= M. L 2. T 3. = = cm 3

Size: px

Start display at page:

Download "= M. L 2. T 3. = = cm 3"

Miles Miles
5 years ago
Views:

1 Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: 1 Q1. Work is defined as the scalar product of force and displacement. Power is defined as the rate of change of work with time. The dimension of power is A) M L T B) M L T C) M L 3 T D) M L T E) M L T - W = F. d [W] = [F]. L = M. L T. L = ML. T P = dw dt [P] = [W] T = ML. T T = M. L. T 3 Q. The density of iron is 7.86 g/cm 3, and the mass of an iron atom is g. How many iron atoms are there in 1.00 in 3 of iron? (1 in =.54 cm) A) atoms 3 B) atoms 1 C) atoms 5 D) atoms 19 E) atoms Volume: V = 1.00 in 3 (.54)3 cm in 3 = 16.4 cm 3 Volume of 1 atom: v = m = = cm 3 ρ 7.86 The number of atoms is: N = V v = = atoms

2 Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: Q3. Two cars are initially at rest. Car A is at x = 0, and car B is at x = m. They start to move, at the same time, along the same line in the positive x direction with constant accelerations: a A = 4.00 m/s and a B = 1.00 m/s. How long does it take car A to overtake car B? A) 0.0 s B) 30.0 s C) 5.0 s D) 34.5 s E) 17.5 s Let d 1 = distance moved by A Let d = distance moved by B When they meet: d 1 = d (1) 0 d 1 = v 0 t + 1 a At = t = t () d = v 0 t + 1 a Bt = t = 1 t (3) From ()and (3), back to (1): t = 1 t t = 600 t = 600 = 400 t = 0.0 s 3 Q4. Which of the following statements is WRONG? A A) A body can have constant velocity and still have a varying speed. Not Possible B) A body can have a constant speed and still have a varying velocity. uniform circular motion C) A body can have zero velocity and still be accelerating. free fall at max. height D) A body can have a constant acceleration and a variable velocity. by definition E) A body can have an upward velocity while experiencing a downward acceleration. free fall

3 Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: 3 Q5. FIGURE 1 shows the motion of a particle moving along an x axis with constant acceleration. What is the magnitude of the acceleration of the particle? A) 4 m/s B) 6 m/s C) 3 m/s D) m/s E) 5 m/s x 0 = x(0) = m Figure # 1 x x 0 = v 0 t + 1 at x + = v 0 t + 1 at at t = 1 s: 0 + = v 0 + a v 0 = a at t = 6 s: 6 + = a () + a 8 = 4 a + a 4 = a a = 4 m/s Q6. An object is thrown vertically upward from the ground. When it reaches half of its maximum height, it has a speed of 19.6 m/s. What is the maximum height reached? A) 39. m B) 49.4 m C) 44.0 m D) 3.0 m E) 30.7 m v = v i g(y y i ) Apply this equation to points 1 and : v = v 1 g(y y 1 ) h/ h/ = (19.6) (19.6) h h h = 19.6 = 39. m

4 Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: 4 Q7. The two vectors shown in FIGURE lie in the xy plane. Which of the following vectors has positive x and y components? Figure # A) B A B) A B second quadrant C) B + A fourth quadrant D) A B second quadrant E) A B second quadrant A Q8. A vector A is defined by A =1. 50ˆi ˆj. Find a vector B that makes an angle of 60.0 o with A in the counterclockwise direction, and has a magnitude of 4.00 units. A) B = 1. 04ˆi ˆj B) B = 3. 86ˆi ˆj C) B =1. 04ˆi ˆj D) B = 1. 04ˆi 3. 86ˆj E) B = 3. 86ˆi 1. 04ˆj Since A x = A y θ A = 45 (irst quadrant) θ B = = 105 B x = B. cosθ B = 4.00 cos 105 = 1.04 B y = B. sinθ B = 4.00 cos 105 = +3.86

5 Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: 5 Q9. Three vectors are given by A= 1.0ˆi+ 1.0ˆj, B = 1. 0ˆj kˆ, andc = 1. 0ˆi kˆ. Find A (B C). ı A).0 B) 4.0 C) +.0 D) E) 0 B C = ȷ + k ı + k = (ȷ ı ) ȷ k k ı k ȷ = k ı ȷ A. B C = (ı + ȷ ). ı ȷ k = =.0 Q10. At time t = 0, a particle moving in the xy plane leaves the origin with a velocity of v o = 3. 0ˆi ˆ j (m/s). The particle moves with a constant acceleration of a =1. 0ˆi 5. 0ˆj (m/s ). What is the position vector of the particle at t = 4.0 s? A) 0 ˆ i 0 ˆj m B) 1. 5ˆi +. 5ˆj m C) 7. 0ˆi 15ˆj m D) 0. 75ˆi ˆj m E) 4.0î m Constant acceleration: r = r 0 + v 0 t + 1 a t = 0 + (3ı + 5ȷ )(4) + (ı 5ȷ )(8) = 1ı + 0ȷ + 8ı 40ȷ = 0ı 0ȷ m

Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: 6 Q11. A ball is thrown horizontally from the edge of the top of a building of height 39.6 m.

6 Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: 6 Q11. A ball is thrown horizontally from the edge of the top of a building of height 39.6 m. The ball just reaches the top of another building of height 0.0 m that is at a horizontal distance of 5.0 m away (see FIGURE 3). What is the initial speed of the ball? A) 1.5 m/s B) 10. m/s C) 50.4 m/s D) 9.80 m/s E) 19.9 m/s Figure # 3 Vertical: y y 0 = v 0y t 1 gt = t t =.00 s Horizontal: x = v 0x t v 0 = v 0x = x t = 5.0 = 1.5 m/s.00 Q1. A particle is moving with constant speed around a circle that is centered at the origin. At time t = 0, the particle is at (3.0, 0) m. At t = 3.0 s, the particle reaches the point (-3.0, 0) m for the first time. What is the magnitude of the acceleration of the particle? A) 3.3 m/s B) 1.0 m/s C) 13 m/s D) 8.8 m/s E) 14 m/s In the given time, the particle has moved half a circle. Period = T = 3.0 = 6.0 s a = v R (1) T = πr v v = πr T () From () (1): a = 1 R. 4π R T = 4π R = 3.3 m/s T

7 Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: 7 Q13. A train moves due East at 6.00 m/s along a level, straight track. A boy on the train rolls a ball along the floor with a speed of 3.00 m/s relative to the train. The ball is rolled directly across the width of the train from South to North. What is the speed of the ball relative to a stationary observer on the ground? v TG A) 6.71 m/s N B) 3.46 m/s v C) 5.0 m/s BT v BG W E D) 9.00 m/s E) 3.00 m/s S T: train; B: ball; G: ground V BG = v TG + v BT = = 6.71 m/s Q14. A block of mass M is hung by ropes as shown in FIGURE 4. The system is in equilibrium. Which of the following statements is CORRECT concerning the magnitudes of the three forces? A) F 1 = F = F3 B) F = F3 C) F < F3 D) F1 = F = F 3 / E) F1 > F 3 x component: F 1x F x = 0 F 1x = F x F 1. cos 30 = F. cos 30 Figure # 4 F 1 = F y component: F 1y + F y F 3 = 0 F 3 = F 1y + F y = F 1y + F 1y = F 1y = F 1. sin30 = F 1

Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: 8 Q15. Two blocks, with m 1 =.5 kg and m = 1.5 kg, are in contact on a horizontal frictionless table.

8 Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: 8 Q15. Two blocks, with m 1 =.5 kg and m = 1.5 kg, are in contact on a horizontal frictionless table. A horizontal force F is applied to the larger block, as shown in FIGURE 7. If the magnitude of the force between the two blocks is 1. N, what is the magnitude of the applied force F? A) 3. N B).0 N C).4 N D) 5.6 N E) 1. N consider m a m F 1 Figure # 7 m a = F 1 a = F 1 = 1. = 0.80 m/s m 1.5 Now, consider the system a M F = M. a = ( )(0.80) = 3. N F Q16. FIGURE 5 shows a system of two masses, with m 1 = 5.00 kg and m =.00 kg. Calculate the angle θ for which the system is at rest. Assume the pulley to be massless and frictionless. The incline is frictionless. Figure # 5 A) 3.6 B) 35. C) 63.5 D) 45.0 E) 50.3 For m 1 : T = m 1 g. sinθ (1) For m : T = m g () F N θ θ m, g T Equate (1)and (): m 1 g. sinθ = m g m sinθ = m =.00 = θ = 3.6 m m g

9 Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: 9 Q17. A 75 kg person lowers himself from rest to the ground by means of a rope that passes over a frictionless pulley and is attached to a 60 kg box, as shown in FIGURE 6. The person and the box move along vertical lines. What is the magnitude of the acceleration of the person? Figure # 6 T A) 1.1 m/s B) 9.8 m/s C) 5.4 m/s D) 3.6 m/s E) 4.6 m/s person: m 1 a = T m 1 g m 1 a = T + m 1 g (1) box: m a = T m g () Add (1)and (): (m 1 + m )a = (m 1 m )g a = m 1 m m 1 + m. g = = 1.1 m/s Person: Box: m 1 m 1 g T m m g a a Q18. A 3.00 kg block starts from rest at the top of a 30.0 incline and slides a distance of.00 m down the incline in 1.50 s. Calculate the magnitude of the frictional force acting on the block. A) 9.37 N Y B) 13.5 N C) 1.50 N F N D) 14.7 N X E) 8.98 N f k a Calculate a: θ 0 x = v 0 t + 1 at θ mg a = x t = = 1.78 m/s.5 Now, apply Newton s nd law: ma = mgsinθ f k f k = m(gsinθ a) = 9.37 N

10 Phys101 First Major-1 Zero Version Sunday, March 03, 013 Page: 10 Q19. Q0. An object is given an initial speed of 10 m/s and slides on a horizontal rough surface. It moves a distance of 40 m before coming to rest. During deceleration, the only horizontal force acting on the object is the force of friction. What is the coefficient of kinetic friction between the object and the surface? A) 0.13 B) 0.83 C) 0.31 D) 1.3 E) 0.04 Calculate a: v 0 f = v i ax a = v i x = 100 = 1.5 m/s 80 Apply Newton s nd law: ma = f k ma = µ k. F N = µ k. mg µ k = a g = = 0.13 A highway curve has a radius of 100 m and is designed for a speed of 50.0 km/h. At what angle should the curve be banked so that a car rounding the curve will not slip? f k mg F N A) 11.1 B) 39.0 C) 15.7 D) 16.3 E) 1.6 v = 50.0 km h. 1 h 3600s. 100 m = 13.9 m/s 1 km F r = F N. Sin θ (1) F N F r θ mg mg = F N. Cos θ () Divide(1)by (): F r mg = tan θ 1 tan θ = mg. mv R = v Rg = (13.9) = θ = 11.1

Phys101 First Major-111 Zero Version Monday, October 17, 2011 Page: 1

Phys101 First Major-111 Zero Version Monday, October 17, 2011 Page: 1 Monday, October 17, 011 Page: 1 Q1. 1 b The speed-time relation of a moving particle is given by: v = at +, where v is the speed, t t + c is the time and a, b, c are constants. The dimensional formulae