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P211 Final Sample 1 19) Two blocks (m = 2 kg and M = 9 kg) and a spring (k = 160 N/m) are arranged on a horizontal, frictionless surface as shown in the figure below. The coefficient of static friction between the two blocks is μ s = 0.4. What is the maximum possible amplitude of simple harmonic motion of the spring-blocks system if no slippage is to occur between the blocks? A) 0.049 m B) 0.221 m C) 0.319 m D) 0.270 m E) 0.156 m 20) An unknown mass M in a spring-mass system is attached to a massless spring of unknown spring constant k. The spring is stretched by a distance of 10 cm, at which point the mass is released from rest and oscillates back-and-forth on a frictionless table. If the period T of the oscillations is measured to be 2s, what is the maximum speed of the mass as it passes through the equilibrium position (at which point the spring is neither stretched nor compressed)? A) 0.052 m/s B) 0.125 m/s C) 0.314 m/s D) 0.781 m/s E) 1.76 m/s Page 10
P211 Final Sample 1 21) A flying saucer deep in interstellar space, free of external forces, rotates about its symmetry axis with angular velocity ω. The aliens inside the saucer rearrange the furniture to make room for a Christmas tree they have just procured from Elvis. As a result the angular velocity of the saucer increases to 2ω. If originally the rotational kinetic energy of the saucer was K, after the rearrangement its new rotational kinetic energy is: A) K/4 B) K/2 C) K D) 2K E) 4K 22) A block of mass M = 4 kg is resting on a frictionless table and is attached to an uncompressed, unstretched spring (k = 460 N/m) fixed to a wall at the other end, as in the figure below. A bullet of mass m = 0.06 kg is fired into the block in a direction parallel to the alignment of the spring, with a speed of v b = 100 m/s, and is embedded in the block. What is the amplitude x m of the resulting simple harmonic motion? (Hint: energy is not conserved during the collision between the bullet and the block, but momentum is; during oscillations, energy is conserved, but not momentum.) x = 0 x = 0 −x m +x m M v b m M+m Initially Finally A) 0.052 m B) 0.139 m C) 0.211 m D) 0.876 m E) 1.74 m Page 11
Page 1 and 2: P211 Final Sample 1 1) The figure b
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