Traditional Practice Test One Solutions

Traditional Practice Test One Solutions - Salamah(jms5723...

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Salamah (jms5723) – Practice Midterm 01 – Yao – (58180) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The velocity of a projectile at launch has a horizontal component v h and a vertical com- ponent v v . Note: Air resistance is negligible. When the projectile is at the highest point of its trajectory, which of the following show the vertical and the horizontal components of its velocity and the vertical component of its acceleration in 3 columns? Vertical Horizontal Vertical Velocity Velocity Acceleration 1. 0 0 g 2. v v 0 0 3. 0 v h 0 4. v v v h 0 5. 0 v h g correct Explanation: Basic concept: Newton’s second law of mo- tion vector F = mvectora The only force on the projectile is the gravita- tional force, which gives projectile a constant vertical acceleration of the magnitude g . There is no acceleration in the horizontal direction, which means at the highest point, the horizontal component of the velocity is the same as the initial value v h . One other obvious thing: the vertical com- ponent of the velocity is zero at the highest point, because from the trajectory, the projec- tile moves horizontally at the highest point. 002 (part 1 of 2) 10.0 points The orbit of a Moon about its planet is ap- proximately circular, with a mean radius of 3 . 3 × 10 8 m. It takes 19 . 7 days for the Moon to complete one revolution about the planet. Find the mean orbital speed of the Moon. Correct answer: 1218 . 19 m / s. Explanation: Dividing the length C = 2 πr of the trajectory of the Moon by the time T = 19 . 7 days = 1 . 70208 × 10 6 s of one revolution (in seconds!), we obtain that the mean orbital speed of the Moon is v = C T = 2 π r T = 2 π (3 . 3 × 10 8 m ) 1 . 70208 × 10 6 s = 1218 . 19 m / s . 003 (part 2 of 2) 10.0 points Find the Moon’s centripetal acceleration. Correct answer: 0 . 00449691 m / s 2 . Explanation: Since the magnitude of the velocity is con- stant, the tangential acceleration of the Moon is zero. The centripetal acceleration is a c = v 2 r = (1218 . 19 m / s ) 2 3 . 3 × 10 8 m = 0 . 00449691 m / s 2 . 004 10.0 points Some enterprising physics students working on a catapult decide to have a water balloon fight in the school hallway. The ceiling is of height 3 . 5 m, and the balloons are launched at a velocity of 12 m / s. The acceleration of gravity is 9 . 8 m / s 2 . At what angle must they be launched to just graze the ceiling? Correct answer: 43 . 6467 . Explanation: Basic Concepts The Kinematics Equa- tions are v = v o + a t
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Salamah (jms5723) – Practice Midterm 01 – Yao – (58180) 2 s = s o + v o t + 1 2 a t 2 v 2 = v 2 o + 2 a s Solution Consider only the vertical mo- tion. When the balloon reaches the ceiling, y = h , and the velocity v y = 0. Gravity acts down and the initial velocity v oy = v o sin θ acts up. Applying the third equation kine- matics equation for the vertical motion, 0 = ( v o sin θ ) 2 - 2 g h v 2 o sin 2 θ = 2 g h sin θ = radicalBigg 2 g h v 2 o = 2 g h v o θ = arcsin braceleftbigg 2 g h v o bracerightbigg = arcsin braceleftbigg radicalbig 2 (9 . 8 m / s 2 ) (3 . 5 m) (12 m / s) bracerightbigg = 43 . 6467 .
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