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Unformatted text preview: homework 08 GADHIA, TEJAS Due: Mar 21 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A rocket is fired vertically upward. At the instant it reaches speed of 281 m / s, it explodes into three fragments with equal masses. Im- mediately after the exposion, the first frag- ment flies vertically with speed 627 m / s, while the second fragment flies horizontally with speed 332 m / s. What is the speed of the third fragment immediately after the exposion? Correct answer: 396 . 081 m / s (tolerance 1 %). Explanation: During the explosion, the net momentum is conserved. Hence, the net momentum of the three fragments immediately after the explo- sion is equal to the whole rockets momentum immediately before the explosion, m 1 vectorv 1 + m 2 vectorv 2 + m 3 vectorv 3 = M r vectorv r . Using m 1 = m 2 = m 3 = 1 3 M r , this equation simplifies to vectorv 1 + vectorv 2 + vectorv 3 = 3 vectorv r . Consequently, given the velocity vector of the whole rocket and of the first and second com- ponents, we can find the velocity vector of the third component according to vectorv 3 = 3 vectorv r- vectorv 1- vectorv 2 . In components, we have v 3 x = 3 v rx- v 1 x- v 2 x = 3 -- v 2 =- 332 m / s , v 3 y = 3 v ry- v 1 y- v 2 y = 3 v r- v 1- = 3(281 m / s)- 627 m / s = 216 m / s . The speed of the third component is the mag- nitude of this velocity vector, v 3 = radicalBig v 2 3 x + v 2 3 y = radicalBig (- 332 m / s) 2 + (216 m / s) 2 = 396 . 081 m / s . Question 2 part 1 of 1 10 points A triangular wedge 7 m high, 12 m base length, and with a 13 kg mass is placed on a frictionless table. A small block with a 7 kg mass (and negligible size) is placed on top of the wedge as shown in the figure below. 13 kg 12 m 7m 7 k g X wedge 13 kg 12 m 7m 7 k g All surfaces are frictionless, so the block slides down the wedge while the wedge slides sidewise on the table. By the time the block slides all the way down to the bottom of the wedge, how far X wedge does the wedge slide to the right? Correct answer: 4 . 2 m (tolerance 1 %). Explanation: Let : M = 13 kg , m = 7 kg , L = 12 m , and H = 7 m . Consider the wedge and the block as a two- body system. The external forces acting on this system the weight of the wedge, the homework 08 GADHIA, TEJAS Due: Mar 21 2007, 4:00 am 2 weight of the block and the normal force from the table are all vertical, hence the net hor- izontal momentum of the system is conserved, P wedge x + P block x = constant . Furthermore, we start from rest = center- of-mass is not moving, and therefore the X coordinate of the center-of-mass will remain constant while the wedge slides to the right and the block slides down and to the left, X cm = mX block + M X wedge m + M = constant ....
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