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Unformatted text preview: practicework 03 BAUTISTA, ALDO Due: Apr 3 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 2. 10 points. A 46 . 9 kg girl is standing on a 83 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless sup porting surface. The girl begins to walk along the plank at a constant speed of 1 . 95 m / s to the right relative to the plank. What is her velocity relative to the ice sur face? Correct answer: 1 . 24596 m / s (tolerance 1 %). Explanation: v g = velocity of girl relative to ice v p = velocity of plank relative to ice v gp = relative velocity of girl. Let : m g = 46 . 9 kg , m p = 83 kg , and v gp = 1 . 95 m / s . By conservation of momentum 0 = m g v g + m p v p , v p = m g v g m p The relative velocity is v gp = v g v p (1) = v g m g v g m p = v g m p + m g v g m p = v g ( m p + m g ) m p . Thus v g = m p v gp m g + m p = (83 kg) (1 . 95 m / s) 83 kg + 46 . 9 kg = 1 . 24596 m / s . Question 2 Part 2 of 2. 10 points. What is the velocity of the plank relative to the ice surface? Correct answer: . 704042 m / s (tolerance 1 %). Explanation: Using Eq. (1), v p = v g v gp = 1 . 24596 m / s 1 . 95 m / s = . 704042 m / s , directed opposite to the girls direction. Question 3 Part 1 of 2. 10 points. A(n) 470 kg mass is sliding on a horizontal frictionless surface with a speed of 11 m / s when it collides with a 87 kg mass initially at rest, as shown in the figure. The masses stick together and slide up a frictionless track at 60 from horizontal. The acceleration of gravity is 9 . 8 m / s 2 . 470 kg 87 kg 9 . 8m / s 2 11 m / s 60 What is the speed of the two blocks imme diately after the collision? Correct answer: 9 . 28187 m / s (tolerance 1 %). Explanation: Let : v 12 = speed of blocks after collision m 1 = 470 kg , m 2 = 87 kg , v 1 = 11 m / s , and g = 9 . 8 m / s 2 . practicework 03 BAUTISTA, ALDO Due: Apr 3 2006, 4:00 am 2 The initial momentum of block 1 is m 1 v 1 . When m 1 sticks with m 2 , the resulting mo mentum p 12 is equal to that initial momen tum; thus, p 12 ( m 1 + m 2 ) v 12 = m 1 v 1 . Therefore, the speed of the combined masses is v 12 = m 1 m 1 + m 2 v 1 = (470 kg) (470 kg) + (87 kg) (11 m / s) = 9 . 28187 m / s . Question 4 Part 2 of 2. 10 points. 557 kg h 9 . 8m / s 2 To what maximum height h above the hor izontal surface will the masses slide? Correct answer: 4 . 39556 m (tolerance 1 %). Explanation: The kinetic energy K of the combined sys tem is K = 1 2 ( m 1 + m 2 ) v 2 12 = m 2 1 2 ( m 1 + m 2 ) v 2 1 At the maximum height, all the kinetic energy is converted to potential energy, and so ( m 1 + m 2 ) g h = K = m 2 1 2 ( m 1 + m 2 ) v 2 1 Solving for h , we get h = v 2 1 2 g m 1 m 1 + m 2 2 = (11 m / s) 2 2 (9 . 8 m / s 2 ) (470 kg) (470 kg) + (87 kg) 2 = 4 . 39556 m ....
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