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# HW 008 - hanna(brh687 – hk 8 – Opyrchal –(11105 1...

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Unformatted text preview: hanna (brh687) – hk 8 – Opyrchal – (11105) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A bullet of mass 7 g moving with an initial speed 500 m / s is fired into and passes through a block of mass 2 kg, as shown in the figure. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 585 N / m. 2 kg 500 m / s 7 g 585 N / m v 4 . 2 cm If the block moves a distance 4 . 2 cm to the right after the bullet passed through it, find the speed v at which the bullet emerges from the block. Correct answer: 294 . 768 m / s. Explanation: Let : k = 585 N / m , x = 4 . 2 cm , m = 7 g , M = 2 kg , v = 500 m / s , and V = Block ′ s velocity From conservation of energy 1 2 M V 2 = 1 2 k x 2 . (1) Hence, V = radicalbigg k x 2 M = radicalBigg (585 N / m) (4 . 2 cm) 2 (2 kg) = 0 . 718311 m / s . Then from momentum conservation mv = M V + mv , or (2) v = v − M V m = (500 m / s) − (2 kg) (0 . 718311 m / s) (7 g) (0 . 001 kg / g) = 294 . 768 m / s . 002 (part 2 of 2) 10.0 points Find the magnitude of the energy lost in the collision. Correct answer: 570 . 375 J. Explanation: Using Eq. 1, the energy lost is Δ E = K f + U f − K i = mv 2 2 + k x 2 2 − mv 2 i 2 = (0 . 007 kg) (294 . 768 m / s) 2 2 + (585 N / m) (0 . 042 m) 2 2 − (0 . 007 kg) (500 m / s) 2 2 = (304 . 109 J) + (0 . 51597 J) − (875 J) = − 570 . 375 J | Δ E | = 570 . 375 J . 003 10.0 points A(n) 21 kg object, initially at rest in free space, “explodes” into three segments. The masses of two of these segments are both 6 kg and their velocities are 5 m / s. The angle between the direction of motion of these seg- ments is 56 ◦ . What is the speed of the third segment? Correct answer: 5 . 88632 m / s. Explanation: hanna (brh687) – hk 8 – Opyrchal – (11105) 2 The mass of the third segment is m 3 = m 123 − m 1 − m 2 = (21 kg) − (6 kg) − (6 kg) = 9 kg . The momentum of the segments are p 1 = m 1 v 1 = (6 kg) (5 m / s) = 30 kg m / s p 2 = m 2 v 2 = (6 kg) (5 m / s) = 30 kg m / s p 3 = m 3 v 3 = (9 kg) v 3 . Momentum is conserved. 0 = vectorp 1 + vectorp 2 + vectorp 3 θ p p p 1 2 3 α α = 180 ◦ − θ = 180 ◦ − 56 ◦ = 124 ◦ The law of cosines gives p 2 3 = p 2 1 + p 2 2 − 2 p 1 p 2 cos(180 ◦ − θ ) (1) = (30 kg m / s) 2 + (30 kg m / s) 2 − 2 (30 kg m / s) (30 kg m / s) cos(124 ◦ ) = (1800 kg 2 m 2 / s 2 ) − ( − 1006 . 55 kg 2 m 2 / s 2 ) = (2806 . 55 kg 2 m 2 / s 2 ) p 3 = radicalBig 2806 . 55 kg 2 m 2 / s 2 = 52 . 9769 kg m / s ....
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HW 008 - hanna(brh687 – hk 8 – Opyrchal –(11105 1...

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