1. Physics 6A Practice Final Solutions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2. 1) A helicopter traveling upwards at 121 m/s drops a package from a height of 500m above the ground. Assuming free-fall, how long does it take to hit the ground? This is a straightforward kinematics problem. We can take up to be positive, so the initial velocity is v 0 =121 m/s and the acceleration will be –g. Initial height is y 0 =500m. Using the basic formula for height in free fall: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB We can use the quadratic formula and take the positive answer. This is answer c.

3. This is a projectile problem with a horizontal initial velocity. Here are the initial and final values that we know: 2) A person skateboarding with a constant speed of 3.5 m/s releases a ball from a height of 1.5m above the ground. Find the speed of the ball as it hits the ground. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB When the ball hits the ground we will have the same x-component of velocity, but the y-component will have increased. We need to use the Pythagorean theorem to find the magnitude of the velocity: Answer c

4. We will need to find the components of the initial velocity: 3) A projectile is launched from the origin with an initial speed of 20.0m/s at an angle of 35.0° above the horizontal. Find the maximum height attained by the projectile. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The vertical component of velocity will be 0 when it reaches the highest point, so we can use a kinematics formula to find the maximum height: Answer b

5. 4) You are standing on a scale in an elevator. When the elevator is at rest the scale reads 750 N. You press the button for the top floor and the elevator begins to accelerate upward at a constant rate. If the scale now reads 850 N what is the acceleration of the elevator? mg F scale Draw a free-body diagram for the person. The original reading on the scale is the person’s weight. Divide by 9.8 to get the mass. Write down Newton’s 2 nd law: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Answer a

6. 5) Two blocks are connected by a string and are pulled vertically upward by a force F = 90N applied to the upper block. Find the tension in the string connecting the two blocks. Draw the force diagram for the system, then use Newton’s 2 nd law twice. The first time is to find the acceleration of the entire system, the second time just use the forces on the 1kg mass to find the tension in the string. This is answer d. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 2 kg 1 kg F=90N 19.6N 9.8N T T These tensions cancel out because they are internal forces to the system. Total mass of system Weight of 2kg mass Weight of 1kg mass

7. 6) A jet plane comes in for a downward dive as shown in Figure 3.39. The bottom part of the path is a quarter circle having a radius of curvature of 350m. According to medical tests, pilots lose consciousness at an acceleration of 5.5g. At what speed will the pilot black out for this dive? The given acceleration is 5.5g. This means 5.5 times the acceleration due to gravity: 350m The path of the airplane is circular, so the given acceleration must be centripetal (toward the center). We have a formula for centripetal acceleration: Answer a

8. 7) A 1000 kg car is driven around a turn of radius 50 m. What is the maximum safe speed of the car if the coefficient of static friction between the tires and the road is 0.75? 50 m F friction The friction force must be directed toward the center of the circle. Otherwise the car will slide off the road. If we want the maximum speed, then we want the maximum static friction force. The road is flat (not banked), so the Normal force on the car is just its weight. This friction force is the only force directed toward the center, so it must be the centripetal force: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Answer c.

9. 8) A car is traveling at a speed of 40 m/s. The brakes are applied, and a constant force brings the car to a complete stop in a time of 6.2 seconds. The tires on the car have a diameter of 70 cm. How many revolutions does each tire make while the car is braking? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Given: v 0 =40 m/s; v f =0; Δt=6.2s; diam=70cm→r=0.35m Answer c.

10. 9) Two blocks of equal mass M are attached by a massless rope, with one block on a frictionless table, and the other block hanging down below, as shown. When the block on the table is moving in a circular path at 1 revolution per second, the hanging block is stationary. Find the radius of the circle. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB a)10 cm b) 25 cm c) 50 cm d) 150 cm Tension in rope = weight of hanging block = Mg Tension in rope is also the centripetal force on the moving block Set these equal and solve for R: use v=Rω we must convert the given speed from revolutions per second to radians per second Plug in to get R: Answer b.

11. 10) A merry-go-round is initially rotating at a rate of 1 revolution every 8 seconds. It can be treated as a uniform disk of radius 2 meters and mass 400 kg. A 50 kg child runs toward the merry-go-round at a speed of 5.0 m/s, jumping on to the rim (tangentially, as shown). Find the child’s linear speed after jumping onto the merry-go-round. a) 1.1 m/s b) 2.3 m/s c) 5.0 m/s d) 7.2 m/s Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB We can use conservation of angular momentum for this one. Initial angular speed of disk

12. 11) A ballistic pendulum consists of a solid block of titanium with mass 5 kg, suspended from a light wire. A bullet of mass 5 g is launched toward the block at an unknown speed. The bullet bounces back at half its original speed, and the block rises to a height of 1.8 cm above its starting point. What was the initial speed of the bullet? a) 100 m/s b) 200 m/s c) 300 m/s d) 400 m/s Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB v0v0 Before Collision 5kg Highest Point 1.8cm 5kg ½v0½v0 v block After Collision 5kg Use conservation of momentum for the collision, then conservation of energy for the swinging to the highest point. Now we can put this value into our momentum formula to get the initial speed of the bullet (Round up to get 400m/s)

13. 12) Two cars are moving toward an intersection. Car A is traveling East at 20 m/s, and Car B is traveling North at 12 m/s. The mass of Car A is 1000 kg and the mass of Car B is 2000 kg. Driver A is applying mascara to her eyelashes, and driver B is reading a text message, so neither of them slows down as they approach the intersection. When the cars crash into each other, they stick together. Find the common velocity of the cars just after the collision. a) 32.0 m/s at an angle of 45° North of East b) 12.0 m/s at an angle of 30° North of East c) 23.3 m/s at an angle of 23° North of East d) 10.4 m/s at an angle of 50° North of East Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB x yA B 20m/s 12m/s A/B v final Momentum is conserved in each direction, so we get two formulas: Combine the components with the Pythagorean theorem to find the final speed, and use tangent to find the angle:

14. 13) The Atwood’s machine system shown is comprised of a block of mass M attached by a massless rope to block of mass 2M. The rope passes over a solid cylindrical pulley of mass M and radius R, and the rope does not slip on the pulley. Find the acceleration of the heavier block. Use g for gravitational acceleration. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB a) 2/7 g b) 2/5 g c) 1/2 g d) 2/3 g M 2M M R positive torque T1T1 T1T1 T2T2 T2T2 2MgMg We can set up force formulas for the masses, and a torque formula for the pulley. Note that acceleration came out negative due to our choice of direction for positive torque. Make sure the signs in all formulas match up with your choice for positive, or you will get the wrong answer.

15. 14) A light uniform ladder of length 5m is leaning against a wall so that the top of the ladder is 4m above the ground and the bottom of the ladder is 3m from the wall, as shown. How high can a person of mass 150 kg walk up the ladder before the ladder slips? Assume the coefficient of static friction between the ladder and the ground is 0.6 and that the wall is frictionless. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB a) 1.0 m b) 2.5 m c) 3.0 m d) 4.0 m 4m 5m 3m d Forces on the ladder are shown in blue (the weight of the ladder is negligible). We need to find the distance d. 3 formulas we can write down – 2 force formulas and one torque (use the ground as the pivot point). N fsfs mg wall F wall ℓ use similar triangles to find the lever arm for the weight:

16. 15) Block A in the figure weighs 60N. The coefficient of static friction between the block and the tabletop is 0.25. Find the maximum weight, w, of the hanging block so that the system remains at rest. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Draw the forces on the diagram. Everything is in equilibrium, so the forces will add up to zero for each object in the system (essentially we have 3 free body diagrams in the picture). Block A is on a horizontal table with no vertical forces other than gravity and normal force, so the normal force must equal the weight. This tells us the max. friction force is (.25)(60N)=15N Since the problem says to find the maximum hanging weight, we can say F friction =15N. This will be the same as the tension in the horizontal rope, so T 1 =15N. Now look at the hanging weight. The only forces are gravity and the tension in the vertical rope. So T 2 = w. Finally, look at the junction where all 3 ropes are tied together. That point is in equilibrium as well, so we can write down the forces in each direction and they should balance out. N 60N F friction T1T1 T1T1 T2T2 T2T2 T3T3 Answer a.

17. 16) Two balls are rolled down a hill. Ball A is a solid sphere with mass M and radius R. Ball B is a hollow sphere with mass M and radius 2R. Compare the speeds of the balls when they reach the bottom of the incline. a) V A = 0.6 V B b) V A = V B c) V A = 1.1 V B d) V A = 1.7 V B Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Moment of Inertia for each ball: Use conservation of energy for each ball:

18. 17) A box of mass M starts from rest at the top of a frictionless incline of height h. It slides down the hill and across a horizontal surface, which is also frictionless, except for a rough patch of length h, with coefficient of kinetic friction 0.25. The box comes into contact with a spring (spring constant = k), compressing it. The spring then unloads, sending the box back in the opposite direction. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB h h E initial =mgh When the box slides across the rough patch, energy is lost to friction. (each trip) The spring just changes the direction of the box – energy is conserved while in contact with the spring. So the total energy lost due to friction is 0.5mgh. E final = 0.5mgh = mgh final h final = 0.5h

19. 18) A diver tucks her body in mid-flight, reducing her moment of inertia by a factor of 2. What happens to her angular speed and kinetic energy? a) Both angular speed and kinetic energy remain the same. b) Angular speed is doubled, and kinetic energy remains the same. c) Angular speed is doubled, and kinetic energy is increased by a factor of 4. d) Both angular speed and kinetic energy are doubled. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Use conservation of angular momentum. Angular speed is doubled Plug this in to the kinetic energy formula. Kinetic Energy is doubled

20. 19) A uniform marble rolls without slipping down the path shown, starting from rest. Find the minimum height required for the marble to make it across without falling into the pit. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB a) 23m b) 14m c) 30m d) 35m 36m 25m 45m h This is a 2-stage problem. When the marble rolls down the hill, we can use conservation of energy. Then it’s projectile motion as it flies across the gap. We need to find the speed v Projectile motion – initial speed is horizontal, and the marble drops 20m. Find time: Horizontal distance is 36m:

21. 20) A school yard teeter-totter with a total length of 5.2 m and a mass of 36 kg is pivoted at its center. A child of mass 18-kg sits on one end of the teeter-totter. Where should the parent push downward with a force of 210 N to balance the teeter totter? a) 0.5 m from the center b) 2.2 m from the center c) 1.3 m from the center d) 1.9 m from the center Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 210 N 5.2 m 176.4 N 352.8 N d The torques must balance out. Measuring from the center, we have: The weight of the teeter-totter is at the center, so it produces no torque.