Physics Exam Solutions: Problems on Vectors, Kinetic Energy, and Forces, Exams of Physics

Download Physics Exam Solutions: Problems on Vectors, Kinetic Energy, and Forces and more Exams Physics in PDF only on Docsity! Final exam Solutions, Physics 2a, 2010 Double check that you bubble in your code number correctly. If there’s a mistake, your score will be lowered –as a penalty. In case you forgot to write these down in your formula sheet: I =    1 12 ML2 rod through center 1 3 ML2 rod through end 1 2 MR2 solid cylinder 2 5 MR2 solid sphere 2 3 MR2 thin walled hollow sphere. 1 VERSION A 1. Let ~v1 = 3̂i + 4ĵ and ~v2 = 3̂i − 4k̂. What is the angle between these two vectors? (a) cos−1(3/5). (b) sin−1(9/16). (c) cos−1(9/25). (d) sin−1(3/4). Soln: ~v1 · ~v2 = 9 = v1v2 cos θ, with v1 = v2 = 5, so θ = cos−1(9/25). 2. A student is on the top of a building with two eggs and a stopwatch. The first egg is dropped with zero initial velocity and takes 2s to hit the ground. The second egg is thrown downward and is timed to take 1s to hit the ground. What was the approximate initial downward velocity of the second egg? (a) 10 m/s (b) 15 m/s (c) 20 m/s (d) 5m/s Soln: h = 1 2 g(2s)2 = 1 2 g(1s)2 + v(1s), so v ≈ 1 2 g(4 − 1) ≈ 15m/s. 3. Superman is flying along with a velocity of 30m/s, at a constant height of 30m above the ground. He packed his lunch in his backpack, but forgot to zip it up and now the lunch is spilling out without his noticing it. Approximately much kinetic energy will his apple, which weighs 1N, have when it hits the ground (ignoring air resistance). (a) 30J (b) 45 J (c) 120J (d) 75J Soln: The initial energy is E = mgh + 1 2 mv2 which for mg = 1 and h = 30 and v = 30 gives E = mg(30) + 1 2 m(30)2 ≈ 30 + 45 = 75. Ignoring air resistance, this same energy is all kinetic energy when it hits the ground. 4. A particle is moving along a path, with position ~r(t). At a certain instant, it is noticed that d~r dt 6= 0, and also d2~r dt2 6= 0, but d~r dt · d2~r dt2 < 0 and d~r dt × d2~r dt2 6= 0. Which is true at that instant. (a) The particle is going straight, with changing speed. (b) The particle is turning direction, but with constant speed. 2 15. As you know, Bruce and Henrietta like playing the game where she “forgets” her bagels, and then tries to catch them when she’s down on street level. Using the Bagel-BazookaTM , Bruce launches a bagel with speed 25 m/s from the window of their apartment, 20 meters above the ground. The launch angle is θ0 = tan −1( 3 4 ) above the horizon. You can approximate 9.8 ≈ 10. Approximately how long after launch does the bagel reach its maximum, peak height above the ground? Soln: Since v0x = 20 and v0y = 15, the bagel’s velocity at time t after launch is (vx, vy) = (20, 15 − 10t), and the bagel’s location at time t after launch is (x(t), y(t)) = (20t, 20 + 15t − 5t2). The peak occurs when vy = 0, so t = 1.5s. 16. Same setup. Approximately what is the bagels’ range, i.e. how far down the street from the building will the bagel land? Soln: It hits the ground at t = 4s, where (x, y) = (80, 0). 17. A 10 kg watermelon will be launched, straight up like a rocket, using a spring gun. The spring is compressed 4m, which requires a force of 800N , and is then locked in place. The watermelon is then put inside. Someone then pulls the trigger and up the watermelon goes. The average friction force between the watermelon and the inside of the spring gun is 100N . How much energy was stored in the spring before the trigger was pulled? Soln: Using Fx = −kx for the spring, with x = 4 and |F | = 800, we get k = 200N/m. So the stored energy is U = 1 2 (200)(4)2 = 1600J . 18. Same setup. Approximately how high does the watermelon go (taking its initial posi- tion on the compressed spring to be y = 0)? Soln: The friction converts 100N × 4m = 400J of the initial energy into heat. So the maximum potential energy is 1200 = mgh, which taking m = 10 and g ≈ 10 gives h ≈ 12m. 19. Consider a loop-the-loop track, with negligible friction. The loop has radius R. Sup- pose that, at the instant where the car is at the top of the loop, the normal force between the car and the track is three times its weight, N = 3mg. What is the velocity of the car at the top of the loop? Soln: The inward forces at the instant at the top are Fin = −Fy = mg +N = 4mg = marad = mv 2/R. So v2top = 4gR. 5 20. A rifle bullet with mass 0.01 kg strikes and embeds itself in a block of wood of mass 0.99 kg that rests on a frictionless horizontal surface and is attached to a coil spring. The impact causes the spring to compress 0.2m. The spring is such that a force of 3.6N is required to compress the spring by 0.001m. What was the bullet’s initial speed? Soln: 1 2 (m + M)v2f = 1 2 kx2, with m + M = 1kg and x = 0.1 and k = 3.6N/.001m = 3600N/m. So the block’s velocity was v2f = 3600(.1) 2 = 36, so vf = 6m/s. Conservation of momentum gives the bullet’s initial speed: mvi = (m + M)vf , so vi = M+m m vf , with m = .01 and M = .99 and vf = 6 gives vi = 600m/s. 21. A bowling ball rolls without slipping up a ramp, whose angle is sin−1(1/5). Let î be a unit vector pointing in the horizontal direction of the ball’s motion, and ĵ be a unit vector in the vertical direction, pointing up, and k̂ = î × ĵ. Define ω = |~ω| and α = |~α| (where, as usual, |~v| denotes the magnitude of a vector ~v). Which of the following is true? (a) ~ω = ωk̂, ~α = αk̂. (b) ~ω = −ωk̂, ~α = αk̂. (c) ~ω = ωk̂, ~α = −αk̂. (d) ~ω = −ωk̂, ~α = −αk̂. Soln: The answer is (b). Drawing the setup, k̂ points out of the paper. The ball rolls clockwise, so ~ω points into the paper. It’s slowing, so ~α points in the opposite direction. 22. Same setup. The ball’s acceleration has magnitude Soln: The downward acceleration is given by ma = mg sin α − Ff with Ff = τ/R = Iα/R = Ia/R2. So a(m + I/R2) = mg sin α. Using sinα = 1/5 and I = 2 5 mR2, this gives a = g/7. 23. A frictionless turntable of mass 4kg and radius 1m is initially spinning at 30 radians per second. A 4kg lump of clay is dropped from above, and it hits the turntable at a distance of 0.5m from the rotational axis, where it sticks. What is the final angular speed of the turntable and clay? Soln: Angular momentum is conserved, so L = Iiωi = (2kgm 2)(30) = 60kgm2/s = Ifωf . Here Ii = 1 2 (4)11 = 2kgm2. The lump of clay contributes an additional moment of inertia of (4kg)( 1 2 m)2 = 1kgm2, so If = (2 + 1) = 3kgm 2. So ωf = 20s −1. 24. Romeo’s mass is 100kg. His 10m long ladder also has mass 100kg. He props his ladder against a wall, with an angle of θ = sin−1(4/5). The wall is frictionless, while the 6 ground (fortunately) has friction. When he is way up the ladder, what friction force must the ground provide to avoid having the ladder and Romeo slide out? Soln: Setting to zero the torque around the bottom of the ladder gives n2(8) = (1600N)(3), so the normal force from the wall is n2 = 600N , which is also the friction force provided by the ground. 25. Planet X has mass MX , radius RX , and an escape velocity of 100m/s. Planet Y has mass MY = 9MX and radius RY = 1 4 RX . What is the escape velocity on planet Y ? Soln: ve ∼ √ M/R, so the escape velocity on planet Y is bigger by a factor of √ 36 = 6. 26. The moon rotates around the earth with period approximately 27 days. The moon is a distance R1 = 384, 000km from the center of the earth. Suppose that a satellite is rotating around the earth at a distance R2 = 1 9 R1 from the center of the earth. How long does it take this satellite to rotate around the earth? Soln: Since T ∼ r3/2, the answer is 27 days ×(1/9)3/2 = 1day. 27. A 1 kg mass is on a spring, oscillating. At time t = 0, the spring is stretched to its maximum distance from equilibrium, which is 0.5m. The spring oscillates back to that position at time t = 3.14s. What is the mass’ approximate maximum speed? Soln: We are given A = 0.5 and T ≈ π, so ω ≈ 2. The maximum speed is ωA ≈ 1. 28. A 1kg mass on a spring is oscillating without friction, with amplitude A and maximum speed v. At the moment when the mass is at its equilibrium position, a 1kg lump of putty is dropped from above and sticks. What is the amplitude of oscillation thereafter? Soln: By momentum conservation, the speed afterwards is 1 2 v, so the energy is Ef = (2)( 1 2 )2Ei = 1 2 Ei, and thus the amplitude is Af = Ai/ √ 2. 29. A small mass m is on a string of length L, with small oscillations of period 1s. What is the period of oscillation of a uniform rod of the same mass m, and length L, pivoted at its end? Soln: The period in the first case is 2π √ L/g and in the second is 2π √ I/mgd, where I = 1 3 mL2 is the moment of inertia and d = L/2 is the distance to the center of mass. So the period is multiplied by a factor of √ 2/3. 7