# Solution to phys101-t112-final Exam

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1 Solution to phys101-t112-final Exam Q1. An 800-N man stands halfway up a 5.0-m long ladder of negligible weight. The base of the ladder is.0m from the wall as shown in Figure 1. Assuming that the wall-ladder contact is frictionless, then the magnitude of normal force of the wall on the ladder is: The moment about the lower contact point gives: mg (1.5) = N (4) N = 00 N A) 00 N B) 150N C) 400 N D) 600 N E) 800 N Q2. A cube with edges exactly 2.0 m long is made of material with a bulk modulus of N/m 2. When it is subjected to a pressure of Pa its change in the volume is: B 5 P PV 7.0 x 10 8 = V = = = m 9 V / V B.5x10 A) m B) m C) m D) m E) m 1

2 Q. A uniform beam of length 6.0 m and mass 150 kg is pivoted to a vertical wall at point O and is suspended horizontally by a rope of negligible mass making an angle θ = 60 with the wall as shown in Figure 2. An unknown mass M is hanged at point P, 4.0 m away from the pivot point O. If the system is in equilibrium as shown with the tension in the rope equal to N, what is the value of mass M? The moment at O implies: A) 52 kg B) 9 kg C) 127 kg D) 100 kg E) 2 kg o cos mg (4) + Mg () = T cos θ(6) m = = 52.0 kg

3 Q4. A picture is to be hung from the ceiling by means of two wires as shown in Figure. Order the following arrangements of the wires according to the tension in wire B, from least to greatest. A) II, I, III B) I, II, III C) III, I, II D) I, III, II E) III, II, I Q5. A satellite is put in a circular orbit about Earth with radius = 8 R E and period of T 1. The satellite had been moved to another circular orbit of radius 2 R E, and its period became T 2. The ratio T 1 / T 2 will be equal to: Kepler s law implies: A) 8.00 B) C) D) 0.50 E) 4.00 T T /2 /2 r 8 = = = = r 2 2 8

4 Q6. A planet X has radius and mass equal to R E M and 4 8E respectively, where RE and M E are Earth s radius and mass. If the escape velocity of an object from the surface of Earth is 11.2 km/s, then escape velocity of the same object from the surface of the planet X would be: 2GM M 2GM V escape ( Moon) R M M M R E 14 1 V escape = = = = = R V ( ) 2 escape Earth GM E M E R M 81 2 R E 1 1 V escape ( Moon ) = V escape ( Earth ) = (11.2) = 7.92 km/s 2 2 A) 7.92 km/s B) 15.8 km/s C) 5.60 km/s D) 22.4 km/s E) 1.01 km/s Q7. In space, sphere A of mass 20.0 kg is located at the origin of an x axis and sphere B of mass 10.0 kg is located on the axis at x = 0.80 m. Sphere B is released from rest while sphere A is held at the origin. What is the kinetic energy of B when it has moved 0.20 m toward A? Conservation of the total energy implies: mm mm G = G + K K = GMm ( ) = J A) J B) J C) J D) J E) J Q8. If the radius of a star were to reduce by 50%, while it s mass remain the same, the acceleration due to gravity on the star s surface would: A) increase by a factor of 4 B) decrease by a factor of 4 C) increase by a factor of 8 D) decrease by a factor of 8 E) decrease by a factor of 16 4

5 Q9. A column of oil of height 70.0 cm supports a column of an unknown liquid as suggested in the Figure 4 (not drawn to scale). Assume that both liquids are at rest and that the density of the oil is kg/m. Determine the density of the unknown liquid. 2 ρ LhL Po +ρ LghL = Po +ρrghr ρ R = = = kg / m h 27 A) kg/m B) kg/m C) kg/m D).6 10 kg/m E) kg/m R Q10. For the hydraulic lift systems shown in Figure 5, rank in order from largest to smallest, the magnitudes of the forces F a, F b and F c required to balance the masses? The masses are in kilograms. F > F = F F > F > F F = F = F F < F < F F > F > F A) b a c B) a b c C) a b c D) b a c E) c b a 5

6 Q11. Figure 5 shows a 2.00-kg block tied, by string, to a bottom of a container filled to the rim with water. If the displaced water has a mass of 5.00 kg, find the tension in the string. Draw the free body diagram, then we can have mg + T = m g T = m m g = (5 2)9.8 = 29.4 N w ( ) w A) 29.4 N B) 10.2 N C) 22.8.N D) 7.00 N E) 100 N Q12. A bucket with m is to be filled through a pipe with m radius. If the water flows through the pipe end with a speed of m/s, how long does it take to fill the bucket completely? dv dv ( ) Rv = va = dt 162 s. dt = va = 0.61 ( ) = π 2 A) 162 s B) 170 s C) 119 s D) 280 s E) 490 s 6

7 Q1. A glass tube has several different cross-sectional areas with the values indicated in the Figure 6. A piston at the left end of the tube exerts pressure so that mercury within the tube flows from the right end with a speed of 8.0 m/s. Three points within the tube are labeled A, B, and C. What is the total pressure at point A? Atmospheric pressure is N/m 2 ; and the density of mercury is kg/m. First use the continuity equation AC 8 v AAA = vcac v A = vc = 6 = 4.0 m/s. A A 12 The pressures on either side of the junction must be equal. This requires: p + ρv A = po + ρvc p = po + ρ ( vc v A) = ( ) (8 4 ) = Pa 2 2 A) Pa B) Pa C) Pa D) Pa E) Pa 7

8 Q14. In Figure 7, the horizontal block-spring system has a kinetic energy of K = 5.0 J and an elastic potential energy of U =.0 J, when the block is at x = cm. What are the kinetic and elastic potential energy when the block is at x = x m? Conservation of total energy implies: ( K + U) = ( K + U) A) K = 0 and U = 8J B) K = 5J and U = J C) K = 5J and U = - J D) K = 8J and U = 0 E) K = 0 and U = -8 J x = 2 x =± 2 5+ = 0+ U U = 8 Q15. A thin rod of length L = 1.5 m and mass M is pivoted at one end of the rod and is made to oscillate as a physical pendulum with frequency f. The value of f is: f A) 0.50 Hz B) 2.5 Hz C) 0.25 Hz D) 1.0 Hz E) 2.0 Hz ( /2) 1 1 mgh 1 mg L 1 g = = = = = = 0.5 Hz T 2π I 2π 1 2 ml 2π 2L 2π

9 Q16. At t = 0, a particle is located at x = 25.0 m and has a velocity of 12.5 m/s in the positive x direction. The acceleration (a x ) of the particle varies with time (t) as shown in Figure 8. What is the velocity of the particle at t = 5.00 s? A) +0 m/s B) 15 m/s C) +15 m/s D) 0 E) 1.2 m/s 1 v f v i = area(0 5 s ) = (1 + 6)5 = v = v = =+ 0.0 m/s f i 9

10 Q17. Figure 9 shows the trajectory of ball in Planet X. The ball s position is shown at 1.0 s 2.00 ˆi ˆj m/s. It reaches the intervals until t =.0 s. At t = 1.0 s, the ball s velocity is ( ) maximum height at t = 2.0 s. What is the value of g (in 2 m/s ) on this planet? Answer A) 2 B) C) 4 D) 9.8 E) 1 v x = 2.0 m/s; v ( t ) = v gt v (2) = 2.0 g (1) = 0 y oy y g = 2 m/s 2 Q18. A 2.00-kg stone is tied to a m string and whirled at a constant speed of 4.00 m/s in a vertical circle. The tension in the string at the bottom of the circle is: At the bottom of the circle the free body diagram implies: v v 4 m = T mg T = mg + m = 2(9.8 + ) = 8.6 N,up A) 8.6 N, up B) 44.4 N, up C) 9.80 N, up D) 44.4 N, down E) 8.6 N, down r r

11 Q19. A 6.00-kg block is in contact with a 4.00-kg block on a frictionless surface as shown in Figure 10. The 6.00-kg block is being pushed by a 20.0-N force toward the 4.00-kg block. What is the magnitude of the force of the 6.00-kg block on the 4.00-kg block ma = F 1 0= = F = 4a= 8N 4,6 2 a 2 0 a 2m/s, A) 8.00 N B) 12.0 N C) 6.00 N D) 4.00 N E) 10.0 N Q20. A 16-kg fish is weighed with two identical, massless, spring scales, each of negligible weight, as shown in Figure 11. What will be the readings on the scales? A) The sum of the two readings will be 2 kg B) The bottom scale will read 16 kg, and the top scale will read zero C) The top scale will read 16 kg, and the bottom scale will read zero. D) Each scale will show a reading greater than zero and less than 16 kg, but the sum of the two readings will be 16 kg E) Each scale will read 8 kg. 11

12 Q21. When you travel, you always exert less force to pull a block M instead of pushing it, see Figure 12. That is F(pull) < F(push). Why? A) Because the normal force becomes less while pulling B) Because the normal force becomes more while pulling C) Because the normal force becomes zero while pulling D) Because the gravitational forces decreases while pulling E) No scientific reason it is just a habit. 12

13 Q22. A 1-N weight and a 12-N weight are connected by a massless string over a massless, frictionless pulley as shown in Figure 1. The downward acceleration of 1-N weight is: ( g is acceleration due to gravity) ( m2 m1) ( m + m ) 1 a = g = g A) g/25 B) g/12 C) g/1 D) g E) g/2 Q2. At t = 0, a force F acts on 2-kg particle that has an initial velocity of ( iˆ jˆ) velocity became ( 2.0iˆ.0 jˆ) m/s force was: A) 12 J B) 4 J C) 18 J D) 40 J 4.0 iˆ+ 6.0 jˆ J E) ( ) m/s. The + at t = s. During this time interval the work done by the external m W = K = ( v f v i ) = ( 1 25 ) = 12 J 2 2

14 Q24. A 2.0-kg block is attached to a horizontal ideal spring with a spring constant of 200 N/m. When the spring in its equilibrium position, the block is given a speed of 5.0 m/s. What is the maximum extension of the spring? m m 2 ( v i ) = kx x =± v = 5 = 0.5 m 2 2 k 200 A) 0.50 m B) 0.05 m C) 0.25 m D) 10 m E) 0.12 m Q25. At t = 0, a horse pulls a cart with a force of 180 N at an angle of 0 above the horizontal and moves horizontally at a speed of 1.20 m/s. What is the instantaneous power (in Watts) of the force at t = 0? o P = Fv. = cos0 = 187 Watts A) 187 B) 150 C) 108 D) 216 E) 0 Q26. Particle 1 with mass 2.0 kg and velocity v1 ( 5.0 ˆ i i ) m/s elastic collision with particle 2 with mass 2.0 kg and velocity v 2 ( ˆ i i ) collision, the final velocities of particle 1 ( v 1f ) and particle 2 (v 2f ) are: Use the equations in the formula sheet, we can have: v1f = (- 6.0 i) m/s, v2f = (5.0 i) m/s A) v ( 1f = i ) v 2f = ( iˆ) B) v ( ˆ 1f = i ) v 2f = ( iˆ) C) v ( ˆ 1f = i ) v 2f = ( iˆ) D) v ( ˆ 1f = i ) v 2f = ( iˆ) E) v ( ˆ 1f = i ) v 2f = ( iˆ) 6.0 m/s, 5.0 m/s 6.0 m/s, 5.0 m/s 11.0 m/s, 0.0 m/s 0.0 m/s, 11.0 m/s 6.0 m/s, 5.0 m/s = m/s undergoes a one-dimensional = 6.0 m/s. After the 14

15 Q27. A 2.00 kg particle has the xy coordinates (-1.20 m, m), and a 4.00 kg particle has the xy coordinates (0.600 m, -1.0 m). Both lie on a horizontal plane. At what (x,y) coordinates must you place a.00 kg particle such that the center of mass of the three-particle system has the coordinates (0.00 m, 0.00 m)? mx 1 1+ mx mx 2 ( 1.2) + 4( 0.6) + x X com = 0= x = 0 m + m + m y A) (0.0, 1.0) m B) (1.0, 2.0) m C) (0.5, -1.0) m D) (-1.2, 1.0) m E) (-0.5, -1.5) m 1 2 my + my + my ( ) + ( ) + y y 1 m com = = = m1+ m2 + m Q28. Five objects of mass m are under a force F at a distance from an axis of rotation perpendicular to the page through the point A, as shown in Figure 14. The one (or ones) that has zero torque about the axes through A is: A) c, d B) b, a C) a, e D) d E) e 15

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