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

Transcription

1 Physics 1111 Quiz 1 August 27, 2007 Name SOLUTION 1. If the displacement of the object, x, is related to time, t, according to the relation x = A t, the constant, A, has the dimension of which of the following? a. Acceleration. b. Velocity. c. Time. d. Length. 2. The meter is currently defined as a. The distance between two etched lines in a platinum-iridium bar kept in Serves, France. b. One ten-millionth of the distance between the North Pole and the equator. c. The distance traveled by light in 1/299,792,458 of a second. d. 1,553,146.1 wavelengths of red cadmium light in dry air at 25 o C. 3. Given that a = 3.50 m and b = 6.50 m, find:

2 c 2 = a 2 + b 2 c = (a 2 + b 2 ) 1/2 a. The length of side c. c = ((3.50 m) 2 + (6.50 m) 2 ) 1/2 = 7.38 m b. The tangent of angle tan = a/b = (3.50 m)/(6.50 m) = c. The angle = tan -1 (0.538) = 28.3 o Physics 1111 Quiz 2 September 5, 2007 Name SOLUTION 1. If you start from the Art Gallery, travel to the Café, and then to the Bakery, what is your displacement? a km. b km. c km. d km.

3 2. If you start from the Bakery, travel to the Art Gallery, and then to the Café, in 1.00 hour, what is your average speed? a km/hr. b km/hr. c km/hour. d km/hr. 3. Given the position-versus-time graph for a basket ball player traveling up and down the courting a straight-line path find the instantaneous velocity of the player a. At t = 6.00 s. V = 0 (tangent is horizontal at t = 6.00 s) b. At t = 9.00 s V = rise/run = (-6.00 m)/(4.00 s) = m/s

4 Physics 1111 Quiz 3 September 12, 2007 Name SOLUTION A boy on a bridge throws a stone vertically downward with an initial speed of 14.7 m/s toward the river below. a. If the stone hits the water 2.00 s later, what is the height of the bridge above the water? V 0 = m/s t = 2.00 s y = 0 y = y 0 + V 0 t ½ g t 2 0 = y 0 + V 0 t ½ g t 2 y 0 = - V 0 t + ½ g t 2 y 0 = - (-14.7 m/s) (2.00s) + ½ (9.81 m/s 2 ) (2.00s) 2 y 0 = 49.0 m b. What is the velocity of the stone just before it hits the water? V = V 0 - g t V = (-14.7 m/s) - (9.81 m/s 2 ) (2.00s) = 34.3 m/s c. What is the acceleration of the stone at half distance between the bridge and the water? g = 9.81 m/s 2, downward

5 October 9, 2007 Physics 1111 Quiz 5 Name _SOLUTION A crate of 50.0 kg mass containing new physics textbooks is dragged by enthusiastic physics students a distance of 30.0 m along a 25 o incline to a physics lab. The students maintain a constant pull of magnitude 250 N applied to the crate by means of a rope parallel to the incline. a. Calculate the work done by each force on the crate. Ignore friction. W A = A x cos () = (250 N)(30.0m) = 7500 J W w = (mg) x cos ( ) = (50.0 kg)(9.81 m/s 2 )(30.0m)cos ( ) = J W n = n x cos ( ) = 0 b. Calculate the total work done on the crate. W net = W A + W w + W n = 1281 J c. If initially the crate was at rest, what is the final speed of the crate as it reaches the lab? W net = K f - K i = ½ m V f 2 V f 2 = 2 W net /m V f = (2 W net /m) 1/2 V f = 7.16 m/s Physics 1111 Quiz 6 October 17, 2007

6 Name _SOLUTION A 2.00-kg block sliding along a smooth horizontal ice with a speed of 10.0 m/s encounters a rough patch 10.0 m in length. After the block passes over it, its speed is reduced to 7.50 m/s. a. Calculate the work done by friction on the block. W f = E f - E i = K f - K i = ½ m V f 2 - ½ m V i 2 W f = ½ (2.00 kg) (7.50 m/s) 2 - ½ (2.00 kg) (10.0 kg) 2 W f = J b. What is the magnitude of the friction force? W f = f x cos ( ) = - f x f = W f /x f = (-43.8 J) /m 4.38 N c. What is the coefficient of kinetic friction between the block and the rough patch? f k = k n = k (m g) k = f k /(m g) = Clayton College & State University Physics 1111 Quiz 7 October 24, 2007 Name _SOLUTION

7 An 8.00-kg cart floating on an air track is traveling east at 15.0 cm/s. Coming the other way at 25.0 cm/s is a 2.00-kg cart on which is affixed a wad of bubblegum. The two slam into each other and stick together. a. Find their velocity after impact. m 1 v 1i + m 2 v 2i = (m 1 + m 2 ) v f v f = (m 1 v 1i + m 2 v 2i )/(m 1 + m 2 ) v f = ((8.00 kg)(15.0 cm/s) + (2.00 kg)(-25.0 cm/s))/(10.0 kg) = 7.00 cm/s east b. How much energy was lost during the collision? K i = ½ m 1 v 2 2 1i + ½ m 2 v 2i = ½ (8.00 kg)(0.150 m/s) 2 + ½ (2.00 kg)(0.250 m/s) 2 = J K f = ½ (m 1 + m 2 ) v 2 f = ½ (10.0 kg)(0.0700) 2 = J K = J Physics 1111 Quiz 8 November 5, 2007 Name SOLUTION 1. Express 27.5 o angle in radians. (27.5 o ) /(180 o ) = rad 2. As the wind dies, a windmill that was rotating at 3.60 rad/s comes to a full stop in 5.00 s. t a. Find the average angular acceleration of the windmill. ( t = (0 (3.60 rad/s))/(5.00 s) = rad/s 2

8 b. Through what angle did the windmill rotate before coming to the full stop? ) ) = ( ) ) = ( rad/s rad/s 2 ) = 9.00 rad c. How many rotations does it correspond to? (9.00 rad)(1 rev)/(2 rad) = 1.43 rev

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