<p>January 13, 2009 PHY2054 Discussion-Spring ‘09 Warm Up Quiz (Physics 1 Review)</p><p>Name: UFID:</p><p>**1. (0pt) Tim Tebow throws a football toward Percy Harvin with an initial speed of 25.0 m/s at an angle of 30.0º above the horizontal. At that instant, Percy is 40.0 m from Tim. In what direction and with what constant speed should Percy run in order to catch the football at the level at which it was thrown? </p><p>The horizontal and vertical components of the initial velocity are given by v0x = v0cosθ = 25cos30º = 21.7 m/s, v0y = v0sinθ = 25sin30º = 12.5 m/s When the ball comes back to its original height, vertical displacement is zero. Thus the flight time and the horizontal range are </p><p>2 0 = v0yt-(1/2)gt ⇒ 0 = t(v0y-(1/2)gt) ⇒ t = 2v0y/g = 2×12.5/9.8 = 2.55 s</p><p>Δx = v0xt = 21.7×2.55 = 55.3 m Percy needs to run a distance of d = 55.3 – 40 = 15.3 m away from Tim with a constant speed of v = d/t = 15.3/2.55 = 6.00 m/s</p><p>***2. (0pt) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its surface slopes downwards toward the outside, making an angle of 15.0º with the horizontal. A 20.0-kg travel bag is placed on the carousel, 8.00 m from the axis of rotation. The bag goes around once in 30.0 s. What minimum coefficient of friction between the bag and carousel is needed to keep the bag from slipping? </p><p>The angular velocity of the bag is ω = 2π/T = 2×3.14/30 = 0.209 rad/s Taking +x axis horizontally toward the center and +y axis vertically upward, we apply Newton’s 2nd law. Since we need to calculate the minimum coefficient of friction, we assume the bag is on the verge of slipping. We have mrω2 = fcosθ-nsinθ = n(μcosθ-sinθ) ⇒ n = mrω2/(μcosθ-sinθ) 0 = fsinθ+ncosθ-mg ⇒ n= mg/(μsinθ+cosθ) ⇒mrω2/(μcosθ-sinθ) = mg/(μsinθ+cosθ) ⇒ μ = (gsinθ+rω2cosθ)/(gcosθ- rω2sinθ) = (9.8sin15º+8×0.2092cos15º)/(9.8cos15º-8×0.2092sin15º) = 0.307 Constants & Formulas</p><p>2 2 2 2 g = 9.80 m/s Kinetic Energy: KE = (1/2)mv Centripetal accel.: ac = rω =v /r</p><p>Wave eqn: v = fλ Grav. Potential Energy: PEg = mgy Kinematics eqns: ASK</p><p>**3. (0pt) A block of mass m1 = 0.500 kg is released from rest at the top of a curved wedge of mass m2 = 2.50 kg, which sits on a frictionless horizontal surface. What is the velocity of the block when it leaves the wedge if the height of the wedge is h = 1.00 m? </p><p>Since the friction between the block and the wedge is an internal force and does not do work on the system. Both the mechanical energy and momentum are conserved. Thus we have</p><p>2 2 Ei = Ef ⇒ m1gh = (1/2)m1v1 +(1/2)m2v2 pi = pf ⇒ 0 = m1v1+m2v2 ⇒ v2 = (m1/m2)v1 Plugging the 2nd equation into the 1st, the final velocity of the block is </p><p>2 2 2 m1gh = (1/2)m1v1 +(1/2)(m1 /m2)v1 ⇒ v = √(2m2gh/(m1+m2)) = √(2×2.5×9.8×1/(0.5+2.5)) = 4.04 m/s</p><p>*4. (0pt) A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 40.0 vibrations in 30.0 s. Also, a given crest travels 2.00 m along the rope in 5.00 s. What is the wavelength?</p><p>The frequency and speed of the wave are f = 40 vibration/30 sec = 1.333 Hz v = 2 m/5 sec = 0.400 m/s Applying the wave equation, the wavelength is λ = v/f = 0.4/1.333 = 0.300 m</p>