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

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1 Physics 1111 Quiz 1 June 4, 007 Name _SOLUTION 1. Solve for x: (.5 x + 5) 1/ = 4.5 x + 5 = 16.5 x = 11 x = Solve for t: 3 t 8t + 15 = 0 t 1, = 8 +/- (8-4(3)(15)) 1/ t 1, = 8 +/- (-116) 1/ No real solutions Given that a = 1.50 m and b =.50 m, find: c = a + b c = (a + b ) 1/ a. The length of side c. c = ((1.50 m) + (.50 m) ) 1/ =.9 m b. The sine of angle sin = a/c = (1.50 m)/(.9 m) = 0.514

2 c. The angle = sin -1 (0.514) = 31.0 o 4. Which one of the lines below has a negative slope? The one in the upper left corner. Name SOLUTION 1. If you start from the Bakery, travel to the Café, and then to the Art Gallery, what distance did you travel? a km. b km. c km. d km.. If you start from the Bakery, travel to the Art Gallery, and then to the Café, in 1.00 hour, what is your average velocity? a..50 km/hr. b km/hr. c km/hour. d km/hr.

3 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 =.00 s. V = (6.00 m)/(4.00s) = 1.50 m/s b. At t = 8.00 s V = 0 Physics 1111 Quiz 3 June 11, 007 Name SOLUTION Coasting due east on your bicycle at 8.00 m/s, you encounter a sandy patch of road 7.0 m across. When you leave the sandy patch your speed has been reduced to 6.50 m/s. V 0 = 8.00 m/s V = 6.50 m/s x - x 0 = 7.0 m a. Assuming the bicycle slows with constant acceleration, what was its acceleration in the sandy patch? Give both magnitude and direction. V = V 0 + a (x - x 0 ) V - V 0 = a (x - x 0 ) (V - V 0 )/ ((x - x 0 )) = a a = ((6.50 m/s) (8.00 m/s) )/ ((7.0 m)) = m/s (Negative value means that the

4 direction of acceleration is west) b. What time did it take you to cross the patch? V = V 0 + a t V - V 0 = a t t = (V - V 0 ) / a t = ((6.50 m/s) (8.00 m/s))/(-1.51 m/s ) = s c. What was your average velocity on the sand? V av = (x - x 0 )/(t - t 0 ) = (7.0 m)/(0.993 s) = 7.5 m/s June 13, 007 Physics 1111 Quiz 4 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 ) = (450.0 m) cos(40.0 o ) = 345 m A y = A sin( A ) = (450.0 m) sin(40.0 o ) = 89 m B x = B cos( B ) = (90 m) cos(-35.0 o ) = 38 m B y = B sin( B ) = (90 m) sin( o ) = m R x = A x + B x = (345 m) + (38 m) = 583 m R y = A y + B y = (89 m) + (- 166 m) = 13 m

5 R = ( R x + R y ) 1/ = ( (583 m) + (13 m) ) 1/ = 596 m R = tan -1 (R y /R x ) = tan -1 ((13 m)/(583 m)) = 11.9 o June 18, 007 Physics 1111 Quiz 5 Name SOLUTION A soccer ball is kicked with a speed of 9.50 m/s at an angle of 40.0 o above the horizontal direction. Some time later the ball lands at the same level from which it was kicked. a. Find the components of the ball s initial velocity. V x0 = V 0 cos( 0 ) = (9.50m/s) cos(40.0 o ) = 7.8 m/s V y0 = V 0 sin( 0 ) = (9.50m/s) sin(40.0 o ) = 6.11 m/s b. What are the components of the ball s velocity at the top of the trajectory? V x = V x0 = 7.8 m/s V y = 0 m/s c. How long does it take the ball to reach the top of the trajectory? V y = V y0 gt (V y0 - V y ) /g = t t = (6.11 m/s 0 m/s)/(9.80 m/s ) = 0.63 s d. How far does the ball go in the horizontal direction? x = x 0 + V x0 ( t) = 0 m + (7.8 m/s) (1.5 s) = 9.07 m June 7, 007

6 Physics 1111 Quiz 6 Name SOLUTION A crate of 50.0 kg mass containing a new lab instrument is dragged by enthusiastic physics students a distance of 30.0 m along a straight, level corridor to a physics lab. The students maintain a constant pull of magnitude 00 N applied to the crate by means of a rope inclined to the horizontal at angle of 40.0 o. The velocity of the crate along the straight path is not constant. a. Calculate the work done by each force on the crate. W = F x cos ( ) W A = A x cos( ) = (00.N) (30.0 m) cos(40.0 o ) = 4596 J W n = n x cos( ) = n (30.0 m) cos(90.0 o ) = 0 J W w = w x cos( ) = w (30.0 m) cos(90.0 o )= 0 J b. Calculate the total work done on the crate. W NET = W A + W n + W w = 4596 J + 0 J + 0 J = 4596 J c. If initially the crate was at rest, what is the final speed of the crate as it reaches the lab? 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) = 13.6 m/s Physics 1111 Quiz 7 July, 007 Name SOLUTION A skier with a mass of 70.0 kg glides certain distance up the slope (30.0 o above the horizontal) before coming to rest. If the speed of the skier at the bottom of the slope was 15.0 m/s and friction can be neglected, how far did he move along the slope?

7 E i = m g y i + ½ mv i corresponds to U g = 0) = ½ mv i (We assume that the bottom of the slope E f = m g y f + ½ mv f = m g y f (since V f = 0) E i = E i ½ mv i = m g y f ½ V i = g y f (½ V i )/g = y f y f = 11.5 m d = y f / sin(30.0 o ) = 3.0 m July 9, 007 Physics 1111 Quiz 8 Name SOLUTION 1. You are tooling along in a 1.10 x 10 3 kg sports car in a dense fog. Uh oh. You suddenly collide into the rear end of a 1.90 x 10 3 kg behemoth that was moving at 1.0 m/s in the same direction. Your cars stick together and move at 15.0 m/s immediately after collision. Consider the motion of all objects in this problem to be in a straight line only with only collision forces relevant. a. Were you exceeding the legal speed limit of 30.0 m/s (roughly 110 km/h) before the crash? m 1 V 1i + m V i = (m 1 + m )V f m 1 V 1i = - m V i + (m 1 + m )V f V 1i = (- m V i + (m 1 + m )V f ) /m 1 V 1i = 0. m/s b. How much of the initial kinetic energy of these two cars was lost in the collision? K i = ½ m 1 V 1i + ½ m V i = ½ (1.10 x 10 3 kg) (0. m/s) + ½ (1.90 x 10 3 kg) (1.0 m/s) = 361, J = 361 kj K f = ½ (m 1 + m )V f = ½ (3.00 x 10 3 kg) (15.0 m/s) = 337,500 J = 338 kj K = 338 kj 361 kj = -3.0 kj

8 July 11, 007 Physics 1111 Quiz 9 Name SOLUTION 1. Express 47.5 o angle in radians. (47.5 o x rad)/180 o = 0.89 rad. As the wind dies, a windmill that was rotating at.50 rad/s comes to a full stop in 6.00 s. Find the average angular acceleration of the windmill. av = t = (0.50 rad/s)/(6.00 s) = rad/s 3. A disk is rotating about its center with a constant angular velocity of 10.3 rad/s. Through what angular displacement does a point on the rim of the disk go during 15.0 s time interval? av = t av t = (10.3 rad/s)(15.0 s) = 155 rad Physics 1111 Quiz 11 July 3, 007 Name SOLUTION 4. A person weighting 800 N stands with one foot on each of two bathroom scales. Which statement is definitely true? a. Each scale will read 800 N. b. Each scale will read 400 N. c. If one scale reads 500 N, the other will read 300 N. d. None of the above is definitely true.. What condition or conditions are necessary for static equilibrium?

9 a. F x = 0. b. F x = 0, F y = 0, = 0. c. = 0. d. F x = 0, F y = The disk in the Figure below has two forces of equal magnitudes but opposite directions acting on it. The disk is in a. Translational equilibrium. b. Rotational equilibrium. c. Static equilibrium. d. The disk is not in equilibrium.

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

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