Test 3 Review-solutions - pandey(sp34566 Test 3 Review...

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pandey (sp34566) – Test 3 Review – swinney – (57305) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A mass of 0.2 kg is attached to a massless spring, which provides the only force acting on the mass as it executes simple harmonic os- cillation with an angular frequency of 2 rad/s. If x (0) = - 0 . 5 m and v x (0) = - 1 . 0 m/s for this oscillator, what is A ? The general solution for a simple harmonic oscillation is x ( t ) = A cos( ω t + φ ) . 1. 2.0 m 2. 1.70 m 3. 0.25 m 4. 2 m 5. 1.25 m 6. Zero 7. 1.5 m 8. 1.0 m 9. 0.707 m correct 10. 0.5 m Explanation: The easiest solution involves conservation of energy, noting that k = m ω 2 . E = 1 2 k A 2 = 1 2 m ω 2 A 2 and E = 1 2 m v x (0) 2 + 1 2 m ω 2 x (0) 2 . Equating these, A = radicalBigg x (0) 2 + parenleftbigg v x (0) ω parenrightbigg 2 = radicalBigg ( - 0 . 5 m) 2 + parenleftbigg - 1 . 0 m / s 2 s 1 parenrightbigg 2 = 0 . 707 m . 002 10.0points A harmonic oscillator consists of a mass M on a horizontal frictionless surface attached to a spring of negligible mass and with force constant k . The oscillator has its maximum kinetic energy (KE) at x = 0, and its KE goes to zero at x = A . When it is at x = A 4 , how does its KE compare to its maximum value; i.e. , what is K ( x = A/ 4) K max ? 1. K K max = 7 9 2. K K max = 15 16 correct 3. K K max = 1 2 4. K K max = 5 16 5. K K max = 7 16 6. K K max = 1 4 7. K K max = 13 16 8. K K max = 5 8 9. K K max = 3 4 10. K K max = 1 8 Explanation: E = K + U s = 1 2 k A 2 = K max is conserved as the oscillator moves. At x = A 4 , U s = 1 2 k parenleftbigg A 4 parenrightbigg 2 = E 16 , so
pandey (sp34566) – Test 3 Review – swinney – (57305) 2 K = E - E 16 = 15 16 E K K max = 15 16 . 003 10.0points A student performs a ballistic pendulum experiment using an apparatus similar to that shown in the figure. Initially the bullet is fired at the block while the block is at rest (at its lowest swing point). After the bullet hits the block, the block rises to its highest position, see dashed block in the figure, and continues swinging back and forth. The following data is obtained: the maximum height the pendulum rises is 5 cm, at the maximum height the pendulum sub- tends an angle of 48 . 2 , the mass of the bullet is 93 g, and the mass of the pendulum bob is 793 g. The final velocity is v f = radicalbig 2 g h . Using conservation of momentum m 1 v i = ( m 1 + m 2 ) v f , so v i = bracketleftbigg ( m 1 ) + ( m 2 ) m 1 bracketrightbigg v f = bracketleftbigg ( m 1 ) + ( m 2 ) m 1 bracketrightbigg radicalbig 2 g h = bracketleftbigg (93 g) + (793 g) (93 g) bracketrightbigg × radicalBig 2 (9 . 8 m / s 2 ) (0 . 05 m) = 9 . 43113 m / s . 004(part1of4)10.0points A 8 kg block slides down a frictionless incline making an angle of 25 with the horizontal.
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