Exam_3_09s

# Exam_3_09s - Exam 3 Physics 130 Short Answer Section....

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Unformatted text preview: Exam 3 Physics 130 Short Answer Section. Please answer all of the questions. 1 . A 200,000 kg locomotive has a linear momentum of 1,000,000 kg m s ˆ i . What is the locomotive’s velocity vector and what kinetic energy does it have? p = m v v = p m = 5 m / s ˆ i KE = 1 2 mv 2 = 2.5 × 10 6 J 2. How much work does the gravitational force due to the sun do on the earth in its orbit? Assume that the earth moves in a perfect circle at constant speed (uniform circular motion) with the sun at the exact center of the circle. No work is done. The force is perpendicular to the displacement 3. A potential has the form U = 2 x 5 . What force results from this potential? F x = − 10 x 4 4. A spring with spring constant k=50N/m is compressed by 0.1 m. This spring relaxes, and the block (mass 10 kg) moves across a horizontal surface and then up an incline. No friction is present. What energy is initially stored in the spring? What gravitational potential energy does the block have when it stops at the top? How far up the incline does it slide before stopping? 30 d? U s = 1 2 kx 2 = 1 2 ⋅ 50 N / m ⋅ (0.1 m ) 2 = 0.25 J U g = U s = 0.25 J mgd sin 30 = U g = 0.25 d = U g mg sin 30 = 0.0051 m 5. Mass 1 has a value of 20kg and is located at position (0,5), Mass 2 has a value of 40 kg and is located at (5,5). Compute the x and y positions of the center of mass. If the masses suddenly begin to attract each other, where will they meet? x cm = 20 kg ⋅ + 40 kg ⋅ 5 m 20 kg + 40 kg = 3.33 m y cm = 20 kg ⋅ 5 m + 40 kg ⋅ 5 m 20 kg + 40 kg = 5.0 m They will meet at the cm. 6. A ball (m=1 kg) has a velocity v i = − 15 m / s ˆ j just before it hits the ground. It bounces, and immediately after the bounce, it has a velocity of v f = + 10 m / s ˆ j What was the ball’s change in momentum? If this change in momentum occurred over a time of 0.05 s, what average force did the ball experience? Δ p = p f − p i = m ( v f − v i ) = 1 kg (10 m / s ˆ j − ( − 15 m / s ˆ j )) = 25 kgm / s ˆ j F = Δ p Δ t = 25 kgm / s ˆ j 0.05 s = 500 N ˆ j 7. A 5 kg mass initially moves to the right at 50m/s. It collides with and sticks to a mass of...
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## This note was uploaded on 09/27/2010 for the course ASD 2456 taught by Professor Jojo during the Spring '10 term at Brooklyn Law School.

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Exam_3_09s - Exam 3 Physics 130 Short Answer Section....

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