Physics 1 Third Exam Practice w/answers Equations: x = x + v t + ½a t 2 v = v + a t v 2 = v 2 + 2a(x-x ) v = (v v )/2 ΣF = ma F 12 = - F 21 F(G) = mg F(f) k = µ k F N F(f) s µ s F N v = x/ t a = v/ t a r = v 2 /r 1. Define the following quanties: a) Work p = mv F = p/ t Impulse = F t = p mv 1 +mv 2 =mv 1 +mv 2 v x = v cos(θ) v y = v sin(θ) v 2 = v 2 2 x + v y tan(θ) = v y /v x W = Fd cos(θ) KE t = ½mv 2 GPE = mgy KE + PE + W = KE f + PE f SPE = ½kx 2 b) Impulse c) Potential d) Momentum 2. In which of the following is work done (by the physics definition) on the object? Explain why or why not. a) You push hard on a cement wall No work since the wall does not move: no distance. b) You carry your lunch tray across the cafeteria at the same height. No work since the force is perpendicular to the direction of motion. c) You draw a circle on a sheet of paper with a pencil Yes, although you come back to the same place, friction has opposed the motion the entire time. Physics 1 practice exam 3 Page 1
3) A glass ball collides head-on with a steel ball (about the same size, but heavier) initially not moving. In the space below, draw the momentum diagrams for the system before and after the collision. p glass before p steel before p total p glass after p steel after 4) A soapbox car has an initial energy distribution as seen in the first chart. In the remaining charts, draw the amount of kinetic energy there must be in the car and the raceway. (Thermal = Internal energy.) Diagram Diagram Potential Kinetic Thermal Potential Kinetic Thermal Diagram Diagram Potential Kinetic Thermal Potential Kinetic Thermal Physics 1 practice exam 3 Page 2
5) An 8 kg skier travels m down a slope that starts out at a o angle and gradually flattens out to a o angle, resulting in a total vertical drop of meters. The initial push gives her a speed of 3.5 m/s, and she reaches the bottom traveling at 21.5 m/s. What was the average force of friction? GPE KE IE GPE KE IE m = 8 kg h = m d = m v t = 3.5 m/s v b = 21.5 m/s g = 9.8 m/s 2 F(f) =? GPE t + KE t = KE b + IE b mgh + ½mv t 2 = ½mv b 2 + F(f)d F(f)* m = 8kg*9.8m/s 2 *m + ½(8kg)(3.5m/s) 2 ½(8kg)(21.5m/s) 2 F(f) = 347 N 6) Tarzan (mass of 75 kg) rescues Jane by swing in at 4.5 m/s and grabbing her so that they then swing off at 2.7 m/s. What must be Jane s mass? m t v b = m t v a + m j v a 75kg * 4.5 m/s = 75kg * 2.7 m/s + m j *2.7 m/s m t = 75 kg m j =? v b = 4.5 m/s v a = 2.7 m/s m j = 75 kg * (4.5 m/s 2.7 m/s)/2.7 m/s = kg Physics 1 practice exam 3 Page 3
7) A g Velcro dart is launched from a spring-loaded gun. It sticks to a fuzzy die (mass g) hanging from the ceiling. The dart and the die swing back, reaching a maximum height of cm above where the die started. How fast was the dart traveling before it hit the die? Two parts: first a collision, then a change in height m 1 = g m 2 = g v 1 =? v 2 =? h =. m KE GPE KE GPE p 1b = p 1a + p 2a m 1 v 1 = (m 1 +m 2 )v 2 KE = GPE ½(m 1 +m 2 )v 2 2 = (m 1 +m 2 )gh v 2 = sqrt(2gh) = sqrt(2*9.8m/s 2 *.1 m) = 1.96 m/s v 1 = (m 1 +m 2 )v 2 /m 1 = (1g/g)*1.96 m/s = 6.86 m/s Physics 1 practice exam 3 Page 4
8) Real world problem. Fill in ONLY the indicated steps of the FOCUS and DESCRIBE steps. DO NOT SOLVE THE PROBLEM As a concerned citizen, you have volunteered to serve on a committee investigating injuries to Middle School students participating in sports programs. Currently your committee is investigating the high incidence of ankle injuries on the basketball team. You are watching the team practice, looking for activities which can result in large horizontal forces on the ankle. Observing the team practice jump shots gives you an idea, so you try a small calculation. A - kg student jumps 1. m straight up and shoots the.8 kg basketball at his highest point. From the rajectory of the basketball, you deduce that the ball left his hand at from the horizontal at m/s. What is his horizontal velocity when he hits the ground? EVERYDAY LANGUAGE Sketch with Given Information What are you trying to find? final horizontal velocity m/s @ o What are the physics principle(s)? momentum Physics Description Diagram p kid before p kid after (horizontal) p ball before total momentum p ball after (horizontal) Define Variables p kid before = kgm/s p ball before = kgm/s m kid = kg m ball =.8 kg p kid after =? p ball after =? v ball = m/s v ball-x =? v ball-y =? θ = o v kid-x =? Quantitative Relationships (Write down ONLY the equations needed to solve this problem.) = p ball after p kid after p = mv v ball-x = v ball cos(θ) Physics 1 practice exam 3 Page 5