C_Energy_Momentum Practice 2008

Transcription

1 Name: Class: Date: ID: A C_Energy_omentum Practice 8. A -kg block slides down a frictionless ramp and loses meters of elevation in the process. What is the kinetic energy change of the block? J B) -, J C), J D) -, J E), J. A 6-kg skier skies from rest down a m long slope that is at a 3 o angle with the horizontal. What is the potential energy change of the skier? -6J B) -3 J C) 3 J D) -6 J E), J 3. A 6-kg skier skies from rest down a m long slope that is at a 3 o angle with the horizontal. At the bottom of the slope, she is moving at m/s. How much work did friction do on the skier? B) -3 J C) 3 J D) -6 J E) 6, J v o m h 4. For a block of mass m to slide without friction up the rise of height h as shown above, it must have a minimum initial speed v o of gh B) gh C) gh D) gmh E) gmh 5. A man holds a -kg weight at arm s length for minute. His arm is.5 m above the ground. The work done by the man on the weight while he is holding the weight still is: B) 5 J C) J D) 6 J E) 6 J 3 o 5 m 6. A giant man pulls a -kg crate from the bottom to the top of a frictionless 3 o slope that is 5 m high as shown above. Assuming the crate moves up the ramp at constant speed, the work done by gravity is: 3 J D) -5 J B) 5 J E) Gravity does no work; the man does it all. C) -3 J 7. A spring hangs net to meter stick. When a -N weight is attached to the end of the spring, it hangs down to the 4 cm mark on the meter stick. When a -N weight is attached, it hangs down to the 6 cm mark. When an unknown weight X is attached, it hangs down to the 3 cm mark. How much does X weigh? N B) N C) 3 N D) 4 N E) 5 N

2 8. When a certain rubber band is stretched a distance, it eerts a restoring force F = -(a + b ), where a and b are constants. The work by done by an eternal force opposing this restoring force in stretching this rubber band from = to = L is: al + bl 3 C) a + bl E) al + bl3 3 B) al + bl D) bl v v v v W W I II III IV W W 9. A single constant force, F, does work W on a particle, initially at rest. Which of the five graphs above correctly show the relationship between the work done, W, and the particle speed, v? I B) II C) I and III D) II and III E) I and IV. At time t =, a.-kg particle has a velocity in m/s of 4i 3j. Three seconds later, its velocity in m/s is i + 3j. During this time the net work done on the particle was 4. J. B) 4. J. C) J. D) 4 J. E) (4i + 36j) J. The load being lifted by the pulley system shown above has a mass m. What force must be applied to the rope on the left to lift the load? Assume there are no resistive forces, and that the pulleys themselves have no mass. mg/4 B) mg /3 C) mg / D) mg E) 4 mg. An escalator is used to move people (6 kg each) per minute from the first floor of a department store to the second floor, 5 m above. The minimum power required to accomplish this is approimately W. B) W. C) W. D) W. E) 6, W.

3 P h Q h 3. A block at point P is released from rest at height h and slides along the frictionless track shown above. At point Q, which is at height h, its speed is Ê h h g h h. C) Á ˆ Ê g. E) Áh h ˆ g. B) gê Áh h ˆ. D) gê Áh h ˆ. 4. The potential energy function of a body of mass m is given by U() = mg + ½ k. The force acting on the body at position is B) mg mg + k3 6 k3 6. C) mg + k. E) No relationship eists between force and potential energy.. D) mg k. 5. A block starting from rest slides down a frictionless inclined plane of length L. When the block has attained one-half of its final speed, the distance it has traveled along the plane is L Ê ˆ L L 3L B) L 4 Á C) D) E) 4 Force I, II, and II act upon a body such that the body moves. The effect in terms of work performed and/or potential energy change are described below. Force I moves the body and returns it to its original position, doing no net work in the process. Force II moves the body from A to B via several different paths, doing the same work each time. Force III moves the body from A to B and changes the potential energy of the system in the process. 6. Which of the forces described above are conservative? Force I C) Force III E) None of the Forces. B) Force II D) Forces I, II, and III 7. Which of the forces described above, when acting alone, will result in a change in total amount of mechanical energy possessed by the body upon which they act? Force I C) Force III E) None of the Forces B) Force II D) Forces I and II 3

4 8. A cone is balanced on its point as shown in the figure above. This is an eample of stable equilibrium. C) neutral equilibrium. E) none of the above. B) unstable equilibrium. D) all of the above. 9. Which of the following potential energy (U) versus displacement () graphs best represents the cone balanced on its point shown above? Consider the midpoint of (=) to be the point at which the cone is perfectly balanced, and positive and negative values of to be where the cone is tipped to one side or the other. U U U U C) U E) B) D). Consider U to be potential energy and to be position on the -ais. If at a given point on the potential energy curve du/d = and d U/d =, then represents a point of stable equilibrium. C) neutral equilibrium. E) none of the above. B) unstable equilibrium. D) disequilibrium.. A force of (5. i +. j - 3. k) N acts on a particle. During the time the force acts upon it, the particle undergoes a displacement of (.4 i +.5 j +. k) m. How much work does the force do on the particle? (i + j + 6k) J B) (i + j - 6k) J C) -3 J D) 3 J E) 6 J. A man sits in the back of a canoe in still water. He then moves to the front of the canoe and sits down there. Afterwards, the canoe is forward of original position and moving forward. B) is forward of original position and moving backward. C) is rearward of original position and moving forward. D) is rearward of original position and moving backward. E) is rearward of original position and not moving. 3. Two objects, P and Q, have the same momentum, but Q has more kinetic energy than P. Therefore, Q has more mass than P. D) is moving slower than P. B) is moving faster than P. E) is moving the same speed as P. C) has the same mass as P. 4

5 4. A rifle of mass is initially at rest but free to recoil. It fires a bullet of mass m and velocity v (relative to the ground). After firing, the velocity of the rifle (relative to the ground) is: v mv mv mv B) C) D) v E) m F ( 4 ) t ( - ) 5. A rifle of mass is initially at rest but free to recoil. It fires a bullet of mass m and velocity v (relative to the ground). After firing, the velocity of the rifle (relative to the ground) is: v mv mv mv B) C) D) v E) m 6. A kg rock traveling at i m/s is subjected to an impulsive force shown in the graph above, directed in the i direction. What is the resulting velocity of the rock? -6 i B) 6 i C) - i D) -4 i E) 8 i 7. A golf ball of mass m is hit by a golf club so that the ball leaves the tee with speed v. The club is in contact with the ball for time T. The average force of the club on the ball during the time T is: mvt B) mv T C) mv T D) mv T E) mt v 8. An inelastic collision is one in which momentum is not conserved but kinetic energy is. B) total mass is not conserved but momentum is. C) neither kinetic energy nor momentum is conserved. D) momentum is conserved but kinetic energy is not. E) the total impulse is equal to the change in kinetic energy. 9. Blocks A and B are moving toward each other along the -ais. A has a mass of. kg and a velocity of 5 m/s (in the positive direction), while B has a mass of 4. kg and a velocity of 5 m/s (in the negative direction). They suffer an elastic collision and move off along the -ais. After the collision the velocities (in m/s) of A and B, respectively, are 5 and 5. C) 5 and 5. E) -5 and -5 B) 5 and 5. D) 5 and 5. 5

6 3. An unstable nucleus has mass and is initially at rest. It ejects a particle of mass m with speed v. The remaining nucleus recoils in the opposite direction with a speed v o B) mv C) mv ( + m) D) ( + m)v m E) mv ( m) 3. The following graphs, all drawn to the same scale, represent the net force F as a function of displacement for an object that moves along a straight line. Which graph represents the force that will cause the greatest change in the kinetic energy of the object from = to =? F F F F C) F E) B) D) 3. A projectile of mass m is shot straight into the air with speed v, momentum p, and kinetic energy K. It reaches a maimum height of h. In order for the projectile to obtain a maimum height of eactly height h, it is necessary to double v C) double p E) none of the above B) double K D) double m m m 33. When the speed of each object is attached to the pulley above is v, the magnitude of the total linear momentum of the system is (m + m ) v C) ½ (m + m ) v E) m v B) (m - m ) v D) ½ (m m ) v 34. When the object of mass m has descended a distance h, the potential energy of the pulley system illustrated above has decreased by (m m ) g h C) (m + m ) g h E) B) m g h D) ½ (m + m ) g h 6

7 m o v o y 6 o 6 o m o v o 35. Two particles of equal mass m o, moving with equal speeds v o along paths inclined at 6 o to the -ais as shown above, collide and stick together. Their velocity after the collision has magnitude v. B) v. C) 4 o o v o. D) 3 v o. E) v o. L A B C D E 36. The center of mass of a uniform wire, bent in the shape shown above, is located closest to point A B) B C) C D) D E) E L 37. ass is moving with speed v toward stationary mass. The speed of the center of mass of the system is a. ( / ) v b. ( + / ) v c. ( + / ) v d. ( - / ) v e. ( / ( + ) ) v Ê ˆ Á v C) + Á Ê B) + ˆ Ê Á v D) Á Ê ˆ Ê v E) ˆ + Á v ˆ v 38. A 6.-kilogram brick is dropped straight down on a 4.-kilogram cart moving horizontally at 5. meters/second. What is the change in velocity of the cart? -3. m/s B) -. m/s C) D). m/s E) 3. m/s 7

8 39. A student stands on the back end of a giant skateboard that is initially at rest. Assume the frictional forces in the wheels of the skateboard can be ignored. As the student walks toward the front of the skateboard, which of the following statements about the center of mass of the student-skateboard system are true? I. The position of the center of mass is unchanged. II. The velocity of the center of mass is unchanged. III. The momentum of the center of mass is unchanged. none C) III only E) all B) I only D) II and III only 4. An air track car with mass m and velocity v to the right collides elastically with a second air track car with mass m and initial velocity zero. What is the velocity of the m car after the collision? B) v to the right C) v to the right E) v to the right 3 3 v to the right D) v to the right 4. At time t =, a 4-kg particle has a velocity in m/s of 4i 3j. At t = 3 its velocity in m/s is i + 3j. During this time the work done on it was: 8 J B) 8 J C) 4 J D) 8 J E) 8i + 7j J 4. A particle of mass m moves along a straight path with speed v defined by the function v = ct 3, where c is a constants and t is time. What is the magnitude F of the net force on the particle at time t = t? 3ct B) 3mct C) mct D) mct 4 E) ¼ mct 4 + v o 8

9 ID: A C_Energy_omentum Practice 8 Answer Section ULTIPLE CHOICE. ANS: E K = U = (U f U i ) = ( mgh) = mgh = ()()() =,J. ANS: D U = U f U i = mgh = (6)()(sin3) = (6)()() = 6J 3. ANS: B W nc = K + U = K f K i + U f U i = mv + mgh f = (6)() (6)()(sin3) = 3 6 = 3J 4. ANS: C K = U K f K i = (U f U i ) ( mv ) = (mgh ) o mv = mgh o v o = gh 5. ANS: A His force is directed up, but the block is not moving up or down, so no work. 6. ANS: C W g = U g = (U f U i ) = (mgh ) = ()()(5) = 3J

10 ID: A 7. ANS: E F = k N = k(4 o ) N = k(6 ); dividing the top equation by the bottom gives: = 4 6 which can be used to show that By substitution of o = cm into one of the equations above, we get k = 5N / m X = (5)(3 ) = 5() = 5N 8. ANS: E L L Ê W applied = F()d = (a + b )d = a + b 3 3 Á ˆ L = al + bl3 3 Note that the force is the opposite of the given restoring force, since we are asking about the applied force eerted on the rubber band -- 3rd Law! 9. ANS: E W net = K = mv The net work varies linearly with v, and quadratically with v.. ANS: C W net = K = K f K i = mv mv = m(v v f i f i ) v i = 4 + ( 3) = = 5m / s v f = + (3) = = 3m / s W net = ()(3 5) = J. ANS: A Four strings attached to the load must be shortened. The string is pulled four times further than the load is lifted. Increasing the distance by a factor of four reduces the force by a factor of four.. ANS: C P = W t = mgh t = È () ÎÍ (6)()(5) 6 = W

11 ID: A 3. ANS: D K = U mv = mg(h f h ) v f = g(h h ) 4. ANS: D F = du d( mg + d = k ) d = ( mg + k) = mg k 5. ANS: A If it goes all the way down, the speed is given by v = gh where h is the vertical height lost. To get / of this speed v = gh = g h 4 Therefore, this speed is attained when /4 of the height has been lost. This will occur when /4 of the distance L has been covered. 6. ANS: D Each statement describes a characteristic of a conservative force. 7. ANS: E All are conservative. Only non-conservative forces change mechanical energy by increasing or reducing the sum of potential plus kinetic energy. 8. ANS: B If a force is eerted to unbalance the system, it will topple. This is characteristic of unstable equilibium. 9. ANS: A Unstable equilibrium is a maimum potential energy situation.. ANS: C The first and second derivatives are both zero, which is a condition for a flat potential energy surface. 3

12 ID: A. ANS: C W = F R = (5)(.4) + ()(.5) + ( 3)() = + 6 = 3J. ANS: E After the movement, the canoe must have no momentum, since it had none before the movement started. To maintain the same center of mass, the mass of the canoe must move backward to account for the movement of the man s center of mass forward. 3. ANS: B Kinetic energy depends upon m and v, whereas momentum depends on m and v. Therefore, any reduction in m with a proportional increase in v necessary to have the same momentum will increase the kinetic energy. 4. ANS: C = mv + v r v r = mv the negative sign indicates a direction opposite that of the bullet. 5. ANS: C = mv + v r v r = mv the negative sign indicates a direction opposite that of the bullet. 6. ANS: A J = Area = (base)(height) = (.s)(8,n) = 8Ns J = p = p f p i p f = p i + J = mv i + J = (kg)(m / s) 8N = 6Ns v f = 6Ns kg = 6m / s 7. ANS: B Ft = p f p i = mv f F = mv f t = mv T 4

13 ID: A 8. ANS: D Definitional. 9. ANS: A Using momentum conservation (no units for simplicity) ()(5) + (4)( 5) = v A + 4V B = v A + 4v B = v A + v B v A = v B Looking at the answer choices, the first two are possible, but only a) allows for a rebound of A. The other would have required A to travel through B. 3. ANS: E = mv ( m)v v = mv m (the nucleus goes down in mass by m when it ejects a particle of mass m) 3. ANS: E Greatest area under the curve means greatest work. 3. ANS: B You need to double the initial kinetic energy to double the final potential energy, which happens if you multiply the velocity by. 33. ANS: B The smaller block is going up at the same speed the larger block is going down. So P = p + p = m v m v = (m m )v (This is the magnitude; if up is considered positive, to correctly show the sign of the linear magnitude would require (m m )v which would be negative indicating a downward vector momentum. 34. ANS: A One block s potential energy goes up by m gh while the other s potential energy goes down by m gh. Therefore, the decrease in potential energy has a magnitude given by (m m ) g h. 5

14 ID: A 35. ANS: B The y-momentum before the collision is zero, so it will remain zero after the collision. The -momentum problem is: m o v o cos(6 o ) + m o v o cos(6 o ) = (m )v Ê ˆ v o Á = v v = v o 36. ANS: B Assume the length of each wire is 4 units m() + m() + m() y cm = = units or just above the unit mark, but not as far as. units 37. ANS: E v cm = v ANS: A = P B P A mv B = (m + m brick )v A 4(5) = (v A ) v A = m / s v = 5 = 3m / s 39. ANS: E Forces are internal; no change in CO. 6

15 ID: A 4. ANS: A mv = mv + mv AND mv = mv + mv v = v + v AND v = v + v squaring the left-hand equation, and setting it equal to the right-hand equation, yields v = v + 4v v + 4v = v + v and simplifying 4v v + 4v = v which yields v + v = v and then v = v or v = v Putting this in initial momentum equation yields mv = m( v + v ) which yields v = 3 v which yields v = v 3 4. ANS: C W = K = K f K i = (4) È ( + 3 ) (4 + ( 3) ) ÎÍ = 4J 4. ANS: B a = dv dt = 3ct F = ma = 3cmt evaluated at t=t 7