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Homework 7-solutions

# Homework 7-solutions - pels(kcp389 Homework 7 hill(666 This...

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pels (kcp389) – Homework 7 – hill – (666) 1 This print-out should have 49 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points Calculate the magnitude of the linear momen- tum for each of the following cases a) a proton with mass 1 . 67 × 10 27 kg mov- ing with a velocity of 6 × 10 6 m / s. Correct answer: 1 . 002 × 10 20 kg · m / s. Explanation: Let : m = 1 . 67 × 10 27 kg and v = 6 × 10 6 m / s . Momentum is vectorp = mvectorv , so p = ( 1 . 67 × 10 27 kg ) ( 6 × 10 6 m / s ) = 1 . 002 × 10 20 kg · m / s . 002 (part 2 of 4) 10.0 points b) a 1 . 6 g bullet moving with a speed of 358 m / s to the right. Correct answer: 0 . 5728 kg · m / s. Explanation: Let : m = 1 . 6 g and v = 358 m / s . p = (1 . 6 g) (358 m / s) · 1 kg 1000 g = 0 . 5728 kg · m / s . 003 (part 3 of 4) 10.0 points c) a 8 kg sprinter running with a velocity of 13 . 8 m / s. Correct answer: 110 . 4 kg · m / s. Explanation: Let : m = 8 kg and v = 13 . 8 m / s . p = (8 kg) (13 . 8 m / s) = 110 . 4 kg · m / s . 004 (part 4 of 4) 10.0 points d) Earth ( m = 5 . 98 × 10 24 kg) moving with an orbital speed equal to 29700 m / s. Correct answer: 1 . 77606 × 10 29 kg · m / s. Explanation: Let : m = 5 . 98 × 10 24 kg and v = 29700 m / s . p = ( 5 . 98 × 10 24 kg ) (29700 m / s) = 1 . 77606 × 10 29 kg · m / s . 005 10.0 points A 2 kg steel ball strikes a wall with a speed of 14 m / s at an angle of 50 . 7 with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. x y 14 m / s 2 kg 14 m / s 2 kg 50 . 7 50 . 7 If the ball is in contact with the wall for 0 . 105 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 337 . 803 N. Explanation:

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pels (kcp389) – Homework 7 – hill – (666) 2 Let : M = 2 kg , v = 14 m / s , and θ = 50 . 7 . The y component of the momentum is un- changed. The x component of the momentum is changed by Δ P x = 2 M v cos θ . Therefore, using impulse formula, F = Δ P Δ t = 2 M v cos θ Δ t = 2 (2 kg) (14 m / s) cos 50 . 7 0 . 105 s bardbl vector F bardbl = 337 . 803 N . Note: The direction of the force is in negative x direction, as indicated by the minus sign. 006 (part 1 of 3) 10.0 points The force of magnitude F x acting in the x direction on a 2 . 6 kg particle varies in time as shown in the figure. 0 1 2 3 4 5 0 1 2 3 4 F x (N) t (s) Find the impulse of the force. Correct answer: 9 N · s. Explanation: The impulse is the area under the F versus t graph; i.e , the sum of the area of the rectangle plus two triangles, so Impulse = 2 bracketleftbigg (3 N) (2 s) 2 bracketrightbigg + (3 N) (1 s) = 9 N · s 007 (part 2 of 3) 10.0 points Find the final velocity of the particle if it is initially at rest. Correct answer: 3 . 46154 m / s. Explanation: Given : m = 2 . 6 kg and v i = 0 m / s . Impulse = Δ p F Δ t = m ( v f v i ) m v f = F Δ t + m v i v f = F Δ t m + v i = 9 N · s 2 . 6 kg + 0 m / s = 3 . 46154 m / s 008 (part 3 of 3) 10.0 points Find the final velocity of the particle if it is initially moving along the x axis with a velocity of 3 m / s.
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