 N ) iˆ + (−38.0 N ) ˆj Ans: �F⃗1 + �F⃗2 = ma�⃗ 20.0 �i + �F⃗2 = 3.0 �−12.0 Sinθ �i − 12.0 cos θ �j � = −36.0 �0.5 �i + 0.866 �j � N = �− 18.0 �i − 31.2 �j � N �⃗2 = �− 20.0 �i − 18.0 �i − 31.2 �j � N ∴F �⃗2 = (−38.0 N) �i + (−31.2 N) �j F Fig# 1 Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 3 Q5. A block slides with a constant acceleration down a rough plane inclined at an angle of 40.0o with the horizontal. If the block starts from rest and slides a distance of 39.2 m down the plane in 4.00 s, calculate the coefficient of kinetic friction between the block and the plane. A) B) C) D) E) 0.186 0.215 0.125 0.110 0.098 Ans: fk θ mgsinθ − µk mgcosθ = ma a = gsin400 − µk gcos40° = 9.8 × 0.643 − µk × 9.8 × 0.766 ∴ a = 6.299 − 7.51µk 1 ∆x = v0 t + 2 at 2 39.2 = 0 + 1 8 (6.299 − 7.51 µk ) × 16 2 39.2 = 50.39 − 60.08µk ⇒ µk = 0.186 Q6. A 100-kg parachute falls at a constant speed of 1.00 m/s. At what rate is energy being lost? A) 980 W B) 19.8 W C) 89.0 W D) 49.0 W E) 490 W Ans: P = mgv = 100 × 9.8 × 1.0 = 980 W Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 4 Q7. The work done by a conservative force acting on a body A) B) C) D) E) does not change the total energy. does not change the potential energy. does not change the kinetic energy. is always equal to zero. is always equal to the sum of the changes in potential and kinetic energies. Ans: Q8. 𝐀 The velocity of a given body is increased to such an extent that the kinetic energy of the body is increased by a factor of 16. The momentum of the body is increased by a factor of: A) B) C) D) E) 4.0 16 8.0 2.0 1.0 Ans: ki = kf = ⇒ 1 P1 2 mv1 2 = 2m 2 1 P2 2 mv2 2 = 2 2m kf P2 2 = � � ki P1 P2 4= � � P1 P2 = 4.0 P1 Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 5 Q9. A 32.0-kg thin loop has a radius of 2.00 m and is rotating at 280 rev/min about its central axis. What is the average power required to bring it to a stop in 20.0 s? A) B) C) D) E) 2.75 × 103 W 1.32 × 103 W 3.15 × 103 W 1.32 × 102 W 9.38 × 101 W Ans: ωi = �280 rev min � �2π rad rev min � � 60 s� = 29.3 rad/s I = MR2 = 32.0 × (2.0)2 = 128 kg. m2 1 1 2 2 × 128 × (29.3)2 W ∆k �2 I (ωf − ωi )� 2 P= = = = J/S ∆t ∆t 20.5 20 P = 2.75 × 103 W Q10. A uniform solid sphere rolls down an incline. What must be the incline angle if the linear acceleration of the center of the sphere is to have a magnitude of 1.96 m/s2? A) B) C) D) E) 16.3o 45.0o 30.0o 60.0o 25.4o Ans: a= −g sin θ −g sin θ −gsinθ = = 2 I 7 2 MR I+ 1 + 2 2 5 MR 5 MR ∴a= −5 5 gsinθ ⇒ −1.96 = − × 9.8 × sinθ 7 7 sinθ = 0.280 θ = 16.26° = 16.3° Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 6 Q11. A uniform beam is held in a vertical position by a pin at its lower end and a cable at its upper end (see Figure 2). The tension in the cable is 72.0 N. Find the horizontal force F acting on this beam. Fig# 2 A) B) C) D) E) 99.8 N 120 N 65.0 N 135 N 50.3 N Ans: � τ0 = 0 Q12. 5 F = 72 cos 30° × 8 72 cos 30° × 8 = 99.76 N = 99.8 N F= 5 A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.00 g coins stacked over the 10.0 cm mark, the stick is found to balance at the 45.5 cm mark. What is the mass of the meter stick? A) B) C) D) E) 78.9 × 10−3 kg 78.9 kg 98.7 × 10−3 kg 50.0 × 10−3 kg 25.0 × 10−3 kg 10 cm 0 cm 100 cm 2 mg 45.5 Mg Ans: � τ0 = 0 2 mg (45.5 − 0.10) = Mg (50 − 45.5) 2 × 5.0 × 10−3 (35.5) = M (4.5) ⇒ M = 10 × 10−3 × (35.5) = 78.9 × 10−3 kg 4.5 Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 7 Q13. A 12-gram bullet is fired into a 1.0 × 102-gram wooden block initially at rest on a horizontal surface. After impact, the bullet is imbedded inside the wooden block which slides 7.5 m before coming to rest. If the coefficient of friction between the block and the surface is 0.65, what was the speed of the bullet immediately before impact? A) B) C) D) E) 91.2 m/s 49.0 m/s 25.0 m/s 100 m/s 190 m/s Ans: → vb mb VB = 0 100 g MB d = 7.5 m µk = 0.65 mb vb = (mb + MB )v → (1) ∆k + ∆ug = Wnc → (2) 1 0 − 2 (mb + MB )v 2 = −µk (mb + MB )gd ⇒ v = �2µk gd = 9.77 m/s Q14. From equation (1): 12 × 10−3 vb = 112 × 10−3 × 9.77 ⇒ vb = 91.2 m/s A certain wire stretches 1.00 cm when a force F is applied to it. The same force is applied to a second wire of the same material with equal length but with twice the diameter. The second wire stretches: A) B) C) D) E) 0.25 cm 4.0 cm 2.5 cm 0.50 cm 2.0 cm Ans: ∆L1 F1 =E A1 L1 F2 ∆L2 =E A2 L2 F2 A1 ∆L2 L1 ⇒ � �� � = � �� � L F1 A2 ∆L1 2 A1 1 ∴ ∆L2 = � � ∆L1 = ∆L1 = 0.25 cm A2 4 Phys101 Coordinator: Dr. G. Khattak Q15. Final-113 Thursday, August 02, 2012 Zero Version Page 8 A cube of volume 1.0 m3 is made of material with a bulk modulus of 2.0 × 109 N/m2. When it is subjected to a pressure of 2.0 × 105 Pa, the magnitude of the change in its volume is: 1.0 × 10−4 0.25 × 10−4 1.4 × 10−4 3.5 × 10−2 2.4 × 10−3 A) B) C) D) E) m3 m3 m3 m3 m3 Ans: ∆P = −B ∆V = − ∆V V ∆P×V B = 2×105 ×1.0 2.0×109 |∆V| = 1.0 × 10−4 m3 = −1.0 × 10−4 m3 Q16. Let the acceleration due to gravity on the surface of Earth be g E . The acceleration due to gravity (g p ) on the surface of a planet whose mass is equal to that of Earth but whose radius is only 0.100 R E is given by (R E is Earth’s radius) A) B) C) D) E) gp gp gp gp gp = 100 g E = 10.0 g E = 50.0 g E = 25.0 g E = 1.00 g E Ans: ge = gp = GMe R2 GMp Rp2 g p = 100 g e 2 gp Re Re 2 ⇒ = � � = � � ge Rp 0.1R e Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 9 Q17. A solid uniform sphere has a mass of 1.00 × 104 kg and a radius of 1.00 m. What is the magnitude of the gravitational force due to the sphere on a particle of mass m = 1.00 kg located at a distance of 0.500 m from the center of the sphere A) B) C) D) E) 3.34 × 10−7 N 1.17 × 10−5 N 6.68 × 10−3 N 1.50 × 10−5 N 1.67 × 10−8 N m r GmM ′ r2 M′ = Q18. 4 3 M � πR3 � 4 × �3 πr 3 � = R M′ Ans: F= M Mr3 R3 GmMr 6.67 × 10−11 × 1.0 × 1 × 104 × 0.5 F= = = 3.34 × 10−7 N (1.0)3 R3 Figure 3 gives the potential energy function U(r) of a projectile, plotted outward from the surface of a planet of radius R s . What kinetic energy is required of a projectile launched at the surface if the projectile is to “escape” the planet? Fig # 3 A) B) C) D) E) +5.0 × 109 J –5.0 × 109 J +4.0 × 109 J −4.0 × 109 J −9.0 × 108 J Ans: ∆K + ∆U = 0 ⇒ ∆K = − ∆U k f − k i = −(Uf − Ui ) k i = (0 − Ui ) k i = −Ui k = 5 × 109 J Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 10 Q19. A satellite travels in a circular orbit around a certain planet. The orbit has a radius of 10 × 106 m and a period of 8.0 hours. What is the mass of the planet? A) B) C) D) E) 7.1 × 1023 kg 9.1 × 1020 kg 9.8 × 1017 kg 4.9 × 1026 kg 8.5 × 1029 kg Ans: T2 = 4π2 3 r GM 22 2 � × 1021 4π 3 7 r = = 7.1 × 1023 kg M= GT 2 6.67 × 10−11 × (8 × 3600)2 2 (4) � Q20. A 250-kg spaceship is in a circular orbit of radius 3.00 R E about the earth. How much energy is required to transfer the spaceship to a circular orbit of radius 4.00 R E ? (R E is the radius of the earth)? A) 6.52 × 108 J B) 3.52 × 1011 J C) 5.32 × 1014 J D) 4.37 × 105 J E) 8.97 × 101 J Ans: Ei = − Gm ME Gm ME ; Ef = − 2(3.00R E ) 2(4.00R E ) ∆E = − ∆E = Gm ME 1 1 Gm ME 1 � − �= + × 2R E 4 3 2R E 12 Gm ME 6.67 × 10−11 × 250 × 5.98 × 1024 = J 24R E 24 × 6.37 × 106 ⇒ ∆E = 6.52 × 108 J Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 11 Q21. Find the pressure increase in the fluid in a syringe when a nurse applies a force of 31.4 N to the syringe’s circular piston, which has a radius of 1.00 cm. A) B) C) D) E) 1.00 × 105 Pa 1.50 × 104 Pa 2.00 × 103 Pa 3.00 × 106 Pa 5.00 × 103 Pa Ans: P= F 31.4 N = = 9.99 × 104 N/m2 A π(10−2 )2 Q22. In Figure 4, a spring of spring constant 5.00 × 104 N/m is between a rigid beam and the output piston of a hydraulic lever. An empty container with negligible mass sits on the input piston. The input piston has area A, and the output piston has area 20 A. Initially the spring is at its rest length. How many kilograms of sand must be (slowly) poured into the container to compress the spring by 4.00 cm? A) B) C) D) E) 10.2 kg 100 kg 9.80 kg 980 kg 250 kg Fig # 4 Ans: F2 kx F1 = = A2 A1 A2 F1 = kx � A1 � A2 A F1 = 5 × 104 × 4 × 10−2 × �20 A� m= F1 g = 5×104 ×4×10−2 20×9.8 = 102 9.8 kg = 10.2 kg Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 12 Q23. A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 50.0 cm, and the density of iron is 7.87 g/cm3. Find the inner diameter. A) B) C) D) E) 47.8 cm 12.5 cm 15.5 cm 7.87 cm 25.0 cm 𝑟𝑜𝑢𝑡 𝑟𝑖𝑛 Ans: ρf Vsubmerged g = ρ0 V0 g 4 4 1.0 × 3 π(50)3 = 7.87 × 3 π(503 − rin 3 ) (50)3 = 7.87(503 − rin 3 ) ⇒ rin 3 = (50)3 − (50)3 ⇒ rin = 47.8 cm 7.87 Q24. Two streams merge to form a river. One stream has a width of 8.0 m, depth of 4.0 m, and current speed of 2.0 m/s. The other stream is 7.0 m wide and 3.0 m deep, and flows at 4.0 m/s. If the river has a width of 10.0 m and a speed of 4.0 m/s, what is its depth? A) B) C) D) E) 3.7 m 4.0 m 3.0 m 8.0 m 7.0 m Ans: A1 v1 + A2 v2 = A3 v3 32 × 2 + 21 × 4 = 10 d × 4 64 + 84 = 40 d d = 3.7 m Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 13 Q25. A liquid of density 1.00 × 103 kg/m3 flows through a horizontal pipe that has a crosssectional area of 2.00 × 10 − 2 m2 in region A and a cross-sectional area of 10.0 × 10 − 2 m2 in region B. The pressure difference between the two regions is 10.0 × 103 Pa. What is the mass flow rate between regions A and B? A) B) C) D) E) 91.3 kg/s 87.3 kg/s 857 kg/s 576 kg/s 100 kg/s Ans: A1 v1 = A2 v2 2 × 10−2 v1 = 10 × 10−2 v2 ⇒ v1 = 5v2 0 1 1 P1 + 2 ρv1 2 + ρgy1 = P2 + 2 ρv2 2 + ρgy2 0 1 1 P2 − P1 = ρ(v1 2 − v2 2 ) = × 103 �25v2 2 − v2 2 � 2 2 ∴ 10 × 103 = v2 2 = 1 2 × 103 × 24v2 2 10 10 ⇒ v2 = � = 0.9313 m/s 12 12 ∴ Mass �low rate: = A2 v2 ρ = 91.3 kg/s Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 14 Q26. What is the phase constant for the harmonic oscillator with the velocity function v(t) given in Figure 5 if the position function x(t) has the form x(t) = x m cos(ωt + φ)? The vertical axis scale is set by 𝑣 s = 4.00 cm/s. R Fig# 5 A) B) C) D) E) −53.1o +53.1o −5.24o +5.24o −40.0o Ans: From Fig. vs = 4, v = v= 4 × 5 = 5 cm/s 4 5 v 4 s x = xm cos(ωt + ϕ) dx = −xm ωsin(ωt + ϕ) dt vs = −vsin(ωt + ϕ) 4 = −5sin(ϕ); t = 0 sin(ϕ) = − 4 ⇒ ϕ = sin−1 (−4/5) = −53.1° 5 Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 15 Q27. If the phase angle for a block-spring system in SHM is π/6 rad and the block’s position is given by x(t) = x m sin(ωt + φ), what is the ratio of the kinetic energy to the potential energy at time t = 0? A) B) C) D) E) 3 1/3 9 1/9 5 Ans: Q28. x (t) = xm sin(ωt + ϕ) ; v = ωxm cos(ωt + ϕ) 1 1 U = kx 2 = kxm 2 sin2(ϕ) 2 2 1 1 1 k = mv 2 = mω2 xm 2 cos(ϕ) = k xm 2 cos 2 (ϕ) 2 2 2 2 π cos �6� cosϕ k = � ∴ = π � sinϕ U sin �6� k ∴ =3 U Find the mechanical energy of a block-spring system having a spring constant of 2.0 N/cm and an oscillation amplitude of 3.0 cm. A) B) C) D) E) 9.0 × 10−2 J 7.8 × 10−1 J 12 × 10−3 J 13 × 10+2 J 5.0 × 10−3 J Ans: E= 1 2 kxm 2 = 1 �2.0 2 N 102 cm �� cm 1m −2 = 102 × 9 × 10−4 = 9 × 10 E = 9 × 10−2 J � × (3 × 10−2 m)2 J Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 16 Q29. In Figure 6, a stick of length L = 1.73 m oscillates as a physical pendulum. What value of 𝑥 between the stick’s center of mass and its pivot point O gives the least period? Fig# 6 A) B) C) D) E) 0.50 m 0.87 m 1.73 m 0.58 m 0.43 m Ans: T = 2π� I= I ;d = x mgd 1 mL2 + mx 2 12 1 1 2 2 2 mL + mx � T = 2π �12 mgx 1 mL2 + mx 2 12 2 2 T = 4π � � mgx 4π2 1 2 −1 T = � L x + x� g 12 2 2T dt 4π2 1 = �− L2 x −1 + 1� = 0 dx g 12 (1.73)2 L2 L2 2 = 1 ⇒ x = = 12x 2 12 12 or x 2 = 0.249 ⇒ x = 0.499 m Phys101 Coordinator: Dr. G. Khattak Final-113 Thursday, August 02, 2012 Zero Version Page 17 Q30. Figure 7 shows the kinetic energy K of a simple pendulum versus its angle θ from the vertical. The vertical axis scale is set by 𝐾𝑠 = 20.0 mJ. The pendulum bob has mass 0.30 kg. What is the length of the pendulum? Fig# 7 A) B) C) D) E) 2.04 m 1.25 m 2.25 m 0.500 m 3.05 m Ans: θ = 100 × 10−3 rad = 5.73° k max = Umax = mgL(1 − cosθ) ∴ 30 × 10−3 = (0.30)(9.8)(L)(1 − Cos 573°) 30 × 10−3 = (0.30)(9.8)(L)(1 − 0.995) 30 × 10−3 L= (0.30)(9.8)(0.00499) L = 2044 × 10−3 m = 2.04 m