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Unformatted text preview: hernandez (ejh742) – oldhomework 19 – Turner – (58120) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 0.451 kg bead slides on a straight friction less wire with a velocity of 3.39 cm/s to the right, as shown. The bead collides elastically with a larger 0.699 kg bead initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.64 cm/s. . 451 kg 3 . 39 cm / s . 699 kg Find the distance the larger bead moves along the wire in the first 4.4 s following the collision. Correct answer: 11 . 4408 cm. Explanation: Basic Concepts: m 1 vectorv 1 ,i = m 1 vectorv 1 ,f + m 2 vectorv 2 ,f since v 2 ,i = 0 m/s. Δ x = v Δ t Given: Let to the right be positive: m 1 = 0 . 451 kg v 1 ,i = +3 . 39 cm / s m 2 = 0 . 699 kg v 1 ,f = − . 64 cm / s t = 4 . 4 s Solution: v 2 ,f = m 1 v 1 ,i − m 1 v 1 ,f m 2 = (0 . 451 kg)(3 . 39 cm / s) . 699 kg − (0 . 451 kg)( − . 64 cm / s) . 699 kg = 2 . 60019 cm / s to the right. Thus Δ x = (2 . 60019 cm / s)(4 . 4 s) = 11 . 4408 cm 002 (part 1 of 2) 10.0 points Consider the collision of two identical parti cles, with m 1 = m 2 = 10 g. The initial velocity of particle 1 is v 1 and particle 2 is initially at rest, v 2 = 0 m/s.. 1 2 v 1 After an elastic headon collision, the final velocity of particle 2 is v ′ 2 and given by 1. v ′ 2 = v 1 2 2. v ′ 2 = 2 v 1 3 3. v ′ 2 = 2 v 1 4. v ′ 2 = v 1 4 5. v ′ 2 = 0 6. v ′ 2 = 4 v 1 3 7. v ′ 2 = v 1 correct 8. v ′ 2 = 5 v 1 3 9. v ′ 2 = 3 v 1 4 10. v ′ 2 = v 1 3 Explanation: For the final velocity of particle 2 after an elastic collision, we have v ′ 2 = 2 v cm − v 2 . For the present case, v cm = m 1 v 1 + m 2 v 2 m 1 + m 2 = v 1 2 . hernandez (ejh742) – oldhomework 19 – Turner – (58120) 2 So v ′ 2 = 2 parenleftBig v 1 2 parenrightBig − 0 = v 1 ....
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 Spring '08
 Turner
 Physics, Friction, Kinetic Energy, Mass, Momentum, Work, kg

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