Q1. A) 46 m/s B) 21 m/s C) 17 m/s D) 52 m/s E) 82 m/s. Ans: v = ( ( 9 8) ( 98)

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Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: Q. A hot-air balloon is ascending (going up) at the rate of 4 m/s and when the balloon is 98 m above the ground a package is dropped from it, vertically downward. With what speed does the package hit the ground? A) 46 m/s B) 2 m/s C) 7 m/s D) 52 m/s E) 82 m/s v 0 = 4 m s a = 9 8 m s 2 v = 0 y = 98 m v 2 = v 0 2 + 2aΔy v = (4 2 + 2( 9 8) ( 98) v = 46 m/s Q2. A sailor in a small boat sails (moves) 2.00 km east, then 3.50 km southeast, and then an additional distance d. His final position is 5.80 km directly east of his starting point as shown in Figure. Find the distance d. A) 2.8 km B) 3.7 km C) 2.02 km D) 4.0 km E) 5.32 km Figure 2.0 + 3 5 cos 45 0 + d x = 5 8 d x = 3.8 3.5 cos45 3.5 sin45 + d y = 0 d y = 3.5 sin45 d = d x 2 +d y 2 = 2.8 m/s

Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: 2 Q3. An object is in uniform circular motion with radius 3.00 m. At one instant its acceleration is a = (6.00i ˆ 4.00 ˆj ) m/s 2. What is the speed of the object at that instant? a = 6i 4j A) 4.65 m/s B).23 m/s C) 7.5 m/s D) 9.2 m/s E) 2.45 m/s a a = 6 2 + ( 4) 2 r = 3 m a = ω 2 r ω = ar = 4.65 m/s Q4. Figure 2 shows four blocks being pulled across a frictionless floor by a force F = 40 N. Find the magnitude of tension in cord 2. A) 26 N B) N C) 40 N D) 93 N E) 65 N Figure 2 F = Σma 40 = 20 a a = 2 m/s 2 T 2 = 3a = 3 2 = 26N

Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: 3 Q5. In Figure 3, a person drives a car over the top of a hill with a circular arc of radius 250 m. At what greatest speed he can drive without the car leaving the road at the top of the hill? A) 49.5 m/s B) 9.8 m/s C) 2.4 m/s D) 79.2 m/s E) 33.6 m/s F n mg = mv2 r 0 mg = mv 2 max r v max = rg Figure 3 v max = 250 9.8 = 49 5 m/s Q6. A 2.0 kg object, on a frictionless horizontal table, is subjected to three forces that give 2 2 it an acceleration a = (8.0m/s ) i ˆ+ (6.0m/s ) ˆj. If two of the three forces are F (30N) ˆ (6N) ˆ i j and F (2N) ˆ (8.0N) ˆ 2 i j, find the magnitude of the third force. A) 36 N B) 2 N C) 24 N D) 48 N E) 60 N F + F 2 + F 3 = ma 30i + 6j 2i + 8j + F 3 = 2( 8i + 6j ) F 3 = 34i 2 j F 3 = 34 2 + 2 2 = 36N

Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: 4 Q7. A 5.0 kg block initially at rest, slides down the ramp of an inclined plane of angle 30 o, from the height of 5.0 m. Find the speed of the block at the end of the ramp if the coefficient of kinetic friction between the block and the surface of the ramp is 0.23. A) 7.7 m/s B) 4.2 m/s C) 3.5 m/s D).6 m/s E) 2.3 m/s Δk + U = W f Figure 4 30 F N mgcosθ sinθ = h d 2 mv2 mgh = f k d 2 mv2 mgh = μ k mgdcosθ v = 2g (h μ k h sinθ cosθ) v = 2 9.8 5( 0.23 cot30 ) = 7.7 m/s Q8. A box open from the top slides across a frictionless horizontal surface. What happen to the box as water from a rain shower falls vertically downward into the box? A) Its speed decreases B) Its speed increases C) Its speed remains constant D) Its momentum changes E) The vertical falling rain has no effect on the box No externalf force, P i = P f P i = mv 0 P f = (m + m)v P f = P i m v = ( m + m ) v 0 speed decreases

Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: 5 Q9. The average power needed to spin (rotate about its axis) a uniform, solid disk of mass 5.0 kg and radius 0.50 m from rest to a final angular velocity ωf in 3.0 s is 2.6 W. The value of ωf is: P av = k t A) 5.0 rad/s B) 3.9 rad/s C) 2.7 rad/s D).3 rad/s E) 8.5 rad/s P av t = 2 Iω f 2 0 ω f = 2P avt 0.5mR 2 ω f = 2 2.6 3 = 5.0 rad/s 0.5 5 (0.5) 2 Q0. A uniform rod of length L = 40 cm and mass M = 2.0 kg is free to rotate on a frictionless pin (pivot) passing through one end as shown in Figure 4. The rod is released from rest in the horizontal position, determine the angular speed of the rod when it is in the vertical position. Figure 4 A) 8.6 rad/s B) 2.5 rad /s C).4 rad /s D) 5.5 rad /s E) 4.9 rad /s Δk + U = 0 2 Iω2 mg L 2 = 0 ω = 2mg (L 2 ) 3g = L = 3 9.8 = 8.6 rad/s 0.4 3 ml2

Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: 6 Q. What increase in pressure is necessary to decrease the volume of a sphere by 0.23%? Take the bulk modulus of the sphere B = 2.8 0 0 N/m 2. B = p dv V A) 6.4 0 7 N/m 2 B) 6.4 0 9 N/m 2 C).2 0 7 N/m 2 D).2 0 9 N/m 2 E) 5.7 0 0 N/m 2 p = B dv V = 2.8 00 0.23 00 = 6.4 07 N/m 2 Q2. Four particles, each of mass 25 g, that form a square of side length d = 0.60 m as shown in Figure 5. If the particles are forced to form a new square with side length d = 0.20 m, find the work done by the gravitational force. [consider the system as isolated system] Figure 5 A) 7.5 0-3 J B) 2.3 0-3 J C).2 0-3 J D) 3.6 0-3 J E) Zero W = ΔU = U 0 U W = G [ 4m2 d 0 + 2m2 4m2 2d 0 d 2m2 2d ] = Gm 2 [ 4 d 0 + 2 d 0 4 d 2 d ] = Gm 2 [ 4 + 2 d 0 4 + 2 ] d = (4 + 2)Gm 2 [ d d 0 ] m = 25 0 3 kg, d 0 = 0.6 m, d = 0.2 m W = 7.52 0 3 J

Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: 7 Q3. If the mass of Mars is 0. times that of Earth, and its radius is 0.53 that of Earth. Find the gravitational acceleration g at the surface of Mars. A) 3.8 m/s 2 B) 2.2 m/s 2 C) 4.9 m/s 2 D) 5.5 m/s 2 E) 8.0 m/s 2 mg mars = GM mars m r2 mars g mars = G 0.M E (0.53 r E ) 2 g mars = 0. g = 3.8 m/s2 (0.53 ) 2 Q4. A projectile is fired vertically from Earth s surface with an initial speed of 0 km/s. Neglecting air friction, how far above the surface of Earth will the projectile go? A) 2.5 0 7 m B) 3. 0 7 m C).9 0 7 m D) 6.4 0 6 m E).0 0 4 m K + ΔU = 0 (K K 0 ) + (U U 0 ) = 0 2 mv 0 2 2 v 0 2 GMm (r E + h) + GMm = 0 GM ( ) = r E r E r E+h 2 v 0 2 GM = r E r E+h = ( v 2 0 r E+h r E 2GM ) r E+h + h = r E GM 2 2GM v 0 2 r E h = ( r E GM 2 2GM v 0 2 r E ) r E r E = 6.37 0 6 m; M = 9.98 0 4 kg; G = 6.67 0 Nm 2 /kg 2 ; v 0 = 060 m/s h = 2.52 0 7 m

Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: 8 Q5. Two satellites are orbiting the Earth; satellite X is eight times as far from the Earth's center as is satellite Y. The period of the satellite X is: (note Ty is the period of Satellite Y) A) 22.6 Ty B) 52.8 Ty C) 4.4 Ty D).32 Ty E) 72.5 Ty T 2 = 4π2 GM r3 T y 2 T X 2 = 4π 2 GM r y 3 4π 2 GM (8r y) 3 T X 2 = 8 3 T y 2 T X = 52T y T X = 22.6 T y Q6. Which of the following best describes the energy of a satellite rotating about the Earth? A A) The total mechanical energy is constant. B) The kinetic energy is three times larger than its potential energy. C) The kinetic energy is three times smaller than its potential energy. D) Its kinetic energy does not depend on its mass. E) Its potential energy is always zero.

Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: 9 Q7. In Figure 6, a cube of edge length L = 0.600 m and mass 450 kg is suspended by a rope in an open (from the top) tank of liquid of density 030 kg/m 3. Find the tension in the rope. T T + F B = mg A) 2.23 0 3 N B) 7.8 0 3 N C) 4.4 0 3 N D).05 0 3 N E) 3.96 0 4 N mg F B Figure 6 T + m l g = mg T = (m m l )g T = (m ρ l L 3 )g m = 450 kg; ρ l = 030 kg/m 3 ; L = 0.6 m T = 2.23 0 3 N Q8. A cylindrical tank with a large diameter is filled with water to a depth D = 0.30 m. A hole of cross-sectional area A= 6.5 cm 2 in the bottom of the tank allows water to flow out. What is the rate at which water flows out? A).57 0-3 m 3 /s B) 2.42 0-3 m 3 /s C) 8.5 0-4 m 3 /s D) 3.06 0-3 m 3 /s E) 4.5 0-3 m 3 /s P = P 0 ; v = 0 D = 0.3 m ρgh + 2 ρv 2 + P = ρgh 2 + 2 ρv 2 2 + P 2 P 2 = P 0 ; h 2 = 0 ρgd + 0 + P 0 = 0 + 2 ρv 2 2 + P 0 v 2 = 2gD = 2.42 m/s V t = Ax t = Av 2 A = 6.5 0 4 m 2 V t =.57 0 3 m 3 /s

Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: 0 Q9. A circular pool of uniform depth is completely filled with water to a depth of.0 m. If the radius of the pool is.0 m what is the total force at the bottom of the pool? A) 3.5 0 5 N B) 2.3 0 6 N C) 4.2 0 5 N D) 6.5 0 5 N E) 5.7 0 4 N P b = P 0 + ρgh P 0 P b h =.0 m F b = P b A = (P 0 + ρgh) πr 2 P 0 =.0 0 5, h =.0 m, ρ = 000 kg/m 3, r =.0 m F b = 3.5 0 5 N Q20. A hollow spherical iron shell floats and is completely submerged in water. The outer radius is 60.0 cm, and the density of iron is 7.87 g/cm 3. Find the inner radius. mg = ρ w Vg A) 57.3 cm B) 77.5 cm C) 96.0 cm D) 9.2 cm E) 4.2 cm V i ρ = ρ w 4 3 πr 0 2 4 3 π(r 0 3 r 3 4 in )ρ = ρ w 3 πr 0 3 ρr 0 3 r in 3 ρ = ρ w r 0 3 r 3 in ρ = (ρ ρ w) 3 r ρ 0 r in = ( ρ w ρ ) 3 r0 ρ w =.0 g ; ρ = 7.8 g cm3 cm 3, r 0 = 60 cm r in = 57.3 cm

Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: Q2. A 5000-N car on a hydraulic lift rests on a cylinder with a piston of radius r2 = 0.20 m as shown in Figure 7. If a connecting cylinder with a piston of radius r = 0.040 m is driven by compressed liquid, find the force F that must be applied to this smaller piston in order to lift the car. Figure 7 P = P 2 A) 600 N B) 750 N C) 950 N D) 250 N E) 400 N F A = F 2 A 2 F = F 2 A A 2 F = 5000 πr 2 2 πr 2 r = 0.04 m, r 2 = 0.2 m F = 600 N Q22. Which one of the following statements is False about a particle which undergoes simple harmonic motion? A A) The kinetic energy of the particle is minimum at minimum displacement. B) The particle s motion repeats itself after a time period. C) The acceleration of the particle is not constant. D) The displacement of the particle in one complete time period is zero. E) The distance travelled by the particle in one complete time period is 4 times the amplitude.

Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: 2 Q23. A simple harmonic oscillator consists of a block attached to a horizontal spring while the other end is fixed to a wall. The block slides on a frictionless surface with equilibrium point x = 0 and amplitude 0.340 m. A graph of the block s velocity v as a function of time t is shown in Figure 8. Find the angular frequency ω of the block. v max = ωx m ω = v max x m A) 8.5 rad/s B) 23.6 rad/s C) 7.58 rad/s D) 2.2 rad/s E) 6.28 rad/s Figure 8 ω = 2π 0.340 = 8.5 rad/s Q24. The displacement of a particle oscillating along the x-axis is given as a function of time according to the equation: x(t) = 0.40cos(0.20 t+ /2), where x is in m and t in s. What is the total distance traveled by the particle in 35 s: A) 5.6 m B) 4. m C) 3.0 m D).2 m E) 7.4 m T = 2π ω = 2π = 0 s 0.2 π 4x m 0 = x 35 x = 4x m 35 0 x = 5.6 m

Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: 3 Q25. A solid circular disk of radius R is oscillating with time period T in a vertical plane about the pivot point P as shown in Figure 9. If the oscillating disk has to be replaced by a simple pendulum of same time period T, what should be the length of the simple pendulum? Figure 9 A) 3R/2 B) 2R C) R D) R/2 E) 3R T = 2π I mgl T = 2π L g L g = 2 mr2 + mr 2 mgr L g = 3 2 mr2 mgr L = 3R 2