Practice Exam 3

Name ___________________________________ UW Net ID _______________ last first (part before @uw.edu) Physics 121, Spring 2022 Exam 1 1 I. [60 pts] Multiple Choice (5 pts each): Mark your answer on BOTH the Scantron and this page. 1. [5 pts] An object moving in the positive x -direction with constant acceleration goes past point x 0 at time t = 0 s. At some later time t 1 , the object goes past point x 0 again. Which of the velocity vs. time graphs below could describe the motion of the object? 2. [5 pts] A car travels 3.0 miles at a constant speed of 40 miles per hour (mph), then an additional 3.0 miles at a constant speed of 60 mph. Which choice best represents the average speed of the car for the entire 6.0-mile trip? A. 46mph B. 48 mph C. 52mph D. 50 mph E. None of these 3. [5 pts] You throw a ball straight up into the air with an initial velocity, ࠵? ⃑ ࠵? , and you find that it takes a time ∆ ࠵? 1 to rise to its maximum height and then return to your hand. You then throw a ball with twice the inertia at twice the initial velocity and you measure the time it takes to return to your hand, ∆ ࠵? 2 . There is no air resistance. Which choice best represents the relationship between these two time intervals? A. ∆ ࠵? 1 = ∆ ࠵? 2 B. ∆ ࠵? 1 = 4∆ ࠵? 2 C. ∆ ࠵? 1 = 2∆ ࠵? 2 D. 2∆ ࠵? 1 = ∆ ࠵? 2 E. 4∆ ࠵? 1 = ∆ ࠵? 2 t v x t v x t v x t v x t v x t 1 t 1 t 1 t 1 t 1 A. B. C. D. E.

Name ___________________________________ UW Net ID _______________ last first (part before @uw.edu) Physics 121, Spring 2022 Exam 1 2 4. [5 pts] A bowling ball and a ping pong ball collide elastically as shown below. Before the collision, the bowling ball is traveling to the right with speed v , while the ping pong ball is traveling to the left also with speed v . The inertia of the bowling ball m 1 is much larger than the inertia of the ping pong ball m 2 . Which of the following best represents the final velocity v 2f of the ping pong ball? A. –2 v B. –v C. v D. 2 v E. 3 v For the next two questions consider the following collision as observed from Earth’s reference frame: Cart 1 with initial velocity ࠵? ⃑ 1i = v ࠵?̂ collides with cart 2 at rest ࠵? ⃑ 2i = 0. The inertia of cart 2 is six times the inertia of cart 1, i.e. , m 2 = 6 m 1 . The velocity of cart 1 after the collision is: ࠵? ⃑ 1f = – v /2 ࠵?̂ . 5. [5 pts] Observer O observes the collision from the zero-momentum frame. What is the velocity of observer O in the Earth’s reference frame, ࠵? ⃑ EO ? A. – v /7 ࠵?̂ B. – v /6 ࠵?̂ C. 0 D. v /6 ࠵?̂ E. v /7 ࠵?̂ 6. [5 pts] What is the change in kinetic energy of the two-cart system following the collision? A. െ ହ ଵ଺ mv 2 B. െ ଵ ସ mv 2 C. െ ଷ ଵ଺ mv 2 D. െ ଵ ଼ mv 2 E. 0

Name ___________________________________ UW Net ID _______________ last first (part before @uw.edu) Physics 121, Spring 2022 Exam 1 4 10. [5 pts] A student is analyzing the motion of an object in a video using a ruler and a stopwatch. The student’s data are shown in the table at right. Which of the following models for the object’s motion is the best description given this data? A. Motion with constant acceleration in which the object is speeding up B. Motion with constant acceleration in which the object is slowing down C. Motion with constant velocity D. Some other kind of accelerated motion in which the object is speeding up E. Some other kind of accelerated motion in which the object is slowing down 11. [5 pts] A student hangs several objects from a spring, and measures how far the spring stretches. They plot their data as shown. Which of the following is the best estimate for the slope of the best fit line ? A. 0.26 ± 0.08 N/cm B. 0.3 ± 0.08 N/cm C. 0.3 ± 0.1 N/cm D. 0.26 ± 0.04 N/cm E. 0.260 ± 0.038 N/cm 12. [5 pts] A student is investigating the relationship between time elapsed and how fast an object is moving. They produce the graph of time versus speed at right. The slope of the best-fit line is 0.59 ± 0.01 s 2 /m. Which of the following best describes the meaning of the number 0.59? A. The number 0.59 means that on average, the object is moving 0.59 times faster each second. B. The number 0.59 means that on average, the object accelerates by 0.59 m/s each second. C. The number 0.59 means that on average, for each 1 m/s change in the speed of the object, 0.59 seconds elapse. D. The number 0.59 means that on average, the speed of the object doubles every 0.59 seconds. E. None of the above. Time (s) Position (cm) 0.00 12.0 1.00 8.0 2.00 4.2 3.00 0.6 4.00 -2.8

Name ___________________________________ UW Net ID _______________ last first (part before @uw.edu) Physics 121 A , Spring 2022 Exam 1 5 II. Lecture long-answer questions (25 points total) As shown in the ‘initial’ figure, two blocks with inertia m and M are stationary on a frictionless table. There is a small string holding the blocks together, and a massless spring compressed between them. The potential energy stored in the spring is U . Your physics professor cuts the string. You observe that the spring expands until it has zero potential energy, and that the blocks separate. Later, each block is sliding with a constant velocity, as shown in the ‘final’ figure (massless spring not shown). A. [10 pts] Briefly describe in words the step-by-step method that you will use to determine ࠵? ௫ and ࠵? ௫ . You must number your steps. B. [15 pts] Execute the method that you have described in Part A. Number each step, following your answer to part A. You should express your answers only in terms of ࠵?, ࠵?, and U .

Physics 121 Midterm 1 Equation Sheet Spring 2022 Constants Momentum Accel. Due to gravity ! = 9.8 '/) ! Momentum vector * ⃗ = ', ⃗ Impulse - ⃗ = Δ* ⃗ Mathematics Momentum of isolated system *⃗ " = *⃗ # Δ* ⃗ = 0 Quadratic Equation 01 ! + 31 + 4 = 0 Inertia ' $ ' % = − Δ, &,% Δ, &,$ Solution to quad eq’n. 1 = −3 ± √3 ! − 404 20 Difference in variable y: Δ: = : " − : # : (! = : ! − : ( Energy Kinetic Energy ; = 1 2 ', ! 1D Motion Energy of closed system = " = = # Δ= = Δ(; + = #)* ) = 0 Position vector @ ⃗ = 1Â Relative velocity ,⃗ (! = , ⃗ ! − , ⃗ ( Distance between points C = |1 ! − 1 ( | Relative speed , (! = |, ⃗ (! | = |,⃗ ! − , ⃗ ( | 1 − displacement Δ1 = 1 " − 1 # Δ1 = E , & CF * ! * " Coefficient of Restitution G = , (!," , (!,# Displacement vector Δ@⃗ = @⃗ " − @ ⃗ # Δ@⃗ = H1 " − 1 # IÂ = Δ1Â 1-D Coefficient of Restitution G = , !&," − , (&," , !&,# − , (&,# Average velocity vector , ⃗ +, = Δ@ ⃗ ΔF Non-convertible Kinetic Energy ; -. = 1 2 ' */* , -. ! = * */* ! 2' */* Average velocity 1 − component , &,+, = 1 " − 1 # F " − F # = Δ1 ΔF Convertible Kinetic Energy ; 0/), = ; − ; -. ; 0/), = 1 2 J, (! ! Instantaneous velocity ,⃗ = C@⃗ CF Energy Units -KLMG = N! ' ! ) ! Instantaneous velocity 1 − component , & = C1 CF Reduced Mass J = ' ( ' ! ' ( + ' ! Average acceleration vector 0 ⃗ +, = Δ,⃗ ΔF Average acceleration 1 − component 0 &,+, = , " − , # F " − F # = Δ, & ΔF Reference Frames Instantaneous acceleration 0 ⃗ = C, ⃗ CF Velocity of object o in Frame A , ⃗ 1/ Instantaneous accel. 1 − component 0 & = C, & CF = C ! 1 CF ! Velocity of object o in Frame B , ⃗ 2/ = , ⃗ 21 + , ⃗ 1/ Change in velocity 1 − component Δ, & = E 0 & CF * ! * " Position of object o in Frame B @⃗ 2/ = @⃗ 1/ + , ⃗ 21 F 3 Constant acceleration 1(F) = 1 # + , # F + 1 2 0F ! Acceleration of object o in Frame B 0 ⃗ 2/ = 0 ⃗ 1/ , & (F) = , &,# + 0 & F Center of Mass Position @⃗ -. = ' ( @⃗ # + ' ( @⃗ # + ⋯ ' ( + ' ! + ⋯ , &," ! = , &,# ! + 20 & Δ1 Center of Mass Velocity , ⃗ -. = ' ( , ⃗ # + ' ( , ⃗ # + ⋯ ' ( + ' ! + ⋯ Δ1 = , &," − , &,# 2 ΔF , ⃗ -. = * ⃗ %4%*35 ' %4%*35 Accel. on inclined plane 0 & = ! sin S