EPChap09-Solution2 - Chap.9 HW2 Questions 5. Figure 9-28...

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6 Chap.9 HW2 Questions 5. Figure 9-28 shows graphs of force magnitude versus time for a body involved in a collision. Rank the graphs according to the magnitude of the impulse on the body, greatest first. = = tf ti x ix fx dt F p p Impulse This is the area under the F versus t curve. In all three cases it is equal to 12 F o t o so they all tie Problems 35. A 91 kg man lying on a surface of negligible friction shoves a 68 g stone away from himself, giving it a speed of 4.0 m/s. What speed does the man acquire as a result? There are no external forces in the horizontal direction therefore f i P P ) v = Since the man is initially stationary 0 = i P v Let + x be the direction the stone is thrown, then writing out f P explicitly ( m s is the stone, m m is the man) s m v m m v v m v m s m s m m m s s / 10 3 4 91 068 . 0 0 3 × = = = + = The negative sign indicates that the man moves in the -x direction, opposite to the direction of motion of the stone.
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7 43. A 20.0 kg body is moving through space in the positive direction of an x axis with a speed of 200 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 10.0 kg, moves away from the point of explosion with a speed of 100 m/s in the positive y direction. A second part, with a mass of 4.00 kg, moves in the negative x direction with a speed of 500 m/s. (a) In unit-vector notation, what is the velocity of the third part? (b) How much energy is released in the explosion? Ignore effects due to the gravitational force. Hint: set up an x-y coordinate system. Apply conservation of momentum along x. Then apply conservation of momentum along y. Original body be M = 20.0 kg, velocity is r v 0 200 = $ i (m/s); v ox = 200 m/s, v oy = 0 First fragment is m 1 = 10.0 kg, velocity is m/s j ˆ 100 1 = v r v 1x = 0 m/s, v 1y = 100 m/s Second fragment is m 2 = 4.0 kg, velocity r v 2 500 =− $ i (m/s); v 2x = -500 m/s, v 2y = 0 m/s The mass of the third fragment is m 3 = 6.00 kg. a) Conservation of linear momentum Mv m v m v m v r r r s 01 12 23 3 =++ , x -component: x x x x v m v m v m Mv 3 3 2 2 1 1 0 + + = which gives v 3x = 1000 m/s y -component: y y y y v m v m v m Mv 3 3 2 2 1 1 0 + + = which gives v 3y = -167 m/s
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This note was uploaded on 04/07/2008 for the course PHYS 213 taught by Professor Oshea during the Spring '08 term at Kansas State University.

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EPChap09-Solution2 - Chap.9 HW2 Questions 5. Figure 9-28...

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