Physics 211 Fall 2009 xam 3 Version Instructions e sure to answer every question. Follow the rules shown on the screen for filling in the Scantron form. ach problem is worth 10% of the exam. When you are finished, check with r. Mike or his T to be sure you have finished the scantron correctly. 1. Free-ody iagrams [713779] To give a 16 kg child a ride, two teenagers pull on a 3.4 kg sled with ropes, as indicated in Figure 5-23. oth teenagers pull with a force of F = 49 N at an angle of 35 relative to the forward direction, which is the direction of motion. In addition, the snow exerts a retarding force on the sled that points opposite to the direction of motion, and has a magnitude of 57 N. Find the magnitude of the acceleration of the sled and child. Figure 5-23
x*05*0.04 m/s 2 w*07*0.87 m/s 2 y*05*7.08 m/s 2 v*10*1.2 m/s 2 z*03*6.85 m/s 2 Solution or xplanation rawing a free-body diagram, we can find two equations. -2 F x + 57 N = m a x = m a and F y - F y = m a y = 0. The mass is the mass of the boy and the sled. Using the first equation, we get the answer. 2. Momentum and nergy [714510] Two identical cars, one coming from the south and the other from the west, collide and stick together. fter the collision, they move directly northeast. If the speed of each car just before the collision is 59 km/hr, what is their speed just after the collision? (Note: The momentum of both cars after the collision is the sum of momenta of both cars before the collision. In other words, the total momentum of both cars is conserved.) w*07*83.44 km/hr x*05*20.86 km/hr y*00*59.00 km/hr z*00*i need to know the mass of the cars to determine this. v*10*41.72 km/hr
Solution or xplanation The initial momentum of both cars (before the collision) is p 0 = m v 01 + m v 02. If we set the x-axis as east then p 0 = - m (59 km/hr) i - m (59 km/hr) j. fter the collision, we have both cars moving together and their momentum is now p f = 2 m v f. In a collision momentum is always conserved. Thus, p f = p 0. Thus, v f = 1/2 (- 59 km/hr) i - m (59 km/hr) j and v = [1/4 (- 59 km/hr) 2 + 1/4 (- 59 km/hr) 2 ] = 41.72 km/hr. 3. Newton's Laws [716878] Heinz Ward catches a football for a touchdown. If the friction that occurs from the ball when it hits his chest is considered the action force, what is the reaction force? (Yes. Heinz is a real football player who plays for the Pittsburgh Steelers.) x*00*the force of gravity on the ball. y*00*the force of gravity on Heinz. v*10*the friction force Heinz's chest exerts on the ball. z*00*the friction force of the ground on Heinz. w*00*the normal force the ball exerts on Heinz's chest.
Solution or xplanation If the action is a certain type of force from one object acting on another object, then the reaction is the same type of force acting from the other object back on the original object. Note: They are equal and opposite. 4. Free-ody iagrams [713788] Using the free-body diagram below, determine the acceleration of the box. Its mass is m = 17 kg. w*04*(0.12 m/s 2 ) i + (0.24 m/s 2 ) j z*06*(0.41 m/s 2 ) i + (-0.14 m/s 2 ) j v*10*(0.07 m/s 2 ) i + (0.19 m/s 2 ) j y*02*(0.42 m/s 2 ) x*02*(0.71 m/s 2 ) Solution or xplanation You must add all of the forces as vectors to use Newton's second law. To do this you add components. F x = (5 N) cos(40 o ) + (4 N) cos(40 o + 90 o ) + (3 N) cos(270 o ) = 1.26 N = m a x. F y = (5 N) sin(40 o ) + (4 N) sin(40 o + 90 o ) + (3 N) sin(270 o ) = 3.28 N = m a y. You now just divide both components by the mass to get the acceleration.
5. Work-nergy Theorem [1236133] baseball of mass 0.17 kg is thrown from the pitcher's mound to home plate. The pitcher releases the ball from a height of 2.5 meters above the ground and the ball reaches the batter at a height of 1.5 meter above the ground. Ignore air resistance. If the ball has a speed of 43.81 m/s when it reaches the batter, what was the ball's speed when it left the pitcher's hand? w*05*44.03 m/s x*06*30.82 m/s y*04*19.62 m/s z*02*43.81 m/s v*10*43.59 m/s Solution or xplanation There is no nonconservative work and no elastic energy. Thus, 0 = K + U g or K = - U g. Initially, K 0 = 1/2 m v 0 2 U g0 = m g h 0, where v 0 is the speed with which the ball is thrown and h is the height of the ball when it leaves the pitcher's hand. Finally, K f = 1/2 m v f 2 U g0 = m g h f, where v f is the speed with which the ball reaches the batter and h f is the height of the ball at this point.
This gives us U g = m g h and K = 1/2 m (v 2 f - v 2 0 ). Putting these into our first equation, we have 1/2 m (v 2 f - v 2 0 ) = m g h. Solving for v 0, we find that v 0 = (v 2 f - 2 g h). 6. Impulse [714508] Two cars of equal mass (1500 kg) collide head-on at an intersection and stick together. One is initially traveling at a speed of 5 m/s and the other at 10 m/s. The time it takes them to come to a full stop is 50 s. What is the magnitude of the impulse on each car from the time just before they collide until they come to a full stop? y*01*3.8 10 5 kg m/s for both cars z*04*it is nearly zero because p = 0 for the system. w*05*1.5 10 4 kg m/s for both cars x*05*4.5 10 4 kg m/s for both cars v*10*1.5 10 4 kg m/s for one car and 7.5 10 3 kg m/s for the other Solution or xplanation Here we just use the fact that J = p = m v 0.
7. Force and entripetal cceleration [305256] 0.80 kg stone attached to a string is whirled in a horizontal circle of radius 44 cm as in the conical pendulum of xample 5-10. The string makes an angle of 30 with the vertical. Find the speed of the stone. Figure 5-10. x*07*5.26 m/s z*04*2.49 m/s v*10*1.58 m/s w*07*2.73 m/s y*03*0.44 m/s
Solution or xplanation Since the stone travels in a perfect circle, its acceleration is centripetal. Thus, a = v 2 / r and v = (a r). We must now find the acceleration. To do this, we use Newton's second law. F = T + m g = m a. Setting the axes with x pointing to the right and y pointing up, gives us the x-component equation F x = T cos(90 o + ) = - m a. The y-component equation is F y = T sin(90 o + ) - m g = 0. Thus, T = m g / sin (90 o + ) = (0.80 kg)(9.81 m/s 2 ) / sin (120 o ) = 9.06 N and a = -T cos (120 o ) / m = +5.66 m/s 2. This is the centripetal acceleration we were looking for. nd so, v = (a r) = [(+5.66 m/s 2 )(44 cm)(1 m / 100 cm)] = 1.58 m/s.
8. Work-nergy Theorem [714503] ailey. Wonderdog, in an attempt to save a raft full of puppies ready to plunge over a waterfall, uses a bungee cord from a bridge high over a river. ailey has a mass of 30 kg. The bungee cord has a spring constant of 221 N/m and the maximum stretch (from equilibrium) of the cord when ailey picks up each pup is 12 m. From what height above the cords equilibrium length did she dive? (Note: I checked this one VRY carefully. The right answer is among the answers given. If you do not get one of these answers, reread the problem.) w*03*9.01 m x*05*4.51 m z*05*530.40 m y*01*15.34 m v*10*42.07 m Solution or xplanation We will set the initial time to be just as ailey leaves the bridge. The final time is when the cord is stretched to maximum. oth ailey's initial and final kinetic energy is zero since she is not yet in motion initially and she is momentarily at rest when the cord is fully stretched. ailey's initial elastic (spring) potential energy is zero since initially the cord is unstretched. The cord stretches x = 12 m at the maximimum. So, her final elastic potential energy is U sp = 1/2 k x 2 = 15912 J. If we set as zero height the position of ailey when the bungee cord is fully stretched, then her initial height h i is h, the height above the equilibrium of the cord (what we are looking for) added to the 12 m of the stretch. Her final height is h f = 0 m. Thus, her change in height is -(12 m + h). nd her change in gravitational potential energy is U g = - m g (12 m + h). There is no nonconservative work done here since all of the forces on ailey are conservative. Thus, 0 = K + U g + U sp = 0J + 15912 J - m g (12 m + h). Solving for h, we will get the correct answer.
9. Nonconservative Work [716869] r. Mike pushes a 21 kg box a distance of 10 m with a constant speed across a level floor. If the kinetic coefficient of friction between the box and the floor is 0.36, what is the work done by friction on the box? y*02*-3.6 J w*03*0 J z*07*i would need to know the path taken by the box to calculate this. v*10*-741.636 J x*03*-2060.1 J Solution or xplanation For a nonconservative force (friction), we know that W = F r cos. Friction has the magnitude f k = k N Since the floor is level and the box is moving with a constant speed, we can say that N = mg. Thus, W = (0.36) (21 kg) (9.81 m/s 2 ) cos(180 o ) = -741.636 J 10. Impulse [714507] ball experiences an impulse from the floor with a magnitude of 128 kg m/s. The impulse acts for a duration of 0.02 s. What is the magnitude of the average force acting on the ball from the floor? x*01*0.03 N w*05*2.56 N z*02*it depends on the speed of the ball. v*10*6400.00 N y*02*it depends on the mass of the ball.
Solution or xplanation From either the textbook, notes or website, we have that J = F av t. You then solve for F av.