Department of Natural Sciences Clayton State University. Physics 1111 Quiz 1

Transcription

1 Physics 1111 Quiz 1 January 14, 008 Name SOLUTION 1. If the velocity of the object, v, is related to time, t, according to the relation v = A + B t, the constant, A, has the dimension of which of the following? a. Acceleration. b. Velocity. c. Time. d. Length.. The kilogram is currently defined as a. The mass of a platinum-iridium cylinder kept in Sevres, France. b. The mass of x 10 6 protons. c. The mass of one liter of pure water, free of air, at standard temperature and pressure. d. The mass of a cube of pure water, free of air, 10.0 cm on each side, at standard temperature and pressure. 3. Given that a = 1.40 m and c = 3.50 m, find:

2 a. The length of side b. a + b = c b = (c - a ) 1/ b = ((3.50 m) - (1.40 m) ) 1/ = 3.1 m b. The cosine of angle cos = b/c = (3.1 m)/(3.50 m) = c. The angle = cos -1 (0.917) = 3.5 o Physics 1111 Quiz January 3, 008 Name SOLUTION 1. If you start from the Bakery, travel to the Café, and then to the Art Gallery, what is your displacement? a km. b..50 km. c km. d km.. If you start from the Cafe, travel to the Bakery, and then to the Art Gallery, in 1.00 hour, what is your average velocity? a..50 km/hour.

3 b km/hour. c km/hour. d km/hour. 3. Given the position-versus-time graph for a basketball player traveling up and down the courting a straight-line path find the instantaneous velocity of the player a. At t =.00 s, V = rise/run = (6.00 m)/(4.00 s) = 1.50 m/s b. At t = 5.00 s, V = rise/run = (-3.00 m)/(.00 s) = m/s c. At t = 8.00 s. V = 0 (tangent is horizontal at t = 8.00 s)

4 Physics 1111 Quiz 3 January 30, 008 Name SOLUTION A stone is thrown vertically upward with a speed of 18.0 m/s. a. What is the maximum height the stone reaches? V y = V y0 - g (y y 0 ) 0 = V y0 - g (y 0) V y0 = g y y = V y0 /( g) y = 16.5 m b. How long is required to reach this height? V y = V y0 - gt 0 = V y0 - gt V y0 = gt t = V y0 /g t = 1.83 s c. What is velocity and acceleration of the stone at the moment it reaches the heigest point? V y = 0 a = -g d. The stone is caught at the same elevation as it was thrown from. What

5 is its velocity just before it is caught? V y = m/s c. Why are there two answers to (b)? Physics 1111 Quiz 4 February 6, 008 Name _SOLUTION Deep inside an ancient physics text you discover two vectors: A: o B: o Not content with these hoary relics, you are asked to find a new vector R = A + B. Note: Find the magnitude and direction of vector R. A x = A cos( A ) = (150 m) cos(30.0 o ) = 130 m A y = A sin( A ) = (150 m) sin(30.0 o ) = 75.0 m B x = B cos( B ) = (0 m) cos(-45.0 o ) = 156 m B y = B sin( B ) = (0 m) sin( o ) = m

6 R x = A x + B x = (130 m) + (156 m) = 86 m R y = A y + B y = (75.0 m) + (- 156 m) = m R = ( R x + R y ) 1/ = ( (86 m) + (-81.0 m) ) 1/ = 97 m R = tan -1 (R y /R x ) = tan -1 ((-81.0 m)/(86 m)) = o Use the figure above to graphically add these two vectors to check your answer. Physics 1111 Quiz 5 February 13, 008 Name

7 A projectile is fired with an initial speed of 35.0 m/s at an angle of 45.0 o above the horizontal on a long flat firing range. Determine a. The components of the projectile s initial velocity. b. The time of flight. y = y 0 + V y0 t ½ gt 0 = 0 + V y0 t ½ gt 0 = t (V y0 ½ gt ) V y0 ½ gt = 0 t = V y0 /g t = 5.05 s V x0 = V 0 cos( 0 ) = (35.0 m/s) cos(45.0 o ) = 4.7 m/s V y0 = V 0 sin( 0 ) = (35.0 m/s) sin(45.0 o ) = 4.7 m/s c. The projectile s range. x = x 0 + V x0 t x = (0 m) + (4.7 m/s) (5.05 s) = 15 m Physics 1111 Quiz 6 February 7, 008 Name SOLUTION You are pushing a 7.00 kg box up a rough wall with a force of 185 N directed at 30.0 o to the horizontal. The magnitude of the friction force acting on the box is 11.0 N.

8 a. Calculate the work done by each force on the box after you pushed it through m. W = F x cos () W A = A x cos() = (185.N) (0.750 m) cos(60.0 o ) = 69.4 J W n = n x cos() = n (0.750) cos(90.0 o ) = 0 J W w = wx cos() = (7.00 kg)(9.81 m/s ) (0.750 m) cos(180 o )= J W f = fx cos() = (11.0 N)(0.750 m) cos(180 o )= J b. Calculate the total work done on the box. W NET = W A + W n + W w + W f = 69.4 J + 0 J J J = 9.65 J c. If initially the box was at rest, what is the final speed of the box after you pushed it through m? W NET = ½ m V f ½ mv i V i = 0 m/s ½ mv f = W NET V f = sqrt( W NET /m) = 1.66 m/s

9 Physics 1111 Quiz 7 March 1, 008 Name SOLUTION An object of mass.00 kg is held at the top of a triangular wedge as shown in the figure above, and then released. Use the base of the triangle as the reference level for the gravitational potential energy. Neglect friction. 1. What is the gravitational potential energy of the object at the top of the wedge? a J. U g = m g y = (.00 kg)(9.81 m/s )(3.00 m) b N/m. c J. d. 0 J.. What is the kinetic energy of the object at the top of the wedge? a J. b N/m. c J. d. 0 J. V = 0 3. What is the amount of work that gravity does on the object as it comes down to the bottom of the wedge? a J. W g = U gi - U gf = 58.9 J 0 J = 58.9 J b N/m.

10 c J. d. 0 J. Physics 1111 Quiz 8 March 19, 008 Name SOLUTION A 100 kg warthog (Phacochoerus aethiopics) squeals with delight as it slides from rest down the greased (frictionless) slope and plane as pictured below. a. How fast does the hog move as it just reaches the bottom of the slope? E i = m g y i + ½ m V i = m g y i E f = m g y f + ½ m V f = ½ m V f E i = E f m g y i = ½ m V f g y i = ½ V f V f = g y i V f = 9.90 m/s

11 b. As the hog continues along the plane, it encounters a.00 m stretch of level rough ground. The coefficient of friction for the hog on a rough ground is How fast does the hog move after it passes the rough patch? E i = m g y i + ½ m V i = m g y i E f = m g y f + ½ m V f = ½ m V f W nc = E f - E i W nc = ½ m V f - m g y i f = k n = k m g W nc = k mg d cos (180 o ) = - k mg d - k mg d = ½ m V f - m g y i - k g d = V f g y i V f = g y i - k g d V f = 9.9 m/s Physics 1111 Quiz 9 March 6, 008 Name SOLUTION A 5.00 kg object moving to the right with a speed of 4.00 m/s collides head-on with a 10.0-kg object moving toward it with a speed of 3.00 m/s. Two objects stick together after collision. a. What is the final velocity of the combined object? Please specify both magniude and direction. m 1 V 1i + m V i = ( m 1 + m )V f

12 V f = (m 1 V 1i + m V i )/( m 1 + m ) V f = ((5.00 kg)(4.00 m/s) + (10.0 kg)(-3.00 kg))/(5.00 kg kg) = m/s b. Is this collision elastic or inelastic? Please prove your conclusion. K i = ½ m 1 V 1i + ½ m V i = 85.0 J K f = ½ (m 1 + m ) V f = 3.33 J Collision is inelastic, since K i K f Physics 1111 Quiz 10 April 9, 008 Name SOLUTION A dentist s drill starts from rest. After 3.0 s of constant angular acceleration, it turns at a rate of.51 x 10 4 rev/min. a. Find the drill s angular acceleration. (.51 x 10 4 rev/min)( rad/rev)(1 min/60 sec) = 0.63 x 10 4 rad/s t ( t = ((0.63 x 10 4 rad/s) (0))/(3.0 s) = 81 rad/s b. Determine the angle (in radians) through which the drill rotates during this period. ) ) = ( ) ) = (0.63 x 10 4 rad/s 0 rad/s ) = 41 rad = 670 rev

13 Physics 1111 Quiz 11 April 16, 008 Name SOLUTION You are located on the outer rim of a merry-go-round of diameter 0.0 m and are orbiting at 6.00 rev/min. a. What is your angular speed in rad/s? = (6.00 rev/min)( rad/rev)(1 min/60 sec) = 0.68 rad/s b. What is your tangential speed? V t = r = (0.68 rad/s)(10.0 m) = 6.8 m/s c. What is the magnitude of your centripetal acceleration? a cp = V t / r = (6.8 m/s) /(10.0 m) = 3.95 m/s d. Assuming that your mass is 60.0 kg, how much of the centripetal force is necessary to keep you on merrry-go-round? F cp = m a cp = (60.0 kg)(3.95 rad/s ) = 37 N Physics 1111 Quiz 1 April 3, 008 Name SOLUTION Pictured below is a very light wooden plank with two masses, 10.0 kg each, on top of it.

14 Find the reaction forces at points A and B. N Ax = N A cos (90.0 o ) = 0 N Ay = N A sin (90.0 o ) = N A N Bx = N B cos (90.0 o ) = 0 N By = N B sin (90.0 o ) = N B w 1x = m g cos (-90.0 o ) = 0 w 1y = m g sin (-90.0 o ) = - m g = - (10.0 kg)(9.81 m/s ) = N w x = m g cos (-90.0 o ) = 0 w y = m g sin (-90.0 o ) = - m g = - (10.0 kg)(9.81 m/s ) = N F x = 0 F y = 0 N A + N B 98.1 N 98.1 M = 0 N A + N B = 196 N NA = 0 NB = N B (3.00 m) w1 = 0 w = -w (.00 m) = -(98.0 N) (.00 m) = -196 N m

15 = 0 N B (3.00 m) -196 N m = 0 N B = 65.4 N N A + N B = 196 N N B = 131 N

16